Download GATE GATE EE Question Paper With Solutions

Download GATE (Graduate Aptitude Test in Engineering) GATE EE Question Paper With Solutions

2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =
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2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct

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2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]
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2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that

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2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2
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2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem
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2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).

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2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6
FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits
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2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?
FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter
FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?


FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?





14

/
( ) sin( )
sin( )
t M
m
V
i t Ae t
Z
V
DC offset A
z
?
??
??
?
= + ?
?
= = ?

For Maximum value of DC offset A
? ? ? = ? 90
1
3
1
tan 90
377 10 10
tan 90
3.77
45 90
45
L
R
?
?
?
?
?
?
?
?
??
? = ?
??
??
??
??
? = ?
??
??
??
? ? = ? ?
= ? ?


25. M is a 2 ? 2 Matrix with Eigen value 4 and 9 If has ? 1, ? 2 _ _ _ _ _ ? n Eigen values
M
n
? ? 1
n
, ? 2
n
_ _ _ ? n
n
Eigen values
M
2
? 4
2
, 9
2

? M
2
has Eigen values as 16 and 81

26. V S = 400 KV
l = 300 km
L 1 = 1 mH / km / phase
C 1 = 0.01 ?F / km / phase
5
36
11
5
22
11
3.16 10 /
1 10 0.01 10
2 2 50 300
' 0.29
3.16 10
(0.29)
1 1 0.955
22
400
418.85KV
0.955
s
R
v km s
LC
fl
v
A
V
V
A
??
?
?
??
= = = ?
? ? ?
??
= = =
?
= ? = ? =
= = =


27. According to Mill man?s Theorem, the equivalent circuit of the given circuit is

FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?





14

/
( ) sin( )
sin( )
t M
m
V
i t Ae t
Z
V
DC offset A
z
?
??
??
?
= + ?
?
= = ?

For Maximum value of DC offset A
? ? ? = ? 90
1
3
1
tan 90
377 10 10
tan 90
3.77
45 90
45
L
R
?
?
?
?
?
?
?
?
??
? = ?
??
??
??
??
? = ?
??
??
??
? ? = ? ?
= ? ?


25. M is a 2 ? 2 Matrix with Eigen value 4 and 9 If has ? 1, ? 2 _ _ _ _ _ ? n Eigen values
M
n
? ? 1
n
, ? 2
n
_ _ _ ? n
n
Eigen values
M
2
? 4
2
, 9
2

? M
2
has Eigen values as 16 and 81

26. V S = 400 KV
l = 300 km
L 1 = 1 mH / km / phase
C 1 = 0.01 ?F / km / phase
5
36
11
5
22
11
3.16 10 /
1 10 0.01 10
2 2 50 300
' 0.29
3.16 10
(0.29)
1 1 0.955
22
400
418.85KV
0.955
s
R
v km s
LC
fl
v
A
V
V
A
??
?
?
??
= = = ?
? ? ?
??
= = =
?
= ? = ? =
= = =


27. According to Mill man?s Theorem, the equivalent circuit of the given circuit is




15




3 1 2 4
1 2 3 4
1 2 3 4
1 1 1 1
eq
E E E E
R R R R
E
R R R R
+++
=
+++

200 160 100 80
5 40 25 20
1 1 1 1
50 40 25 20
? ? ? ?
+ + ? + ?
? ? ? ?
? ? ? ?
=
+ + +

E eq = 0V
So, the current I flowing is 0 A

28. For synchronous motor
E g = V 1 ? IZ
220
()
3
t
V V Phase =
Z = (0.25 + j 2.5)?
I = 10 ? ?36.86 A
220
(0.25 2.5) 10 36.86
3
g
Ej = ? + ? ? ?
E g = 141.658 ??8.7 V (phase)
E g = 245.36 V (line)

FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?





14

/
( ) sin( )
sin( )
t M
m
V
i t Ae t
Z
V
DC offset A
z
?
??
??
?
= + ?
?
= = ?

For Maximum value of DC offset A
? ? ? = ? 90
1
3
1
tan 90
377 10 10
tan 90
3.77
45 90
45
L
R
?
?
?
?
?
?
?
?
??
? = ?
??
??
??
??
? = ?
??
??
??
? ? = ? ?
= ? ?


25. M is a 2 ? 2 Matrix with Eigen value 4 and 9 If has ? 1, ? 2 _ _ _ _ _ ? n Eigen values
M
n
? ? 1
n
, ? 2
n
_ _ _ ? n
n
Eigen values
M
2
? 4
2
, 9
2

? M
2
has Eigen values as 16 and 81

26. V S = 400 KV
l = 300 km
L 1 = 1 mH / km / phase
C 1 = 0.01 ?F / km / phase
5
36
11
5
22
11
3.16 10 /
1 10 0.01 10
2 2 50 300
' 0.29
3.16 10
(0.29)
1 1 0.955
22
400
418.85KV
0.955
s
R
v km s
LC
fl
v
A
V
V
A
??
?
?
??
= = = ?
? ? ?
??
= = =
?
= ? = ? =
= = =


27. According to Mill man?s Theorem, the equivalent circuit of the given circuit is




15




3 1 2 4
1 2 3 4
1 2 3 4
1 1 1 1
eq
E E E E
R R R R
E
R R R R
+++
=
+++

200 160 100 80
5 40 25 20
1 1 1 1
50 40 25 20
? ? ? ?
+ + ? + ?
? ? ? ?
? ? ? ?
=
+ + +

E eq = 0V
So, the current I flowing is 0 A

28. For synchronous motor
E g = V 1 ? IZ
220
()
3
t
V V Phase =
Z = (0.25 + j 2.5)?
I = 10 ? ?36.86 A
220
(0.25 2.5) 10 36.86
3
g
Ej = ? + ? ? ?
E g = 141.658 ??8.7 V (phase)
E g = 245.36 V (line)




16

29.
32
| | 5
32
2
8
2 ( )
2
(8
2 lim (z 2)
2
8 4 8
28
1
z
z
zz
dz j sumof residues
z
zz
j
z
jj
?
?
??
=
?
++
=
+
??
++
= ? +
??
+
??
??
? + +
??
==
??
??
?


30.
V (t) = ? 170 sin (377 )
6
t
?
?
I (t) = 8 cos (377 )
6
t
?
+

V(t) = ? 170 sin (377 )
6
t
?
?
V(t) = 170 cos (377 )
62
t
??
?+
V(t)=170 cos (377 )
3
t
?
+
P = V rms I rms cos?
170 8
cos30
22
P =
P = 588.89 watts

31. Given R 1 = 5.39?, R 2 = 5.72?, X 1 = X 2 = 8.22?
for frequency ? 10 Hz
FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?





14

/
( ) sin( )
sin( )
t M
m
V
i t Ae t
Z
V
DC offset A
z
?
??
??
?
= + ?
?
= = ?

For Maximum value of DC offset A
? ? ? = ? 90
1
3
1
tan 90
377 10 10
tan 90
3.77
45 90
45
L
R
?
?
?
?
?
?
?
?
??
? = ?
??
??
??
??
? = ?
??
??
??
? ? = ? ?
= ? ?


25. M is a 2 ? 2 Matrix with Eigen value 4 and 9 If has ? 1, ? 2 _ _ _ _ _ ? n Eigen values
M
n
? ? 1
n
, ? 2
n
_ _ _ ? n
n
Eigen values
M
2
? 4
2
, 9
2

? M
2
has Eigen values as 16 and 81

26. V S = 400 KV
l = 300 km
L 1 = 1 mH / km / phase
C 1 = 0.01 ?F / km / phase
5
36
11
5
22
11
3.16 10 /
1 10 0.01 10
2 2 50 300
' 0.29
3.16 10
(0.29)
1 1 0.955
22
400
418.85KV
0.955
s
R
v km s
LC
fl
v
A
V
V
A
??
?
?
??
= = = ?
? ? ?
??
= = =
?
= ? = ? =
= = =


27. According to Mill man?s Theorem, the equivalent circuit of the given circuit is




15




3 1 2 4
1 2 3 4
1 2 3 4
1 1 1 1
eq
E E E E
R R R R
E
R R R R
+++
=
+++

200 160 100 80
5 40 25 20
1 1 1 1
50 40 25 20
? ? ? ?
+ + ? + ?
? ? ? ?
? ? ? ?
=
+ + +

E eq = 0V
So, the current I flowing is 0 A

28. For synchronous motor
E g = V 1 ? IZ
220
()
3
t
V V Phase =
Z = (0.25 + j 2.5)?
I = 10 ? ?36.86 A
220
(0.25 2.5) 10 36.86
3
g
Ej = ? + ? ? ?
E g = 141.658 ??8.7 V (phase)
E g = 245.36 V (line)




16

29.
32
| | 5
32
2
8
2 ( )
2
(8
2 lim (z 2)
2
8 4 8
28
1
z
z
zz
dz j sumof residues
z
zz
j
z
jj
?
?
??
=
?
++
=
+
??
++
= ? +
??
+
??
??
? + +
??
==
??
??
?


30.
V (t) = ? 170 sin (377 )
6
t
?
?
I (t) = 8 cos (377 )
6
t
?
+

V(t) = ? 170 sin (377 )
6
t
?
?
V(t) = 170 cos (377 )
62
t
??
?+
V(t)=170 cos (377 )
3
t
?
+
P = V rms I rms cos?
170 8
cos30
22
P =
P = 588.89 watts

31. Given R 1 = 5.39?, R 2 = 5.72?, X 1 = X 2 = 8.22?
for frequency ? 10 Hz



17

12
10
8.22 1.644
50
XX = = ? = ?
Starting phase current at 10 Hz
22
1 2 1 2
22
( ) ( )
100
(5.39 5.72) (1.644 1.644)
pn
pn
V
I
R R X X
=
+ + +
=
++

I Pn = 8.63A
Starting line current = 3
L Ph
II =
3 8.63
14.95
L
L
I
IA
=?
=


32. Given data L = 50mH, C = 0.05 ?F
Critical resistance to avoid current shopping will be given as
3
6
1 1 50 10
22
0.05 10
L
R
C
?
?
?
==
?

R = 500?

33.

X eq = 0.25 + 0.2 +
0.4
2

X eq = 0.65 PU
P = V PU I PV cos?
0.8 = 1 ? I PV ? 0.8
I PU = 1 PU
1 36.86 I = ? ? [as 0.8 pf lagging]
1 1 36.86 0.65 1.484 20.51
eq
E V jIX
E j Pu
=+
= + ? ? ? = ?
-
FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?





14

/
( ) sin( )
sin( )
t M
m
V
i t Ae t
Z
V
DC offset A
z
?
??
??
?
= + ?
?
= = ?

For Maximum value of DC offset A
? ? ? = ? 90
1
3
1
tan 90
377 10 10
tan 90
3.77
45 90
45
L
R
?
?
?
?
?
?
?
?
??
? = ?
??
??
??
??
? = ?
??
??
??
? ? = ? ?
= ? ?


25. M is a 2 ? 2 Matrix with Eigen value 4 and 9 If has ? 1, ? 2 _ _ _ _ _ ? n Eigen values
M
n
? ? 1
n
, ? 2
n
_ _ _ ? n
n
Eigen values
M
2
? 4
2
, 9
2

? M
2
has Eigen values as 16 and 81

26. V S = 400 KV
l = 300 km
L 1 = 1 mH / km / phase
C 1 = 0.01 ?F / km / phase
5
36
11
5
22
11
3.16 10 /
1 10 0.01 10
2 2 50 300
' 0.29
3.16 10
(0.29)
1 1 0.955
22
400
418.85KV
0.955
s
R
v km s
LC
fl
v
A
V
V
A
??
?
?
??
= = = ?
? ? ?
??
= = =
?
= ? = ? =
= = =


27. According to Mill man?s Theorem, the equivalent circuit of the given circuit is




15




3 1 2 4
1 2 3 4
1 2 3 4
1 1 1 1
eq
E E E E
R R R R
E
R R R R
+++
=
+++

200 160 100 80
5 40 25 20
1 1 1 1
50 40 25 20
? ? ? ?
+ + ? + ?
? ? ? ?
? ? ? ?
=
+ + +

E eq = 0V
So, the current I flowing is 0 A

28. For synchronous motor
E g = V 1 ? IZ
220
()
3
t
V V Phase =
Z = (0.25 + j 2.5)?
I = 10 ? ?36.86 A
220
(0.25 2.5) 10 36.86
3
g
Ej = ? + ? ? ?
E g = 141.658 ??8.7 V (phase)
E g = 245.36 V (line)




16

29.
32
| | 5
32
2
8
2 ( )
2
(8
2 lim (z 2)
2
8 4 8
28
1
z
z
zz
dz j sumof residues
z
zz
j
z
jj
?
?
??
=
?
++
=
+
??
++
= ? +
??
+
??
??
? + +
??
==
??
??
?


30.
V (t) = ? 170 sin (377 )
6
t
?
?
I (t) = 8 cos (377 )
6
t
?
+

V(t) = ? 170 sin (377 )
6
t
?
?
V(t) = 170 cos (377 )
62
t
??
?+
V(t)=170 cos (377 )
3
t
?
+
P = V rms I rms cos?
170 8
cos30
22
P =
P = 588.89 watts

31. Given R 1 = 5.39?, R 2 = 5.72?, X 1 = X 2 = 8.22?
for frequency ? 10 Hz



17

12
10
8.22 1.644
50
XX = = ? = ?
Starting phase current at 10 Hz
22
1 2 1 2
22
( ) ( )
100
(5.39 5.72) (1.644 1.644)
pn
pn
V
I
R R X X
=
+ + +
=
++

I Pn = 8.63A
Starting line current = 3
L Ph
II =
3 8.63
14.95
L
L
I
IA
=?
=


32. Given data L = 50mH, C = 0.05 ?F
Critical resistance to avoid current shopping will be given as
3
6
1 1 50 10
22
0.05 10
L
R
C
?
?
?
==
?

R = 500?

33.

X eq = 0.25 + 0.2 +
0.4
2

X eq = 0.65 PU
P = V PU I PV cos?
0.8 = 1 ? I PV ? 0.8
I PU = 1 PU
1 36.86 I = ? ? [as 0.8 pf lagging]
1 1 36.86 0.65 1.484 20.51
eq
E V jIX
E j Pu
=+
= + ? ? ? = ?
-



18

? = 20.51 degrees

34.

A. 600 mV B. 500 mV
C. 400 mV D. 100 mV

2
12
22
1
11
22
21
1 2 1 1
22
21
1 2 1 1
2 [ ]
1
1
1
out x
L
out
L
out
R
Vx V VoltagedivisionRule
RR
RR
V V V
RR
R R R
V V V
R R R R
R R R
V V V
R R R R
=
+
??
= + ?
??
??
??
= + ?
??
+
??
??
= + ?
??
+
??

FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?





14

/
( ) sin( )
sin( )
t M
m
V
i t Ae t
Z
V
DC offset A
z
?
??
??
?
= + ?
?
= = ?

For Maximum value of DC offset A
? ? ? = ? 90
1
3
1
tan 90
377 10 10
tan 90
3.77
45 90
45
L
R
?
?
?
?
?
?
?
?
??
? = ?
??
??
??
??
? = ?
??
??
??
? ? = ? ?
= ? ?


25. M is a 2 ? 2 Matrix with Eigen value 4 and 9 If has ? 1, ? 2 _ _ _ _ _ ? n Eigen values
M
n
? ? 1
n
, ? 2
n
_ _ _ ? n
n
Eigen values
M
2
? 4
2
, 9
2

? M
2
has Eigen values as 16 and 81

26. V S = 400 KV
l = 300 km
L 1 = 1 mH / km / phase
C 1 = 0.01 ?F / km / phase
5
36
11
5
22
11
3.16 10 /
1 10 0.01 10
2 2 50 300
' 0.29
3.16 10
(0.29)
1 1 0.955
22
400
418.85KV
0.955
s
R
v km s
LC
fl
v
A
V
V
A
??
?
?
??
= = = ?
? ? ?
??
= = =
?
= ? = ? =
= = =


27. According to Mill man?s Theorem, the equivalent circuit of the given circuit is




15




3 1 2 4
1 2 3 4
1 2 3 4
1 1 1 1
eq
E E E E
R R R R
E
R R R R
+++
=
+++

200 160 100 80
5 40 25 20
1 1 1 1
50 40 25 20
? ? ? ?
+ + ? + ?
? ? ? ?
? ? ? ?
=
+ + +

E eq = 0V
So, the current I flowing is 0 A

28. For synchronous motor
E g = V 1 ? IZ
220
()
3
t
V V Phase =
Z = (0.25 + j 2.5)?
I = 10 ? ?36.86 A
220
(0.25 2.5) 10 36.86
3
g
Ej = ? + ? ? ?
E g = 141.658 ??8.7 V (phase)
E g = 245.36 V (line)




16

29.
32
| | 5
32
2
8
2 ( )
2
(8
2 lim (z 2)
2
8 4 8
28
1
z
z
zz
dz j sumof residues
z
zz
j
z
jj
?
?
??
=
?
++
=
+
??
++
= ? +
??
+
??
??
? + +
??
==
??
??
?


30.
V (t) = ? 170 sin (377 )
6
t
?
?
I (t) = 8 cos (377 )
6
t
?
+

V(t) = ? 170 sin (377 )
6
t
?
?
V(t) = 170 cos (377 )
62
t
??
?+
V(t)=170 cos (377 )
3
t
?
+
P = V rms I rms cos?
170 8
cos30
22
P =
P = 588.89 watts

31. Given R 1 = 5.39?, R 2 = 5.72?, X 1 = X 2 = 8.22?
for frequency ? 10 Hz



17

12
10
8.22 1.644
50
XX = = ? = ?
Starting phase current at 10 Hz
22
1 2 1 2
22
( ) ( )
100
(5.39 5.72) (1.644 1.644)
pn
pn
V
I
R R X X
=
+ + +
=
++

I Pn = 8.63A
Starting line current = 3
L Ph
II =
3 8.63
14.95
L
L
I
IA
=?
=


32. Given data L = 50mH, C = 0.05 ?F
Critical resistance to avoid current shopping will be given as
3
6
1 1 50 10
22
0.05 10
L
R
C
?
?
?
==
?

R = 500?

33.

X eq = 0.25 + 0.2 +
0.4
2

X eq = 0.65 PU
P = V PU I PV cos?
0.8 = 1 ? I PV ? 0.8
I PU = 1 PU
1 36.86 I = ? ? [as 0.8 pf lagging]
1 1 36.86 0.65 1.484 20.51
eq
E V jIX
E j Pu
=+
= + ? ? ? = ?
-



18

? = 20.51 degrees

34.

A. 600 mV B. 500 mV
C. 400 mV D. 100 mV

2
12
22
1
11
22
21
1 2 1 1
22
21
1 2 1 1
2 [ ]
1
1
1
out x
L
out
L
out
R
Vx V VoltagedivisionRule
RR
RR
V V V
RR
R R R
V V V
R R R R
R R R
V V V
R R R R
=
+
??
= + ?
??
??
??
= + ?
??
+
??
??
= + ?
??
+
??




19

2 2 2
2 1 2 1
1 1 1
()
100
(50 10)
10
400
out
out
out
R R R
V V V V V
R R R
V
V mV
= ? = ?
=?
=


35.



Output XY XY
XY
=+
=?

The above expression is for XOR gate

36. Discharging of capacitor equation
V C (t) = V oe
?t/?

Where ? = RC = (10
3
) (10
?7
) = 10
?4
sec
V o = 100V
V c(t) = 100 e
?104t

V c(t) = 1V
1 = 100 e
?104t

T = 0.46 msec

37.
( )
0
11
cos sin .
nn
nn
f t a a nt b nt
??
==
= + +
??


FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?





14

/
( ) sin( )
sin( )
t M
m
V
i t Ae t
Z
V
DC offset A
z
?
??
??
?
= + ?
?
= = ?

For Maximum value of DC offset A
? ? ? = ? 90
1
3
1
tan 90
377 10 10
tan 90
3.77
45 90
45
L
R
?
?
?
?
?
?
?
?
??
? = ?
??
??
??
??
? = ?
??
??
??
? ? = ? ?
= ? ?


25. M is a 2 ? 2 Matrix with Eigen value 4 and 9 If has ? 1, ? 2 _ _ _ _ _ ? n Eigen values
M
n
? ? 1
n
, ? 2
n
_ _ _ ? n
n
Eigen values
M
2
? 4
2
, 9
2

? M
2
has Eigen values as 16 and 81

26. V S = 400 KV
l = 300 km
L 1 = 1 mH / km / phase
C 1 = 0.01 ?F / km / phase
5
36
11
5
22
11
3.16 10 /
1 10 0.01 10
2 2 50 300
' 0.29
3.16 10
(0.29)
1 1 0.955
22
400
418.85KV
0.955
s
R
v km s
LC
fl
v
A
V
V
A
??
?
?
??
= = = ?
? ? ?
??
= = =
?
= ? = ? =
= = =


27. According to Mill man?s Theorem, the equivalent circuit of the given circuit is




15




3 1 2 4
1 2 3 4
1 2 3 4
1 1 1 1
eq
E E E E
R R R R
E
R R R R
+++
=
+++

200 160 100 80
5 40 25 20
1 1 1 1
50 40 25 20
? ? ? ?
+ + ? + ?
? ? ? ?
? ? ? ?
=
+ + +

E eq = 0V
So, the current I flowing is 0 A

28. For synchronous motor
E g = V 1 ? IZ
220
()
3
t
V V Phase =
Z = (0.25 + j 2.5)?
I = 10 ? ?36.86 A
220
(0.25 2.5) 10 36.86
3
g
Ej = ? + ? ? ?
E g = 141.658 ??8.7 V (phase)
E g = 245.36 V (line)




16

29.
32
| | 5
32
2
8
2 ( )
2
(8
2 lim (z 2)
2
8 4 8
28
1
z
z
zz
dz j sumof residues
z
zz
j
z
jj
?
?
??
=
?
++
=
+
??
++
= ? +
??
+
??
??
? + +
??
==
??
??
?


30.
V (t) = ? 170 sin (377 )
6
t
?
?
I (t) = 8 cos (377 )
6
t
?
+

V(t) = ? 170 sin (377 )
6
t
?
?
V(t) = 170 cos (377 )
62
t
??
?+
V(t)=170 cos (377 )
3
t
?
+
P = V rms I rms cos?
170 8
cos30
22
P =
P = 588.89 watts

31. Given R 1 = 5.39?, R 2 = 5.72?, X 1 = X 2 = 8.22?
for frequency ? 10 Hz



17

12
10
8.22 1.644
50
XX = = ? = ?
Starting phase current at 10 Hz
22
1 2 1 2
22
( ) ( )
100
(5.39 5.72) (1.644 1.644)
pn
pn
V
I
R R X X
=
+ + +
=
++

I Pn = 8.63A
Starting line current = 3
L Ph
II =
3 8.63
14.95
L
L
I
IA
=?
=


32. Given data L = 50mH, C = 0.05 ?F
Critical resistance to avoid current shopping will be given as
3
6
1 1 50 10
22
0.05 10
L
R
C
?
?
?
==
?

R = 500?

33.

X eq = 0.25 + 0.2 +
0.4
2

X eq = 0.65 PU
P = V PU I PV cos?
0.8 = 1 ? I PV ? 0.8
I PU = 1 PU
1 36.86 I = ? ? [as 0.8 pf lagging]
1 1 36.86 0.65 1.484 20.51
eq
E V jIX
E j Pu
=+
= + ? ? ? = ?
-



18

? = 20.51 degrees

34.

A. 600 mV B. 500 mV
C. 400 mV D. 100 mV

2
12
22
1
11
22
21
1 2 1 1
22
21
1 2 1 1
2 [ ]
1
1
1
out x
L
out
L
out
R
Vx V VoltagedivisionRule
RR
RR
V V V
RR
R R R
V V V
R R R R
R R R
V V V
R R R R
=
+
??
= + ?
??
??
??
= + ?
??
+
??
??
= + ?
??
+
??




19

2 2 2
2 1 2 1
1 1 1
()
100
(50 10)
10
400
out
out
out
R R R
V V V V V
R R R
V
V mV
= ? = ?
=?
=


35.



Output XY XY
XY
=+
=?

The above expression is for XOR gate

36. Discharging of capacitor equation
V C (t) = V oe
?t/?

Where ? = RC = (10
3
) (10
?7
) = 10
?4
sec
V o = 100V
V c(t) = 100 e
?104t

V c(t) = 1V
1 = 100 e
?104t

T = 0.46 msec

37.
( )
0
11
cos sin .
nn
nn
f t a a nt b nt
??
==
= + +
??





20

2
11
2
1
1
2
( )cos ( )
2
| sin cos
2
sin cos
sin2 cos2
2 2 2
2
( )sin ( )
T
n
o
x
o
T
o
o
o
T
n
o
a x t n t d t
T
a A t tdt
x
A
t tdt
A t A t
a
ao
b x t n t d t
T
?
?
?
? ?
??
?
??
??
=
=
=
=
=
?
??
==
??
??
=
=
?
?
?
?
?

1
2
1
1
1
2
sin sin
2
sin
1 cos2
()
22
2
o
o
o
b A t t dt
A
b t dt
At
b dt
A
b
?
?
?
?
?
?
=
=
=?
=
?
?
?


38.
FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?





14

/
( ) sin( )
sin( )
t M
m
V
i t Ae t
Z
V
DC offset A
z
?
??
??
?
= + ?
?
= = ?

For Maximum value of DC offset A
? ? ? = ? 90
1
3
1
tan 90
377 10 10
tan 90
3.77
45 90
45
L
R
?
?
?
?
?
?
?
?
??
? = ?
??
??
??
??
? = ?
??
??
??
? ? = ? ?
= ? ?


25. M is a 2 ? 2 Matrix with Eigen value 4 and 9 If has ? 1, ? 2 _ _ _ _ _ ? n Eigen values
M
n
? ? 1
n
, ? 2
n
_ _ _ ? n
n
Eigen values
M
2
? 4
2
, 9
2

? M
2
has Eigen values as 16 and 81

26. V S = 400 KV
l = 300 km
L 1 = 1 mH / km / phase
C 1 = 0.01 ?F / km / phase
5
36
11
5
22
11
3.16 10 /
1 10 0.01 10
2 2 50 300
' 0.29
3.16 10
(0.29)
1 1 0.955
22
400
418.85KV
0.955
s
R
v km s
LC
fl
v
A
V
V
A
??
?
?
??
= = = ?
? ? ?
??
= = =
?
= ? = ? =
= = =


27. According to Mill man?s Theorem, the equivalent circuit of the given circuit is




15




3 1 2 4
1 2 3 4
1 2 3 4
1 1 1 1
eq
E E E E
R R R R
E
R R R R
+++
=
+++

200 160 100 80
5 40 25 20
1 1 1 1
50 40 25 20
? ? ? ?
+ + ? + ?
? ? ? ?
? ? ? ?
=
+ + +

E eq = 0V
So, the current I flowing is 0 A

28. For synchronous motor
E g = V 1 ? IZ
220
()
3
t
V V Phase =
Z = (0.25 + j 2.5)?
I = 10 ? ?36.86 A
220
(0.25 2.5) 10 36.86
3
g
Ej = ? + ? ? ?
E g = 141.658 ??8.7 V (phase)
E g = 245.36 V (line)




16

29.
32
| | 5
32
2
8
2 ( )
2
(8
2 lim (z 2)
2
8 4 8
28
1
z
z
zz
dz j sumof residues
z
zz
j
z
jj
?
?
??
=
?
++
=
+
??
++
= ? +
??
+
??
??
? + +
??
==
??
??
?


30.
V (t) = ? 170 sin (377 )
6
t
?
?
I (t) = 8 cos (377 )
6
t
?
+

V(t) = ? 170 sin (377 )
6
t
?
?
V(t) = 170 cos (377 )
62
t
??
?+
V(t)=170 cos (377 )
3
t
?
+
P = V rms I rms cos?
170 8
cos30
22
P =
P = 588.89 watts

31. Given R 1 = 5.39?, R 2 = 5.72?, X 1 = X 2 = 8.22?
for frequency ? 10 Hz



17

12
10
8.22 1.644
50
XX = = ? = ?
Starting phase current at 10 Hz
22
1 2 1 2
22
( ) ( )
100
(5.39 5.72) (1.644 1.644)
pn
pn
V
I
R R X X
=
+ + +
=
++

I Pn = 8.63A
Starting line current = 3
L Ph
II =
3 8.63
14.95
L
L
I
IA
=?
=


32. Given data L = 50mH, C = 0.05 ?F
Critical resistance to avoid current shopping will be given as
3
6
1 1 50 10
22
0.05 10
L
R
C
?
?
?
==
?

R = 500?

33.

X eq = 0.25 + 0.2 +
0.4
2

X eq = 0.65 PU
P = V PU I PV cos?
0.8 = 1 ? I PV ? 0.8
I PU = 1 PU
1 36.86 I = ? ? [as 0.8 pf lagging]
1 1 36.86 0.65 1.484 20.51
eq
E V jIX
E j Pu
=+
= + ? ? ? = ?
-



18

? = 20.51 degrees

34.

A. 600 mV B. 500 mV
C. 400 mV D. 100 mV

2
12
22
1
11
22
21
1 2 1 1
22
21
1 2 1 1
2 [ ]
1
1
1
out x
L
out
L
out
R
Vx V VoltagedivisionRule
RR
RR
V V V
RR
R R R
V V V
R R R R
R R R
V V V
R R R R
=
+
??
= + ?
??
??
??
= + ?
??
+
??
??
= + ?
??
+
??




19

2 2 2
2 1 2 1
1 1 1
()
100
(50 10)
10
400
out
out
out
R R R
V V V V V
R R R
V
V mV
= ? = ?
=?
=


35.



Output XY XY
XY
=+
=?

The above expression is for XOR gate

36. Discharging of capacitor equation
V C (t) = V oe
?t/?

Where ? = RC = (10
3
) (10
?7
) = 10
?4
sec
V o = 100V
V c(t) = 100 e
?104t

V c(t) = 1V
1 = 100 e
?104t

T = 0.46 msec

37.
( )
0
11
cos sin .
nn
nn
f t a a nt b nt
??
==
= + +
??





20

2
11
2
1
1
2
( )cos ( )
2
| sin cos
2
sin cos
sin2 cos2
2 2 2
2
( )sin ( )
T
n
o
x
o
T
o
o
o
T
n
o
a x t n t d t
T
a A t tdt
x
A
t tdt
A t A t
a
ao
b x t n t d t
T
?
?
?
? ?
??
?
??
??
=
=
=
=
=
?
??
==
??
??
=
=
?
?
?
?
?

1
2
1
1
1
2
sin sin
2
sin
1 cos2
()
22
2
o
o
o
b A t t dt
A
b t dt
At
b dt
A
b
?
?
?
?
?
?
=
=
=?
=
?
?
?


38.



21

2 2 2
3 2 2 2 3 2
2 2 3
3 2 2 2 3 2
2 2 3
? ??
2 3 4 ,
??
(2 2 2 ) (3 3 3 )
?
(4 4 4 )
( ) (2 2 2 ) (3 3 3 )
4
(4 4 4 )
A xi yj zk U x y z
UA x xy xz i x y y yz j
x z y z z k
dd
div UA x xy xz x y y yz
dx d
d
x z y z z
dz
= + + = + +
= + + + + +
+ + +
= + + + + +
+ + +

div (UA) = (6x
2
+ 2y
2
+ 2z
2
) + (3x
2
+ 9y
2
+ 3z
2
) + (4x
2
+ 4y
2
+ 12z
2
)
at (1, 1, 1) ? x = 1, y = 1, z = 1
div (UA) = 45

39. PMMC Instrument
I fs = 10 mA
R m = 10?

100 = I fs (R m + R se)
100 = 10 ? 10
?3
(10 + R sc)
R se = 10000 ? 10 = 9990?

40.

11
22
1
2
0 1 0
2
10
xx
r
xx
x
y
x
? ? ?
? ? ? ? ? ? ? ?
=+
? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ? ?
??
=??
??
??
??


We know
X AX Bu
Y CX Du
=+
=+

Comparing the above equation with the given problem
0 1 0
2
AB
? ? ?
? ? ? ?
==
? ? ? ?
??
? ? ? ?

C = (1 0)
Characteristic equation is
|SI ? A| = 0
FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?





14

/
( ) sin( )
sin( )
t M
m
V
i t Ae t
Z
V
DC offset A
z
?
??
??
?
= + ?
?
= = ?

For Maximum value of DC offset A
? ? ? = ? 90
1
3
1
tan 90
377 10 10
tan 90
3.77
45 90
45
L
R
?
?
?
?
?
?
?
?
??
? = ?
??
??
??
??
? = ?
??
??
??
? ? = ? ?
= ? ?


25. M is a 2 ? 2 Matrix with Eigen value 4 and 9 If has ? 1, ? 2 _ _ _ _ _ ? n Eigen values
M
n
? ? 1
n
, ? 2
n
_ _ _ ? n
n
Eigen values
M
2
? 4
2
, 9
2

? M
2
has Eigen values as 16 and 81

26. V S = 400 KV
l = 300 km
L 1 = 1 mH / km / phase
C 1 = 0.01 ?F / km / phase
5
36
11
5
22
11
3.16 10 /
1 10 0.01 10
2 2 50 300
' 0.29
3.16 10
(0.29)
1 1 0.955
22
400
418.85KV
0.955
s
R
v km s
LC
fl
v
A
V
V
A
??
?
?
??
= = = ?
? ? ?
??
= = =
?
= ? = ? =
= = =


27. According to Mill man?s Theorem, the equivalent circuit of the given circuit is




15




3 1 2 4
1 2 3 4
1 2 3 4
1 1 1 1
eq
E E E E
R R R R
E
R R R R
+++
=
+++

200 160 100 80
5 40 25 20
1 1 1 1
50 40 25 20
? ? ? ?
+ + ? + ?
? ? ? ?
? ? ? ?
=
+ + +

E eq = 0V
So, the current I flowing is 0 A

28. For synchronous motor
E g = V 1 ? IZ
220
()
3
t
V V Phase =
Z = (0.25 + j 2.5)?
I = 10 ? ?36.86 A
220
(0.25 2.5) 10 36.86
3
g
Ej = ? + ? ? ?
E g = 141.658 ??8.7 V (phase)
E g = 245.36 V (line)




16

29.
32
| | 5
32
2
8
2 ( )
2
(8
2 lim (z 2)
2
8 4 8
28
1
z
z
zz
dz j sumof residues
z
zz
j
z
jj
?
?
??
=
?
++
=
+
??
++
= ? +
??
+
??
??
? + +
??
==
??
??
?


30.
V (t) = ? 170 sin (377 )
6
t
?
?
I (t) = 8 cos (377 )
6
t
?
+

V(t) = ? 170 sin (377 )
6
t
?
?
V(t) = 170 cos (377 )
62
t
??
?+
V(t)=170 cos (377 )
3
t
?
+
P = V rms I rms cos?
170 8
cos30
22
P =
P = 588.89 watts

31. Given R 1 = 5.39?, R 2 = 5.72?, X 1 = X 2 = 8.22?
for frequency ? 10 Hz



17

12
10
8.22 1.644
50
XX = = ? = ?
Starting phase current at 10 Hz
22
1 2 1 2
22
( ) ( )
100
(5.39 5.72) (1.644 1.644)
pn
pn
V
I
R R X X
=
+ + +
=
++

I Pn = 8.63A
Starting line current = 3
L Ph
II =
3 8.63
14.95
L
L
I
IA
=?
=


32. Given data L = 50mH, C = 0.05 ?F
Critical resistance to avoid current shopping will be given as
3
6
1 1 50 10
22
0.05 10
L
R
C
?
?
?
==
?

R = 500?

33.

X eq = 0.25 + 0.2 +
0.4
2

X eq = 0.65 PU
P = V PU I PV cos?
0.8 = 1 ? I PV ? 0.8
I PU = 1 PU
1 36.86 I = ? ? [as 0.8 pf lagging]
1 1 36.86 0.65 1.484 20.51
eq
E V jIX
E j Pu
=+
= + ? ? ? = ?
-



18

? = 20.51 degrees

34.

A. 600 mV B. 500 mV
C. 400 mV D. 100 mV

2
12
22
1
11
22
21
1 2 1 1
22
21
1 2 1 1
2 [ ]
1
1
1
out x
L
out
L
out
R
Vx V VoltagedivisionRule
RR
RR
V V V
RR
R R R
V V V
R R R R
R R R
V V V
R R R R
=
+
??
= + ?
??
??
??
= + ?
??
+
??
??
= + ?
??
+
??




19

2 2 2
2 1 2 1
1 1 1
()
100
(50 10)
10
400
out
out
out
R R R
V V V V V
R R R
V
V mV
= ? = ?
=?
=


35.



Output XY XY
XY
=+
=?

The above expression is for XOR gate

36. Discharging of capacitor equation
V C (t) = V oe
?t/?

Where ? = RC = (10
3
) (10
?7
) = 10
?4
sec
V o = 100V
V c(t) = 100 e
?104t

V c(t) = 1V
1 = 100 e
?104t

T = 0.46 msec

37.
( )
0
11
cos sin .
nn
nn
f t a a nt b nt
??
==
= + +
??





20

2
11
2
1
1
2
( )cos ( )
2
| sin cos
2
sin cos
sin2 cos2
2 2 2
2
( )sin ( )
T
n
o
x
o
T
o
o
o
T
n
o
a x t n t d t
T
a A t tdt
x
A
t tdt
A t A t
a
ao
b x t n t d t
T
?
?
?
? ?
??
?
??
??
=
=
=
=
=
?
??
==
??
??
=
=
?
?
?
?
?

1
2
1
1
1
2
sin sin
2
sin
1 cos2
()
22
2
o
o
o
b A t t dt
A
b t dt
At
b dt
A
b
?
?
?
?
?
?
=
=
=?
=
?
?
?


38.



21

2 2 2
3 2 2 2 3 2
2 2 3
3 2 2 2 3 2
2 2 3
? ??
2 3 4 ,
??
(2 2 2 ) (3 3 3 )
?
(4 4 4 )
( ) (2 2 2 ) (3 3 3 )
4
(4 4 4 )
A xi yj zk U x y z
UA x xy xz i x y y yz j
x z y z z k
dd
div UA x xy xz x y y yz
dx d
d
x z y z z
dz
= + + = + +
= + + + + +
+ + +
= + + + + +
+ + +

div (UA) = (6x
2
+ 2y
2
+ 2z
2
) + (3x
2
+ 9y
2
+ 3z
2
) + (4x
2
+ 4y
2
+ 12z
2
)
at (1, 1, 1) ? x = 1, y = 1, z = 1
div (UA) = 45

39. PMMC Instrument
I fs = 10 mA
R m = 10?

100 = I fs (R m + R se)
100 = 10 ? 10
?3
(10 + R sc)
R se = 10000 ? 10 = 9990?

40.

11
22
1
2
0 1 0
2
10
xx
r
xx
x
y
x
? ? ?
? ? ? ? ? ? ? ?
=+
? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ? ?
??
=??
??
??
??


We know
X AX Bu
Y CX Du
=+
=+

Comparing the above equation with the given problem
0 1 0
2
AB
? ? ?
? ? ? ?
==
? ? ? ?
??
? ? ? ?

C = (1 0)
Characteristic equation is
|SI ? A| = 0



22

0 0 1
0
02
1
2
S
S
S
S
??
??
? ? ? ?
?=
? ? ? ?
??
? ? ? ?
?
=
+

s
2
+ 2S? + ? = 0 (1)
s
2
+ 2?? ns + ? n
2
= 0 (2)
Comparing (1) and (2)
? n
2
= ?
n
?? =
2?? n = 2?
n
??
?
?
?
==

41.


( )
2
6 6 2
2
1
( C ) ( )
2
1
5 10 (100 10 ) (10) ( 0.5)
2
0.5 0.01
0.6V 600mV
D n ox gs t
out
out
out
w
I V V
L
V
V
V
?
??
??
=?
??
??
? = ? ? ? ?
?=
==


42.
FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?





14

/
( ) sin( )
sin( )
t M
m
V
i t Ae t
Z
V
DC offset A
z
?
??
??
?
= + ?
?
= = ?

For Maximum value of DC offset A
? ? ? = ? 90
1
3
1
tan 90
377 10 10
tan 90
3.77
45 90
45
L
R
?
?
?
?
?
?
?
?
??
? = ?
??
??
??
??
? = ?
??
??
??
? ? = ? ?
= ? ?


25. M is a 2 ? 2 Matrix with Eigen value 4 and 9 If has ? 1, ? 2 _ _ _ _ _ ? n Eigen values
M
n
? ? 1
n
, ? 2
n
_ _ _ ? n
n
Eigen values
M
2
? 4
2
, 9
2

? M
2
has Eigen values as 16 and 81

26. V S = 400 KV
l = 300 km
L 1 = 1 mH / km / phase
C 1 = 0.01 ?F / km / phase
5
36
11
5
22
11
3.16 10 /
1 10 0.01 10
2 2 50 300
' 0.29
3.16 10
(0.29)
1 1 0.955
22
400
418.85KV
0.955
s
R
v km s
LC
fl
v
A
V
V
A
??
?
?
??
= = = ?
? ? ?
??
= = =
?
= ? = ? =
= = =


27. According to Mill man?s Theorem, the equivalent circuit of the given circuit is




15




3 1 2 4
1 2 3 4
1 2 3 4
1 1 1 1
eq
E E E E
R R R R
E
R R R R
+++
=
+++

200 160 100 80
5 40 25 20
1 1 1 1
50 40 25 20
? ? ? ?
+ + ? + ?
? ? ? ?
? ? ? ?
=
+ + +

E eq = 0V
So, the current I flowing is 0 A

28. For synchronous motor
E g = V 1 ? IZ
220
()
3
t
V V Phase =
Z = (0.25 + j 2.5)?
I = 10 ? ?36.86 A
220
(0.25 2.5) 10 36.86
3
g
Ej = ? + ? ? ?
E g = 141.658 ??8.7 V (phase)
E g = 245.36 V (line)




16

29.
32
| | 5
32
2
8
2 ( )
2
(8
2 lim (z 2)
2
8 4 8
28
1
z
z
zz
dz j sumof residues
z
zz
j
z
jj
?
?
??
=
?
++
=
+
??
++
= ? +
??
+
??
??
? + +
??
==
??
??
?


30.
V (t) = ? 170 sin (377 )
6
t
?
?
I (t) = 8 cos (377 )
6
t
?
+

V(t) = ? 170 sin (377 )
6
t
?
?
V(t) = 170 cos (377 )
62
t
??
?+
V(t)=170 cos (377 )
3
t
?
+
P = V rms I rms cos?
170 8
cos30
22
P =
P = 588.89 watts

31. Given R 1 = 5.39?, R 2 = 5.72?, X 1 = X 2 = 8.22?
for frequency ? 10 Hz



17

12
10
8.22 1.644
50
XX = = ? = ?
Starting phase current at 10 Hz
22
1 2 1 2
22
( ) ( )
100
(5.39 5.72) (1.644 1.644)
pn
pn
V
I
R R X X
=
+ + +
=
++

I Pn = 8.63A
Starting line current = 3
L Ph
II =
3 8.63
14.95
L
L
I
IA
=?
=


32. Given data L = 50mH, C = 0.05 ?F
Critical resistance to avoid current shopping will be given as
3
6
1 1 50 10
22
0.05 10
L
R
C
?
?
?
==
?

R = 500?

33.

X eq = 0.25 + 0.2 +
0.4
2

X eq = 0.65 PU
P = V PU I PV cos?
0.8 = 1 ? I PV ? 0.8
I PU = 1 PU
1 36.86 I = ? ? [as 0.8 pf lagging]
1 1 36.86 0.65 1.484 20.51
eq
E V jIX
E j Pu
=+
= + ? ? ? = ?
-



18

? = 20.51 degrees

34.

A. 600 mV B. 500 mV
C. 400 mV D. 100 mV

2
12
22
1
11
22
21
1 2 1 1
22
21
1 2 1 1
2 [ ]
1
1
1
out x
L
out
L
out
R
Vx V VoltagedivisionRule
RR
RR
V V V
RR
R R R
V V V
R R R R
R R R
V V V
R R R R
=
+
??
= + ?
??
??
??
= + ?
??
+
??
??
= + ?
??
+
??




19

2 2 2
2 1 2 1
1 1 1
()
100
(50 10)
10
400
out
out
out
R R R
V V V V V
R R R
V
V mV
= ? = ?
=?
=


35.



Output XY XY
XY
=+
=?

The above expression is for XOR gate

36. Discharging of capacitor equation
V C (t) = V oe
?t/?

Where ? = RC = (10
3
) (10
?7
) = 10
?4
sec
V o = 100V
V c(t) = 100 e
?104t

V c(t) = 1V
1 = 100 e
?104t

T = 0.46 msec

37.
( )
0
11
cos sin .
nn
nn
f t a a nt b nt
??
==
= + +
??





20

2
11
2
1
1
2
( )cos ( )
2
| sin cos
2
sin cos
sin2 cos2
2 2 2
2
( )sin ( )
T
n
o
x
o
T
o
o
o
T
n
o
a x t n t d t
T
a A t tdt
x
A
t tdt
A t A t
a
ao
b x t n t d t
T
?
?
?
? ?
??
?
??
??
=
=
=
=
=
?
??
==
??
??
=
=
?
?
?
?
?

1
2
1
1
1
2
sin sin
2
sin
1 cos2
()
22
2
o
o
o
b A t t dt
A
b t dt
At
b dt
A
b
?
?
?
?
?
?
=
=
=?
=
?
?
?


38.



21

2 2 2
3 2 2 2 3 2
2 2 3
3 2 2 2 3 2
2 2 3
? ??
2 3 4 ,
??
(2 2 2 ) (3 3 3 )
?
(4 4 4 )
( ) (2 2 2 ) (3 3 3 )
4
(4 4 4 )
A xi yj zk U x y z
UA x xy xz i x y y yz j
x z y z z k
dd
div UA x xy xz x y y yz
dx d
d
x z y z z
dz
= + + = + +
= + + + + +
+ + +
= + + + + +
+ + +

div (UA) = (6x
2
+ 2y
2
+ 2z
2
) + (3x
2
+ 9y
2
+ 3z
2
) + (4x
2
+ 4y
2
+ 12z
2
)
at (1, 1, 1) ? x = 1, y = 1, z = 1
div (UA) = 45

39. PMMC Instrument
I fs = 10 mA
R m = 10?

100 = I fs (R m + R se)
100 = 10 ? 10
?3
(10 + R sc)
R se = 10000 ? 10 = 9990?

40.

11
22
1
2
0 1 0
2
10
xx
r
xx
x
y
x
? ? ?
? ? ? ? ? ? ? ?
=+
? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ? ?
??
=??
??
??
??


We know
X AX Bu
Y CX Du
=+
=+

Comparing the above equation with the given problem
0 1 0
2
AB
? ? ?
? ? ? ?
==
? ? ? ?
??
? ? ? ?

C = (1 0)
Characteristic equation is
|SI ? A| = 0



22

0 0 1
0
02
1
2
S
S
S
S
??
??
? ? ? ?
?=
? ? ? ?
??
? ? ? ?
?
=
+

s
2
+ 2S? + ? = 0 (1)
s
2
+ 2?? ns + ? n
2
= 0 (2)
Comparing (1) and (2)
? n
2
= ?
n
?? =
2?? n = 2?
n
??
?
?
?
==

41.


( )
2
6 6 2
2
1
( C ) ( )
2
1
5 10 (100 10 ) (10) ( 0.5)
2
0.5 0.01
0.6V 600mV
D n ox gs t
out
out
out
w
I V V
L
V
V
V
?
??
??
=?
??
??
? = ? ? ? ?
?=
==


42.



23



From the given Bode plot,
T(S) = Transfer function =
s 1 1
1 20
K
ss
? ? ? ?
++
? ? ? ?
? ? ? ?

It has three poles and no zero
So, statement 1 is false
?T(s) = ? 90 ? tan
?1
w ? tan
?1

20
w

?T(jw) |w ? ? = ? 270
o

So, statement 2 is true


43. Load supplied previously before adding extra load
12 KW at pf of 0.6
S Load = 12 + j16
Now, Let P be extra load added (Q extra = as unity p.f)
S Load = 12 + P + j16
|S Load| =
22
(12 ) 16 P ++
Rated KVA |S rated| = 25
22
2 2 2
25 (12 ) 16
25 (12 ) 16
7.5, 31.2
P
P
P
= + +
= + +
=?

So, 7.20 KW is extra load which is added

44.
M
?1
M = I
FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?





14

/
( ) sin( )
sin( )
t M
m
V
i t Ae t
Z
V
DC offset A
z
?
??
??
?
= + ?
?
= = ?

For Maximum value of DC offset A
? ? ? = ? 90
1
3
1
tan 90
377 10 10
tan 90
3.77
45 90
45
L
R
?
?
?
?
?
?
?
?
??
? = ?
??
??
??
??
? = ?
??
??
??
? ? = ? ?
= ? ?


25. M is a 2 ? 2 Matrix with Eigen value 4 and 9 If has ? 1, ? 2 _ _ _ _ _ ? n Eigen values
M
n
? ? 1
n
, ? 2
n
_ _ _ ? n
n
Eigen values
M
2
? 4
2
, 9
2

? M
2
has Eigen values as 16 and 81

26. V S = 400 KV
l = 300 km
L 1 = 1 mH / km / phase
C 1 = 0.01 ?F / km / phase
5
36
11
5
22
11
3.16 10 /
1 10 0.01 10
2 2 50 300
' 0.29
3.16 10
(0.29)
1 1 0.955
22
400
418.85KV
0.955
s
R
v km s
LC
fl
v
A
V
V
A
??
?
?
??
= = = ?
? ? ?
??
= = =
?
= ? = ? =
= = =


27. According to Mill man?s Theorem, the equivalent circuit of the given circuit is




15




3 1 2 4
1 2 3 4
1 2 3 4
1 1 1 1
eq
E E E E
R R R R
E
R R R R
+++
=
+++

200 160 100 80
5 40 25 20
1 1 1 1
50 40 25 20
? ? ? ?
+ + ? + ?
? ? ? ?
? ? ? ?
=
+ + +

E eq = 0V
So, the current I flowing is 0 A

28. For synchronous motor
E g = V 1 ? IZ
220
()
3
t
V V Phase =
Z = (0.25 + j 2.5)?
I = 10 ? ?36.86 A
220
(0.25 2.5) 10 36.86
3
g
Ej = ? + ? ? ?
E g = 141.658 ??8.7 V (phase)
E g = 245.36 V (line)




16

29.
32
| | 5
32
2
8
2 ( )
2
(8
2 lim (z 2)
2
8 4 8
28
1
z
z
zz
dz j sumof residues
z
zz
j
z
jj
?
?
??
=
?
++
=
+
??
++
= ? +
??
+
??
??
? + +
??
==
??
??
?


30.
V (t) = ? 170 sin (377 )
6
t
?
?
I (t) = 8 cos (377 )
6
t
?
+

V(t) = ? 170 sin (377 )
6
t
?
?
V(t) = 170 cos (377 )
62
t
??
?+
V(t)=170 cos (377 )
3
t
?
+
P = V rms I rms cos?
170 8
cos30
22
P =
P = 588.89 watts

31. Given R 1 = 5.39?, R 2 = 5.72?, X 1 = X 2 = 8.22?
for frequency ? 10 Hz



17

12
10
8.22 1.644
50
XX = = ? = ?
Starting phase current at 10 Hz
22
1 2 1 2
22
( ) ( )
100
(5.39 5.72) (1.644 1.644)
pn
pn
V
I
R R X X
=
+ + +
=
++

I Pn = 8.63A
Starting line current = 3
L Ph
II =
3 8.63
14.95
L
L
I
IA
=?
=


32. Given data L = 50mH, C = 0.05 ?F
Critical resistance to avoid current shopping will be given as
3
6
1 1 50 10
22
0.05 10
L
R
C
?
?
?
==
?

R = 500?

33.

X eq = 0.25 + 0.2 +
0.4
2

X eq = 0.65 PU
P = V PU I PV cos?
0.8 = 1 ? I PV ? 0.8
I PU = 1 PU
1 36.86 I = ? ? [as 0.8 pf lagging]
1 1 36.86 0.65 1.484 20.51
eq
E V jIX
E j Pu
=+
= + ? ? ? = ?
-



18

? = 20.51 degrees

34.

A. 600 mV B. 500 mV
C. 400 mV D. 100 mV

2
12
22
1
11
22
21
1 2 1 1
22
21
1 2 1 1
2 [ ]
1
1
1
out x
L
out
L
out
R
Vx V VoltagedivisionRule
RR
RR
V V V
RR
R R R
V V V
R R R R
R R R
V V V
R R R R
=
+
??
= + ?
??
??
??
= + ?
??
+
??
??
= + ?
??
+
??




19

2 2 2
2 1 2 1
1 1 1
()
100
(50 10)
10
400
out
out
out
R R R
V V V V V
R R R
V
V mV
= ? = ?
=?
=


35.



Output XY XY
XY
=+
=?

The above expression is for XOR gate

36. Discharging of capacitor equation
V C (t) = V oe
?t/?

Where ? = RC = (10
3
) (10
?7
) = 10
?4
sec
V o = 100V
V c(t) = 100 e
?104t

V c(t) = 1V
1 = 100 e
?104t

T = 0.46 msec

37.
( )
0
11
cos sin .
nn
nn
f t a a nt b nt
??
==
= + +
??





20

2
11
2
1
1
2
( )cos ( )
2
| sin cos
2
sin cos
sin2 cos2
2 2 2
2
( )sin ( )
T
n
o
x
o
T
o
o
o
T
n
o
a x t n t d t
T
a A t tdt
x
A
t tdt
A t A t
a
ao
b x t n t d t
T
?
?
?
? ?
??
?
??
??
=
=
=
=
=
?
??
==
??
??
=
=
?
?
?
?
?

1
2
1
1
1
2
sin sin
2
sin
1 cos2
()
22
2
o
o
o
b A t t dt
A
b t dt
At
b dt
A
b
?
?
?
?
?
?
=
=
=?
=
?
?
?


38.



21

2 2 2
3 2 2 2 3 2
2 2 3
3 2 2 2 3 2
2 2 3
? ??
2 3 4 ,
??
(2 2 2 ) (3 3 3 )
?
(4 4 4 )
( ) (2 2 2 ) (3 3 3 )
4
(4 4 4 )
A xi yj zk U x y z
UA x xy xz i x y y yz j
x z y z z k
dd
div UA x xy xz x y y yz
dx d
d
x z y z z
dz
= + + = + +
= + + + + +
+ + +
= + + + + +
+ + +

div (UA) = (6x
2
+ 2y
2
+ 2z
2
) + (3x
2
+ 9y
2
+ 3z
2
) + (4x
2
+ 4y
2
+ 12z
2
)
at (1, 1, 1) ? x = 1, y = 1, z = 1
div (UA) = 45

39. PMMC Instrument
I fs = 10 mA
R m = 10?

100 = I fs (R m + R se)
100 = 10 ? 10
?3
(10 + R sc)
R se = 10000 ? 10 = 9990?

40.

11
22
1
2
0 1 0
2
10
xx
r
xx
x
y
x
? ? ?
? ? ? ? ? ? ? ?
=+
? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ? ?
??
=??
??
??
??


We know
X AX Bu
Y CX Du
=+
=+

Comparing the above equation with the given problem
0 1 0
2
AB
? ? ?
? ? ? ?
==
? ? ? ?
??
? ? ? ?

C = (1 0)
Characteristic equation is
|SI ? A| = 0



22

0 0 1
0
02
1
2
S
S
S
S
??
??
? ? ? ?
?=
? ? ? ?
??
? ? ? ?
?
=
+

s
2
+ 2S? + ? = 0 (1)
s
2
+ 2?? ns + ? n
2
= 0 (2)
Comparing (1) and (2)
? n
2
= ?
n
?? =
2?? n = 2?
n
??
?
?
?
==

41.


( )
2
6 6 2
2
1
( C ) ( )
2
1
5 10 (100 10 ) (10) ( 0.5)
2
0.5 0.01
0.6V 600mV
D n ox gs t
out
out
out
w
I V V
L
V
V
V
?
??
??
=?
??
??
? = ? ? ? ?
?=
==


42.



23



From the given Bode plot,
T(S) = Transfer function =
s 1 1
1 20
K
ss
? ? ? ?
++
? ? ? ?
? ? ? ?

It has three poles and no zero
So, statement 1 is false
?T(s) = ? 90 ? tan
?1
w ? tan
?1

20
w

?T(jw) |w ? ? = ? 270
o

So, statement 2 is true


43. Load supplied previously before adding extra load
12 KW at pf of 0.6
S Load = 12 + j16
Now, Let P be extra load added (Q extra = as unity p.f)
S Load = 12 + P + j16
|S Load| =
22
(12 ) 16 P ++
Rated KVA |S rated| = 25
22
2 2 2
25 (12 ) 16
25 (12 ) 16
7.5, 31.2
P
P
P
= + +
= + +
=?

So, 7.20 KW is extra load which is added

44.
M
?1
M = I



24

1
12
2
1 1 1 2
2 1 2 2
10
[]
01
10
01
T
T
VV
Vv
U
VV
U
U T U T
U T U T
??
??
= ??
??
?? ??
??
? ? ? ?
=
? ? ? ?
? ? ? ?

1 1 1 2
2 1 2 2
10
01
TT
TT
U V U V
U V U V
==
==

Statement 1 and 2 are both correct

45.
V sr I sr cos ? = V oI o
For single phase fully ? controlled converter
I o = I sr = 10A
180
cos 0.78
230
o
sr
V
V
? = = =

46. Given that
Switch frequency, f s = 250Hz
Load resistance R L = 24?
Supply voltage V s = 48V
T ON = 1 msec
22 2
1
4
0.25
() (0.25 48)
24
s
ON
os
T ms
f
T
T
VV
Loadpower
RR
?
?
==
==
?
= = =

P = 6 watts

47. P o = 120w, Vs = 24V, V o = 48V
1
24
1
48
s
o
V
V
?
?
=
?
?=

? = 0.5 [Duty cycle]
FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?





14

/
( ) sin( )
sin( )
t M
m
V
i t Ae t
Z
V
DC offset A
z
?
??
??
?
= + ?
?
= = ?

For Maximum value of DC offset A
? ? ? = ? 90
1
3
1
tan 90
377 10 10
tan 90
3.77
45 90
45
L
R
?
?
?
?
?
?
?
?
??
? = ?
??
??
??
??
? = ?
??
??
??
? ? = ? ?
= ? ?


25. M is a 2 ? 2 Matrix with Eigen value 4 and 9 If has ? 1, ? 2 _ _ _ _ _ ? n Eigen values
M
n
? ? 1
n
, ? 2
n
_ _ _ ? n
n
Eigen values
M
2
? 4
2
, 9
2

? M
2
has Eigen values as 16 and 81

26. V S = 400 KV
l = 300 km
L 1 = 1 mH / km / phase
C 1 = 0.01 ?F / km / phase
5
36
11
5
22
11
3.16 10 /
1 10 0.01 10
2 2 50 300
' 0.29
3.16 10
(0.29)
1 1 0.955
22
400
418.85KV
0.955
s
R
v km s
LC
fl
v
A
V
V
A
??
?
?
??
= = = ?
? ? ?
??
= = =
?
= ? = ? =
= = =


27. According to Mill man?s Theorem, the equivalent circuit of the given circuit is




15




3 1 2 4
1 2 3 4
1 2 3 4
1 1 1 1
eq
E E E E
R R R R
E
R R R R
+++
=
+++

200 160 100 80
5 40 25 20
1 1 1 1
50 40 25 20
? ? ? ?
+ + ? + ?
? ? ? ?
? ? ? ?
=
+ + +

E eq = 0V
So, the current I flowing is 0 A

28. For synchronous motor
E g = V 1 ? IZ
220
()
3
t
V V Phase =
Z = (0.25 + j 2.5)?
I = 10 ? ?36.86 A
220
(0.25 2.5) 10 36.86
3
g
Ej = ? + ? ? ?
E g = 141.658 ??8.7 V (phase)
E g = 245.36 V (line)




16

29.
32
| | 5
32
2
8
2 ( )
2
(8
2 lim (z 2)
2
8 4 8
28
1
z
z
zz
dz j sumof residues
z
zz
j
z
jj
?
?
??
=
?
++
=
+
??
++
= ? +
??
+
??
??
? + +
??
==
??
??
?


30.
V (t) = ? 170 sin (377 )
6
t
?
?
I (t) = 8 cos (377 )
6
t
?
+

V(t) = ? 170 sin (377 )
6
t
?
?
V(t) = 170 cos (377 )
62
t
??
?+
V(t)=170 cos (377 )
3
t
?
+
P = V rms I rms cos?
170 8
cos30
22
P =
P = 588.89 watts

31. Given R 1 = 5.39?, R 2 = 5.72?, X 1 = X 2 = 8.22?
for frequency ? 10 Hz



17

12
10
8.22 1.644
50
XX = = ? = ?
Starting phase current at 10 Hz
22
1 2 1 2
22
( ) ( )
100
(5.39 5.72) (1.644 1.644)
pn
pn
V
I
R R X X
=
+ + +
=
++

I Pn = 8.63A
Starting line current = 3
L Ph
II =
3 8.63
14.95
L
L
I
IA
=?
=


32. Given data L = 50mH, C = 0.05 ?F
Critical resistance to avoid current shopping will be given as
3
6
1 1 50 10
22
0.05 10
L
R
C
?
?
?
==
?

R = 500?

33.

X eq = 0.25 + 0.2 +
0.4
2

X eq = 0.65 PU
P = V PU I PV cos?
0.8 = 1 ? I PV ? 0.8
I PU = 1 PU
1 36.86 I = ? ? [as 0.8 pf lagging]
1 1 36.86 0.65 1.484 20.51
eq
E V jIX
E j Pu
=+
= + ? ? ? = ?
-



18

? = 20.51 degrees

34.

A. 600 mV B. 500 mV
C. 400 mV D. 100 mV

2
12
22
1
11
22
21
1 2 1 1
22
21
1 2 1 1
2 [ ]
1
1
1
out x
L
out
L
out
R
Vx V VoltagedivisionRule
RR
RR
V V V
RR
R R R
V V V
R R R R
R R R
V V V
R R R R
=
+
??
= + ?
??
??
??
= + ?
??
+
??
??
= + ?
??
+
??




19

2 2 2
2 1 2 1
1 1 1
()
100
(50 10)
10
400
out
out
out
R R R
V V V V V
R R R
V
V mV
= ? = ?
=?
=


35.



Output XY XY
XY
=+
=?

The above expression is for XOR gate

36. Discharging of capacitor equation
V C (t) = V oe
?t/?

Where ? = RC = (10
3
) (10
?7
) = 10
?4
sec
V o = 100V
V c(t) = 100 e
?104t

V c(t) = 1V
1 = 100 e
?104t

T = 0.46 msec

37.
( )
0
11
cos sin .
nn
nn
f t a a nt b nt
??
==
= + +
??





20

2
11
2
1
1
2
( )cos ( )
2
| sin cos
2
sin cos
sin2 cos2
2 2 2
2
( )sin ( )
T
n
o
x
o
T
o
o
o
T
n
o
a x t n t d t
T
a A t tdt
x
A
t tdt
A t A t
a
ao
b x t n t d t
T
?
?
?
? ?
??
?
??
??
=
=
=
=
=
?
??
==
??
??
=
=
?
?
?
?
?

1
2
1
1
1
2
sin sin
2
sin
1 cos2
()
22
2
o
o
o
b A t t dt
A
b t dt
At
b dt
A
b
?
?
?
?
?
?
=
=
=?
=
?
?
?


38.



21

2 2 2
3 2 2 2 3 2
2 2 3
3 2 2 2 3 2
2 2 3
? ??
2 3 4 ,
??
(2 2 2 ) (3 3 3 )
?
(4 4 4 )
( ) (2 2 2 ) (3 3 3 )
4
(4 4 4 )
A xi yj zk U x y z
UA x xy xz i x y y yz j
x z y z z k
dd
div UA x xy xz x y y yz
dx d
d
x z y z z
dz
= + + = + +
= + + + + +
+ + +
= + + + + +
+ + +

div (UA) = (6x
2
+ 2y
2
+ 2z
2
) + (3x
2
+ 9y
2
+ 3z
2
) + (4x
2
+ 4y
2
+ 12z
2
)
at (1, 1, 1) ? x = 1, y = 1, z = 1
div (UA) = 45

39. PMMC Instrument
I fs = 10 mA
R m = 10?

100 = I fs (R m + R se)
100 = 10 ? 10
?3
(10 + R sc)
R se = 10000 ? 10 = 9990?

40.

11
22
1
2
0 1 0
2
10
xx
r
xx
x
y
x
? ? ?
? ? ? ? ? ? ? ?
=+
? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ? ?
??
=??
??
??
??


We know
X AX Bu
Y CX Du
=+
=+

Comparing the above equation with the given problem
0 1 0
2
AB
? ? ?
? ? ? ?
==
? ? ? ?
??
? ? ? ?

C = (1 0)
Characteristic equation is
|SI ? A| = 0



22

0 0 1
0
02
1
2
S
S
S
S
??
??
? ? ? ?
?=
? ? ? ?
??
? ? ? ?
?
=
+

s
2
+ 2S? + ? = 0 (1)
s
2
+ 2?? ns + ? n
2
= 0 (2)
Comparing (1) and (2)
? n
2
= ?
n
?? =
2?? n = 2?
n
??
?
?
?
==

41.


( )
2
6 6 2
2
1
( C ) ( )
2
1
5 10 (100 10 ) (10) ( 0.5)
2
0.5 0.01
0.6V 600mV
D n ox gs t
out
out
out
w
I V V
L
V
V
V
?
??
??
=?
??
??
? = ? ? ? ?
?=
==


42.



23



From the given Bode plot,
T(S) = Transfer function =
s 1 1
1 20
K
ss
? ? ? ?
++
? ? ? ?
? ? ? ?

It has three poles and no zero
So, statement 1 is false
?T(s) = ? 90 ? tan
?1
w ? tan
?1

20
w

?T(jw) |w ? ? = ? 270
o

So, statement 2 is true


43. Load supplied previously before adding extra load
12 KW at pf of 0.6
S Load = 12 + j16
Now, Let P be extra load added (Q extra = as unity p.f)
S Load = 12 + P + j16
|S Load| =
22
(12 ) 16 P ++
Rated KVA |S rated| = 25
22
2 2 2
25 (12 ) 16
25 (12 ) 16
7.5, 31.2
P
P
P
= + +
= + +
=?

So, 7.20 KW is extra load which is added

44.
M
?1
M = I



24

1
12
2
1 1 1 2
2 1 2 2
10
[]
01
10
01
T
T
VV
Vv
U
VV
U
U T U T
U T U T
??
??
= ??
??
?? ??
??
? ? ? ?
=
? ? ? ?
? ? ? ?

1 1 1 2
2 1 2 2
10
01
TT
TT
U V U V
U V U V
==
==

Statement 1 and 2 are both correct

45.
V sr I sr cos ? = V oI o
For single phase fully ? controlled converter
I o = I sr = 10A
180
cos 0.78
230
o
sr
V
V
? = = =

46. Given that
Switch frequency, f s = 250Hz
Load resistance R L = 24?
Supply voltage V s = 48V
T ON = 1 msec
22 2
1
4
0.25
() (0.25 48)
24
s
ON
os
T ms
f
T
T
VV
Loadpower
RR
?
?
==
==
?
= = =

P = 6 watts

47. P o = 120w, Vs = 24V, V o = 48V
1
24
1
48
s
o
V
V
?
?
=
?
?=

? = 0.5 [Duty cycle]



25

P o = V oI o = 120
120
2.54
48
o
IA ==
V SI S = V oI o

120
5
24
s
IA ==
At boundary of continuous & discontinuous
3
2
25
0.5 24
24
50 10 10
L
LS
S
L
Lc
C
I
II
V
I
f
LH
?
?
?
==
? = = ?
?
==
??


48.


No load
I NL = 3A
220 220
1
220
C
IA
Rf
===
I a = I L ? I f = 2A
Back cmf = Eb N = V ? I aR a
= 220 ? 2 ? 0.5 = 219V
Full load
I FL = 25A N f = 1500 rpm
I f = 1A
I a = I FL ? I f = 24A
EbF = V ? I aR a = 220 ? 24 ? 0.5 = 208 V
We know E ? speed (N)
f
bF
bN N
N
E
EN
= (N N = speed at no load)
FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?





14

/
( ) sin( )
sin( )
t M
m
V
i t Ae t
Z
V
DC offset A
z
?
??
??
?
= + ?
?
= = ?

For Maximum value of DC offset A
? ? ? = ? 90
1
3
1
tan 90
377 10 10
tan 90
3.77
45 90
45
L
R
?
?
?
?
?
?
?
?
??
? = ?
??
??
??
??
? = ?
??
??
??
? ? = ? ?
= ? ?


25. M is a 2 ? 2 Matrix with Eigen value 4 and 9 If has ? 1, ? 2 _ _ _ _ _ ? n Eigen values
M
n
? ? 1
n
, ? 2
n
_ _ _ ? n
n
Eigen values
M
2
? 4
2
, 9
2

? M
2
has Eigen values as 16 and 81

26. V S = 400 KV
l = 300 km
L 1 = 1 mH / km / phase
C 1 = 0.01 ?F / km / phase
5
36
11
5
22
11
3.16 10 /
1 10 0.01 10
2 2 50 300
' 0.29
3.16 10
(0.29)
1 1 0.955
22
400
418.85KV
0.955
s
R
v km s
LC
fl
v
A
V
V
A
??
?
?
??
= = = ?
? ? ?
??
= = =
?
= ? = ? =
= = =


27. According to Mill man?s Theorem, the equivalent circuit of the given circuit is




15




3 1 2 4
1 2 3 4
1 2 3 4
1 1 1 1
eq
E E E E
R R R R
E
R R R R
+++
=
+++

200 160 100 80
5 40 25 20
1 1 1 1
50 40 25 20
? ? ? ?
+ + ? + ?
? ? ? ?
? ? ? ?
=
+ + +

E eq = 0V
So, the current I flowing is 0 A

28. For synchronous motor
E g = V 1 ? IZ
220
()
3
t
V V Phase =
Z = (0.25 + j 2.5)?
I = 10 ? ?36.86 A
220
(0.25 2.5) 10 36.86
3
g
Ej = ? + ? ? ?
E g = 141.658 ??8.7 V (phase)
E g = 245.36 V (line)




16

29.
32
| | 5
32
2
8
2 ( )
2
(8
2 lim (z 2)
2
8 4 8
28
1
z
z
zz
dz j sumof residues
z
zz
j
z
jj
?
?
??
=
?
++
=
+
??
++
= ? +
??
+
??
??
? + +
??
==
??
??
?


30.
V (t) = ? 170 sin (377 )
6
t
?
?
I (t) = 8 cos (377 )
6
t
?
+

V(t) = ? 170 sin (377 )
6
t
?
?
V(t) = 170 cos (377 )
62
t
??
?+
V(t)=170 cos (377 )
3
t
?
+
P = V rms I rms cos?
170 8
cos30
22
P =
P = 588.89 watts

31. Given R 1 = 5.39?, R 2 = 5.72?, X 1 = X 2 = 8.22?
for frequency ? 10 Hz



17

12
10
8.22 1.644
50
XX = = ? = ?
Starting phase current at 10 Hz
22
1 2 1 2
22
( ) ( )
100
(5.39 5.72) (1.644 1.644)
pn
pn
V
I
R R X X
=
+ + +
=
++

I Pn = 8.63A
Starting line current = 3
L Ph
II =
3 8.63
14.95
L
L
I
IA
=?
=


32. Given data L = 50mH, C = 0.05 ?F
Critical resistance to avoid current shopping will be given as
3
6
1 1 50 10
22
0.05 10
L
R
C
?
?
?
==
?

R = 500?

33.

X eq = 0.25 + 0.2 +
0.4
2

X eq = 0.65 PU
P = V PU I PV cos?
0.8 = 1 ? I PV ? 0.8
I PU = 1 PU
1 36.86 I = ? ? [as 0.8 pf lagging]
1 1 36.86 0.65 1.484 20.51
eq
E V jIX
E j Pu
=+
= + ? ? ? = ?
-



18

? = 20.51 degrees

34.

A. 600 mV B. 500 mV
C. 400 mV D. 100 mV

2
12
22
1
11
22
21
1 2 1 1
22
21
1 2 1 1
2 [ ]
1
1
1
out x
L
out
L
out
R
Vx V VoltagedivisionRule
RR
RR
V V V
RR
R R R
V V V
R R R R
R R R
V V V
R R R R
=
+
??
= + ?
??
??
??
= + ?
??
+
??
??
= + ?
??
+
??




19

2 2 2
2 1 2 1
1 1 1
()
100
(50 10)
10
400
out
out
out
R R R
V V V V V
R R R
V
V mV
= ? = ?
=?
=


35.



Output XY XY
XY
=+
=?

The above expression is for XOR gate

36. Discharging of capacitor equation
V C (t) = V oe
?t/?

Where ? = RC = (10
3
) (10
?7
) = 10
?4
sec
V o = 100V
V c(t) = 100 e
?104t

V c(t) = 1V
1 = 100 e
?104t

T = 0.46 msec

37.
( )
0
11
cos sin .
nn
nn
f t a a nt b nt
??
==
= + +
??





20

2
11
2
1
1
2
( )cos ( )
2
| sin cos
2
sin cos
sin2 cos2
2 2 2
2
( )sin ( )
T
n
o
x
o
T
o
o
o
T
n
o
a x t n t d t
T
a A t tdt
x
A
t tdt
A t A t
a
ao
b x t n t d t
T
?
?
?
? ?
??
?
??
??
=
=
=
=
=
?
??
==
??
??
=
=
?
?
?
?
?

1
2
1
1
1
2
sin sin
2
sin
1 cos2
()
22
2
o
o
o
b A t t dt
A
b t dt
At
b dt
A
b
?
?
?
?
?
?
=
=
=?
=
?
?
?


38.



21

2 2 2
3 2 2 2 3 2
2 2 3
3 2 2 2 3 2
2 2 3
? ??
2 3 4 ,
??
(2 2 2 ) (3 3 3 )
?
(4 4 4 )
( ) (2 2 2 ) (3 3 3 )
4
(4 4 4 )
A xi yj zk U x y z
UA x xy xz i x y y yz j
x z y z z k
dd
div UA x xy xz x y y yz
dx d
d
x z y z z
dz
= + + = + +
= + + + + +
+ + +
= + + + + +
+ + +

div (UA) = (6x
2
+ 2y
2
+ 2z
2
) + (3x
2
+ 9y
2
+ 3z
2
) + (4x
2
+ 4y
2
+ 12z
2
)
at (1, 1, 1) ? x = 1, y = 1, z = 1
div (UA) = 45

39. PMMC Instrument
I fs = 10 mA
R m = 10?

100 = I fs (R m + R se)
100 = 10 ? 10
?3
(10 + R sc)
R se = 10000 ? 10 = 9990?

40.

11
22
1
2
0 1 0
2
10
xx
r
xx
x
y
x
? ? ?
? ? ? ? ? ? ? ?
=+
? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ? ?
??
=??
??
??
??


We know
X AX Bu
Y CX Du
=+
=+

Comparing the above equation with the given problem
0 1 0
2
AB
? ? ?
? ? ? ?
==
? ? ? ?
??
? ? ? ?

C = (1 0)
Characteristic equation is
|SI ? A| = 0



22

0 0 1
0
02
1
2
S
S
S
S
??
??
? ? ? ?
?=
? ? ? ?
??
? ? ? ?
?
=
+

s
2
+ 2S? + ? = 0 (1)
s
2
+ 2?? ns + ? n
2
= 0 (2)
Comparing (1) and (2)
? n
2
= ?
n
?? =
2?? n = 2?
n
??
?
?
?
==

41.


( )
2
6 6 2
2
1
( C ) ( )
2
1
5 10 (100 10 ) (10) ( 0.5)
2
0.5 0.01
0.6V 600mV
D n ox gs t
out
out
out
w
I V V
L
V
V
V
?
??
??
=?
??
??
? = ? ? ? ?
?=
==


42.



23



From the given Bode plot,
T(S) = Transfer function =
s 1 1
1 20
K
ss
? ? ? ?
++
? ? ? ?
? ? ? ?

It has three poles and no zero
So, statement 1 is false
?T(s) = ? 90 ? tan
?1
w ? tan
?1

20
w

?T(jw) |w ? ? = ? 270
o

So, statement 2 is true


43. Load supplied previously before adding extra load
12 KW at pf of 0.6
S Load = 12 + j16
Now, Let P be extra load added (Q extra = as unity p.f)
S Load = 12 + P + j16
|S Load| =
22
(12 ) 16 P ++
Rated KVA |S rated| = 25
22
2 2 2
25 (12 ) 16
25 (12 ) 16
7.5, 31.2
P
P
P
= + +
= + +
=?

So, 7.20 KW is extra load which is added

44.
M
?1
M = I



24

1
12
2
1 1 1 2
2 1 2 2
10
[]
01
10
01
T
T
VV
Vv
U
VV
U
U T U T
U T U T
??
??
= ??
??
?? ??
??
? ? ? ?
=
? ? ? ?
? ? ? ?

1 1 1 2
2 1 2 2
10
01
TT
TT
U V U V
U V U V
==
==

Statement 1 and 2 are both correct

45.
V sr I sr cos ? = V oI o
For single phase fully ? controlled converter
I o = I sr = 10A
180
cos 0.78
230
o
sr
V
V
? = = =

46. Given that
Switch frequency, f s = 250Hz
Load resistance R L = 24?
Supply voltage V s = 48V
T ON = 1 msec
22 2
1
4
0.25
() (0.25 48)
24
s
ON
os
T ms
f
T
T
VV
Loadpower
RR
?
?
==
==
?
= = =

P = 6 watts

47. P o = 120w, Vs = 24V, V o = 48V
1
24
1
48
s
o
V
V
?
?
=
?
?=

? = 0.5 [Duty cycle]



25

P o = V oI o = 120
120
2.54
48
o
IA ==
V SI S = V oI o

120
5
24
s
IA ==
At boundary of continuous & discontinuous
3
2
25
0.5 24
24
50 10 10
L
LS
S
L
Lc
C
I
II
V
I
f
LH
?
?
?
==
? = = ?
?
==
??


48.


No load
I NL = 3A
220 220
1
220
C
IA
Rf
===
I a = I L ? I f = 2A
Back cmf = Eb N = V ? I aR a
= 220 ? 2 ? 0.5 = 219V
Full load
I FL = 25A N f = 1500 rpm
I f = 1A
I a = I FL ? I f = 24A
EbF = V ? I aR a = 220 ? 24 ? 0.5 = 208 V
We know E ? speed (N)
f
bF
bN N
N
E
EN
= (N N = speed at no load)



26

208 1500
219
N
N
=
N N = 1579.33 rpm
49.
Ac line current rms = (I s) rms =
22
100 81.65
33
o
IA ==
50.




F (P, Q, R, S) = S QR +

51.
P = 0.02
n = 50
? = np = 50 (0.02) = 1
P (x > 2) = 1 ? P (x < 2)
= 1 ? [P(x =0) + P (x = 1)]
1
1 1 (1 )
01 11
o
ee
e
??
?
??
??
?
??
= ? + = ? +
??
??
??

FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?





14

/
( ) sin( )
sin( )
t M
m
V
i t Ae t
Z
V
DC offset A
z
?
??
??
?
= + ?
?
= = ?

For Maximum value of DC offset A
? ? ? = ? 90
1
3
1
tan 90
377 10 10
tan 90
3.77
45 90
45
L
R
?
?
?
?
?
?
?
?
??
? = ?
??
??
??
??
? = ?
??
??
??
? ? = ? ?
= ? ?


25. M is a 2 ? 2 Matrix with Eigen value 4 and 9 If has ? 1, ? 2 _ _ _ _ _ ? n Eigen values
M
n
? ? 1
n
, ? 2
n
_ _ _ ? n
n
Eigen values
M
2
? 4
2
, 9
2

? M
2
has Eigen values as 16 and 81

26. V S = 400 KV
l = 300 km
L 1 = 1 mH / km / phase
C 1 = 0.01 ?F / km / phase
5
36
11
5
22
11
3.16 10 /
1 10 0.01 10
2 2 50 300
' 0.29
3.16 10
(0.29)
1 1 0.955
22
400
418.85KV
0.955
s
R
v km s
LC
fl
v
A
V
V
A
??
?
?
??
= = = ?
? ? ?
??
= = =
?
= ? = ? =
= = =


27. According to Mill man?s Theorem, the equivalent circuit of the given circuit is




15




3 1 2 4
1 2 3 4
1 2 3 4
1 1 1 1
eq
E E E E
R R R R
E
R R R R
+++
=
+++

200 160 100 80
5 40 25 20
1 1 1 1
50 40 25 20
? ? ? ?
+ + ? + ?
? ? ? ?
? ? ? ?
=
+ + +

E eq = 0V
So, the current I flowing is 0 A

28. For synchronous motor
E g = V 1 ? IZ
220
()
3
t
V V Phase =
Z = (0.25 + j 2.5)?
I = 10 ? ?36.86 A
220
(0.25 2.5) 10 36.86
3
g
Ej = ? + ? ? ?
E g = 141.658 ??8.7 V (phase)
E g = 245.36 V (line)




16

29.
32
| | 5
32
2
8
2 ( )
2
(8
2 lim (z 2)
2
8 4 8
28
1
z
z
zz
dz j sumof residues
z
zz
j
z
jj
?
?
??
=
?
++
=
+
??
++
= ? +
??
+
??
??
? + +
??
==
??
??
?


30.
V (t) = ? 170 sin (377 )
6
t
?
?
I (t) = 8 cos (377 )
6
t
?
+

V(t) = ? 170 sin (377 )
6
t
?
?
V(t) = 170 cos (377 )
62
t
??
?+
V(t)=170 cos (377 )
3
t
?
+
P = V rms I rms cos?
170 8
cos30
22
P =
P = 588.89 watts

31. Given R 1 = 5.39?, R 2 = 5.72?, X 1 = X 2 = 8.22?
for frequency ? 10 Hz



17

12
10
8.22 1.644
50
XX = = ? = ?
Starting phase current at 10 Hz
22
1 2 1 2
22
( ) ( )
100
(5.39 5.72) (1.644 1.644)
pn
pn
V
I
R R X X
=
+ + +
=
++

I Pn = 8.63A
Starting line current = 3
L Ph
II =
3 8.63
14.95
L
L
I
IA
=?
=


32. Given data L = 50mH, C = 0.05 ?F
Critical resistance to avoid current shopping will be given as
3
6
1 1 50 10
22
0.05 10
L
R
C
?
?
?
==
?

R = 500?

33.

X eq = 0.25 + 0.2 +
0.4
2

X eq = 0.65 PU
P = V PU I PV cos?
0.8 = 1 ? I PV ? 0.8
I PU = 1 PU
1 36.86 I = ? ? [as 0.8 pf lagging]
1 1 36.86 0.65 1.484 20.51
eq
E V jIX
E j Pu
=+
= + ? ? ? = ?
-



18

? = 20.51 degrees

34.

A. 600 mV B. 500 mV
C. 400 mV D. 100 mV

2
12
22
1
11
22
21
1 2 1 1
22
21
1 2 1 1
2 [ ]
1
1
1
out x
L
out
L
out
R
Vx V VoltagedivisionRule
RR
RR
V V V
RR
R R R
V V V
R R R R
R R R
V V V
R R R R
=
+
??
= + ?
??
??
??
= + ?
??
+
??
??
= + ?
??
+
??




19

2 2 2
2 1 2 1
1 1 1
()
100
(50 10)
10
400
out
out
out
R R R
V V V V V
R R R
V
V mV
= ? = ?
=?
=


35.



Output XY XY
XY
=+
=?

The above expression is for XOR gate

36. Discharging of capacitor equation
V C (t) = V oe
?t/?

Where ? = RC = (10
3
) (10
?7
) = 10
?4
sec
V o = 100V
V c(t) = 100 e
?104t

V c(t) = 1V
1 = 100 e
?104t

T = 0.46 msec

37.
( )
0
11
cos sin .
nn
nn
f t a a nt b nt
??
==
= + +
??





20

2
11
2
1
1
2
( )cos ( )
2
| sin cos
2
sin cos
sin2 cos2
2 2 2
2
( )sin ( )
T
n
o
x
o
T
o
o
o
T
n
o
a x t n t d t
T
a A t tdt
x
A
t tdt
A t A t
a
ao
b x t n t d t
T
?
?
?
? ?
??
?
??
??
=
=
=
=
=
?
??
==
??
??
=
=
?
?
?
?
?

1
2
1
1
1
2
sin sin
2
sin
1 cos2
()
22
2
o
o
o
b A t t dt
A
b t dt
At
b dt
A
b
?
?
?
?
?
?
=
=
=?
=
?
?
?


38.



21

2 2 2
3 2 2 2 3 2
2 2 3
3 2 2 2 3 2
2 2 3
? ??
2 3 4 ,
??
(2 2 2 ) (3 3 3 )
?
(4 4 4 )
( ) (2 2 2 ) (3 3 3 )
4
(4 4 4 )
A xi yj zk U x y z
UA x xy xz i x y y yz j
x z y z z k
dd
div UA x xy xz x y y yz
dx d
d
x z y z z
dz
= + + = + +
= + + + + +
+ + +
= + + + + +
+ + +

div (UA) = (6x
2
+ 2y
2
+ 2z
2
) + (3x
2
+ 9y
2
+ 3z
2
) + (4x
2
+ 4y
2
+ 12z
2
)
at (1, 1, 1) ? x = 1, y = 1, z = 1
div (UA) = 45

39. PMMC Instrument
I fs = 10 mA
R m = 10?

100 = I fs (R m + R se)
100 = 10 ? 10
?3
(10 + R sc)
R se = 10000 ? 10 = 9990?

40.

11
22
1
2
0 1 0
2
10
xx
r
xx
x
y
x
? ? ?
? ? ? ? ? ? ? ?
=+
? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ? ?
??
=??
??
??
??


We know
X AX Bu
Y CX Du
=+
=+

Comparing the above equation with the given problem
0 1 0
2
AB
? ? ?
? ? ? ?
==
? ? ? ?
??
? ? ? ?

C = (1 0)
Characteristic equation is
|SI ? A| = 0



22

0 0 1
0
02
1
2
S
S
S
S
??
??
? ? ? ?
?=
? ? ? ?
??
? ? ? ?
?
=
+

s
2
+ 2S? + ? = 0 (1)
s
2
+ 2?? ns + ? n
2
= 0 (2)
Comparing (1) and (2)
? n
2
= ?
n
?? =
2?? n = 2?
n
??
?
?
?
==

41.


( )
2
6 6 2
2
1
( C ) ( )
2
1
5 10 (100 10 ) (10) ( 0.5)
2
0.5 0.01
0.6V 600mV
D n ox gs t
out
out
out
w
I V V
L
V
V
V
?
??
??
=?
??
??
? = ? ? ? ?
?=
==


42.



23



From the given Bode plot,
T(S) = Transfer function =
s 1 1
1 20
K
ss
? ? ? ?
++
? ? ? ?
? ? ? ?

It has three poles and no zero
So, statement 1 is false
?T(s) = ? 90 ? tan
?1
w ? tan
?1

20
w

?T(jw) |w ? ? = ? 270
o

So, statement 2 is true


43. Load supplied previously before adding extra load
12 KW at pf of 0.6
S Load = 12 + j16
Now, Let P be extra load added (Q extra = as unity p.f)
S Load = 12 + P + j16
|S Load| =
22
(12 ) 16 P ++
Rated KVA |S rated| = 25
22
2 2 2
25 (12 ) 16
25 (12 ) 16
7.5, 31.2
P
P
P
= + +
= + +
=?

So, 7.20 KW is extra load which is added

44.
M
?1
M = I



24

1
12
2
1 1 1 2
2 1 2 2
10
[]
01
10
01
T
T
VV
Vv
U
VV
U
U T U T
U T U T
??
??
= ??
??
?? ??
??
? ? ? ?
=
? ? ? ?
? ? ? ?

1 1 1 2
2 1 2 2
10
01
TT
TT
U V U V
U V U V
==
==

Statement 1 and 2 are both correct

45.
V sr I sr cos ? = V oI o
For single phase fully ? controlled converter
I o = I sr = 10A
180
cos 0.78
230
o
sr
V
V
? = = =

46. Given that
Switch frequency, f s = 250Hz
Load resistance R L = 24?
Supply voltage V s = 48V
T ON = 1 msec
22 2
1
4
0.25
() (0.25 48)
24
s
ON
os
T ms
f
T
T
VV
Loadpower
RR
?
?
==
==
?
= = =

P = 6 watts

47. P o = 120w, Vs = 24V, V o = 48V
1
24
1
48
s
o
V
V
?
?
=
?
?=

? = 0.5 [Duty cycle]



25

P o = V oI o = 120
120
2.54
48
o
IA ==
V SI S = V oI o

120
5
24
s
IA ==
At boundary of continuous & discontinuous
3
2
25
0.5 24
24
50 10 10
L
LS
S
L
Lc
C
I
II
V
I
f
LH
?
?
?
==
? = = ?
?
==
??


48.


No load
I NL = 3A
220 220
1
220
C
IA
Rf
===
I a = I L ? I f = 2A
Back cmf = Eb N = V ? I aR a
= 220 ? 2 ? 0.5 = 219V
Full load
I FL = 25A N f = 1500 rpm
I f = 1A
I a = I FL ? I f = 24A
EbF = V ? I aR a = 220 ? 24 ? 0.5 = 208 V
We know E ? speed (N)
f
bF
bN N
N
E
EN
= (N N = speed at no load)



26

208 1500
219
N
N
=
N N = 1579.33 rpm
49.
Ac line current rms = (I s) rms =
22
100 81.65
33
o
IA ==
50.




F (P, Q, R, S) = S QR +

51.
P = 0.02
n = 50
? = np = 50 (0.02) = 1
P (x > 2) = 1 ? P (x < 2)
= 1 ? [P(x =0) + P (x = 1)]
1
1 1 (1 )
01 11
o
ee
e
??
?
??
??
?
??
= ? + = ? +
??
??
??




27

P (x > 2) = 1 ? e
?1
(1 + 1) = 0.26
52.

L air = 0.2 cm
L m = 40 cm
Given B o = 1 Tesla at
r
? ? ?
L core = 40 ? 0.2 = 39.8cm
Let a = uniform cross ? sectional area
We know that
Re tan
MMF NI
flux
Total luc ce S
? = = =
S T = S airgap + S core
(1)A
1
air core
o o r
core
air
or
LL
A
L
SL
A
? ? ?
??
=+
??
=+
??
??

Case 1: when ? r ? ?, B = 1T
MMF = NI 1 = B 1 A
1
core
air
ro
L
L
A ??
??
+ ? ?
??
??

NI 1 = 1 (a) [l air]
1
air
oo
L
A ??
?=
1
(1)
air
o
l
NI
?
=
Case 2:
? r = 1000
MMF = Same
FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?





14

/
( ) sin( )
sin( )
t M
m
V
i t Ae t
Z
V
DC offset A
z
?
??
??
?
= + ?
?
= = ?

For Maximum value of DC offset A
? ? ? = ? 90
1
3
1
tan 90
377 10 10
tan 90
3.77
45 90
45
L
R
?
?
?
?
?
?
?
?
??
? = ?
??
??
??
??
? = ?
??
??
??
? ? = ? ?
= ? ?


25. M is a 2 ? 2 Matrix with Eigen value 4 and 9 If has ? 1, ? 2 _ _ _ _ _ ? n Eigen values
M
n
? ? 1
n
, ? 2
n
_ _ _ ? n
n
Eigen values
M
2
? 4
2
, 9
2

? M
2
has Eigen values as 16 and 81

26. V S = 400 KV
l = 300 km
L 1 = 1 mH / km / phase
C 1 = 0.01 ?F / km / phase
5
36
11
5
22
11
3.16 10 /
1 10 0.01 10
2 2 50 300
' 0.29
3.16 10
(0.29)
1 1 0.955
22
400
418.85KV
0.955
s
R
v km s
LC
fl
v
A
V
V
A
??
?
?
??
= = = ?
? ? ?
??
= = =
?
= ? = ? =
= = =


27. According to Mill man?s Theorem, the equivalent circuit of the given circuit is




15




3 1 2 4
1 2 3 4
1 2 3 4
1 1 1 1
eq
E E E E
R R R R
E
R R R R
+++
=
+++

200 160 100 80
5 40 25 20
1 1 1 1
50 40 25 20
? ? ? ?
+ + ? + ?
? ? ? ?
? ? ? ?
=
+ + +

E eq = 0V
So, the current I flowing is 0 A

28. For synchronous motor
E g = V 1 ? IZ
220
()
3
t
V V Phase =
Z = (0.25 + j 2.5)?
I = 10 ? ?36.86 A
220
(0.25 2.5) 10 36.86
3
g
Ej = ? + ? ? ?
E g = 141.658 ??8.7 V (phase)
E g = 245.36 V (line)




16

29.
32
| | 5
32
2
8
2 ( )
2
(8
2 lim (z 2)
2
8 4 8
28
1
z
z
zz
dz j sumof residues
z
zz
j
z
jj
?
?
??
=
?
++
=
+
??
++
= ? +
??
+
??
??
? + +
??
==
??
??
?


30.
V (t) = ? 170 sin (377 )
6
t
?
?
I (t) = 8 cos (377 )
6
t
?
+

V(t) = ? 170 sin (377 )
6
t
?
?
V(t) = 170 cos (377 )
62
t
??
?+
V(t)=170 cos (377 )
3
t
?
+
P = V rms I rms cos?
170 8
cos30
22
P =
P = 588.89 watts

31. Given R 1 = 5.39?, R 2 = 5.72?, X 1 = X 2 = 8.22?
for frequency ? 10 Hz



17

12
10
8.22 1.644
50
XX = = ? = ?
Starting phase current at 10 Hz
22
1 2 1 2
22
( ) ( )
100
(5.39 5.72) (1.644 1.644)
pn
pn
V
I
R R X X
=
+ + +
=
++

I Pn = 8.63A
Starting line current = 3
L Ph
II =
3 8.63
14.95
L
L
I
IA
=?
=


32. Given data L = 50mH, C = 0.05 ?F
Critical resistance to avoid current shopping will be given as
3
6
1 1 50 10
22
0.05 10
L
R
C
?
?
?
==
?

R = 500?

33.

X eq = 0.25 + 0.2 +
0.4
2

X eq = 0.65 PU
P = V PU I PV cos?
0.8 = 1 ? I PV ? 0.8
I PU = 1 PU
1 36.86 I = ? ? [as 0.8 pf lagging]
1 1 36.86 0.65 1.484 20.51
eq
E V jIX
E j Pu
=+
= + ? ? ? = ?
-



18

? = 20.51 degrees

34.

A. 600 mV B. 500 mV
C. 400 mV D. 100 mV

2
12
22
1
11
22
21
1 2 1 1
22
21
1 2 1 1
2 [ ]
1
1
1
out x
L
out
L
out
R
Vx V VoltagedivisionRule
RR
RR
V V V
RR
R R R
V V V
R R R R
R R R
V V V
R R R R
=
+
??
= + ?
??
??
??
= + ?
??
+
??
??
= + ?
??
+
??




19

2 2 2
2 1 2 1
1 1 1
()
100
(50 10)
10
400
out
out
out
R R R
V V V V V
R R R
V
V mV
= ? = ?
=?
=


35.



Output XY XY
XY
=+
=?

The above expression is for XOR gate

36. Discharging of capacitor equation
V C (t) = V oe
?t/?

Where ? = RC = (10
3
) (10
?7
) = 10
?4
sec
V o = 100V
V c(t) = 100 e
?104t

V c(t) = 1V
1 = 100 e
?104t

T = 0.46 msec

37.
( )
0
11
cos sin .
nn
nn
f t a a nt b nt
??
==
= + +
??





20

2
11
2
1
1
2
( )cos ( )
2
| sin cos
2
sin cos
sin2 cos2
2 2 2
2
( )sin ( )
T
n
o
x
o
T
o
o
o
T
n
o
a x t n t d t
T
a A t tdt
x
A
t tdt
A t A t
a
ao
b x t n t d t
T
?
?
?
? ?
??
?
??
??
=
=
=
=
=
?
??
==
??
??
=
=
?
?
?
?
?

1
2
1
1
1
2
sin sin
2
sin
1 cos2
()
22
2
o
o
o
b A t t dt
A
b t dt
At
b dt
A
b
?
?
?
?
?
?
=
=
=?
=
?
?
?


38.



21

2 2 2
3 2 2 2 3 2
2 2 3
3 2 2 2 3 2
2 2 3
? ??
2 3 4 ,
??
(2 2 2 ) (3 3 3 )
?
(4 4 4 )
( ) (2 2 2 ) (3 3 3 )
4
(4 4 4 )
A xi yj zk U x y z
UA x xy xz i x y y yz j
x z y z z k
dd
div UA x xy xz x y y yz
dx d
d
x z y z z
dz
= + + = + +
= + + + + +
+ + +
= + + + + +
+ + +

div (UA) = (6x
2
+ 2y
2
+ 2z
2
) + (3x
2
+ 9y
2
+ 3z
2
) + (4x
2
+ 4y
2
+ 12z
2
)
at (1, 1, 1) ? x = 1, y = 1, z = 1
div (UA) = 45

39. PMMC Instrument
I fs = 10 mA
R m = 10?

100 = I fs (R m + R se)
100 = 10 ? 10
?3
(10 + R sc)
R se = 10000 ? 10 = 9990?

40.

11
22
1
2
0 1 0
2
10
xx
r
xx
x
y
x
? ? ?
? ? ? ? ? ? ? ?
=+
? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ? ?
??
=??
??
??
??


We know
X AX Bu
Y CX Du
=+
=+

Comparing the above equation with the given problem
0 1 0
2
AB
? ? ?
? ? ? ?
==
? ? ? ?
??
? ? ? ?

C = (1 0)
Characteristic equation is
|SI ? A| = 0



22

0 0 1
0
02
1
2
S
S
S
S
??
??
? ? ? ?
?=
? ? ? ?
??
? ? ? ?
?
=
+

s
2
+ 2S? + ? = 0 (1)
s
2
+ 2?? ns + ? n
2
= 0 (2)
Comparing (1) and (2)
? n
2
= ?
n
?? =
2?? n = 2?
n
??
?
?
?
==

41.


( )
2
6 6 2
2
1
( C ) ( )
2
1
5 10 (100 10 ) (10) ( 0.5)
2
0.5 0.01
0.6V 600mV
D n ox gs t
out
out
out
w
I V V
L
V
V
V
?
??
??
=?
??
??
? = ? ? ? ?
?=
==


42.



23



From the given Bode plot,
T(S) = Transfer function =
s 1 1
1 20
K
ss
? ? ? ?
++
? ? ? ?
? ? ? ?

It has three poles and no zero
So, statement 1 is false
?T(s) = ? 90 ? tan
?1
w ? tan
?1

20
w

?T(jw) |w ? ? = ? 270
o

So, statement 2 is true


43. Load supplied previously before adding extra load
12 KW at pf of 0.6
S Load = 12 + j16
Now, Let P be extra load added (Q extra = as unity p.f)
S Load = 12 + P + j16
|S Load| =
22
(12 ) 16 P ++
Rated KVA |S rated| = 25
22
2 2 2
25 (12 ) 16
25 (12 ) 16
7.5, 31.2
P
P
P
= + +
= + +
=?

So, 7.20 KW is extra load which is added

44.
M
?1
M = I



24

1
12
2
1 1 1 2
2 1 2 2
10
[]
01
10
01
T
T
VV
Vv
U
VV
U
U T U T
U T U T
??
??
= ??
??
?? ??
??
? ? ? ?
=
? ? ? ?
? ? ? ?

1 1 1 2
2 1 2 2
10
01
TT
TT
U V U V
U V U V
==
==

Statement 1 and 2 are both correct

45.
V sr I sr cos ? = V oI o
For single phase fully ? controlled converter
I o = I sr = 10A
180
cos 0.78
230
o
sr
V
V
? = = =

46. Given that
Switch frequency, f s = 250Hz
Load resistance R L = 24?
Supply voltage V s = 48V
T ON = 1 msec
22 2
1
4
0.25
() (0.25 48)
24
s
ON
os
T ms
f
T
T
VV
Loadpower
RR
?
?
==
==
?
= = =

P = 6 watts

47. P o = 120w, Vs = 24V, V o = 48V
1
24
1
48
s
o
V
V
?
?
=
?
?=

? = 0.5 [Duty cycle]



25

P o = V oI o = 120
120
2.54
48
o
IA ==
V SI S = V oI o

120
5
24
s
IA ==
At boundary of continuous & discontinuous
3
2
25
0.5 24
24
50 10 10
L
LS
S
L
Lc
C
I
II
V
I
f
LH
?
?
?
==
? = = ?
?
==
??


48.


No load
I NL = 3A
220 220
1
220
C
IA
Rf
===
I a = I L ? I f = 2A
Back cmf = Eb N = V ? I aR a
= 220 ? 2 ? 0.5 = 219V
Full load
I FL = 25A N f = 1500 rpm
I f = 1A
I a = I FL ? I f = 24A
EbF = V ? I aR a = 220 ? 24 ? 0.5 = 208 V
We know E ? speed (N)
f
bF
bN N
N
E
EN
= (N N = speed at no load)



26

208 1500
219
N
N
=
N N = 1579.33 rpm
49.
Ac line current rms = (I s) rms =
22
100 81.65
33
o
IA ==
50.




F (P, Q, R, S) = S QR +

51.
P = 0.02
n = 50
? = np = 50 (0.02) = 1
P (x > 2) = 1 ? P (x < 2)
= 1 ? [P(x =0) + P (x = 1)]
1
1 1 (1 )
01 11
o
ee
e
??
?
??
??
?
??
= ? + = ? +
??
??
??




27

P (x > 2) = 1 ? e
?1
(1 + 1) = 0.26
52.

L air = 0.2 cm
L m = 40 cm
Given B o = 1 Tesla at
r
? ? ?
L core = 40 ? 0.2 = 39.8cm
Let a = uniform cross ? sectional area
We know that
Re tan
MMF NI
flux
Total luc ce S
? = = =
S T = S airgap + S core
(1)A
1
air core
o o r
core
air
or
LL
A
L
SL
A
? ? ?
??
=+
??
=+
??
??

Case 1: when ? r ? ?, B = 1T
MMF = NI 1 = B 1 A
1
core
air
ro
L
L
A ??
??
+ ? ?
??
??

NI 1 = 1 (a) [l air]
1
air
oo
L
A ??
?=
1
(1)
air
o
l
NI
?
=
Case 2:
? r = 1000
MMF = Same



28

NI 1 = B 2 A
1
core
air
ro
L
L
A ??
??
+
??
??

Put NI 1 from (1)
2
2
1
1000
39.8
0.2 0.2
1000
air core
air
o o
LL
BL
B
??
??
=+
??
??
??
=+
??
??

B L = 0.834 Tesla

53. Fault current for SLG fault
l
1 2 0
3
3
FG
V
I
X X X Xn
=
+ + +

Fault current for 3? fault
3
1
1 2 0 1
1 0 2
3
3
2
3
2(0.25) 0.05 0.15
3
f
n
V
I
X
VV
X X X X X
XX
Xn
X
?
=
=
+ + +
? ? ?
=
??
=

X n = 0.1 Pu
X n (in?) =
2
30
0.1
50
?
bass
Z MVA
Zpu
KVL
?
??
=
??
??

X n (in?) = 1.8?

54.
FirstRanker.com - FirstRanker's Choice



2

General Aptitude

1. Newspapers are a constant source of delight and recreation for me. The only (what bother?s) trouble is
that I read too (a lot/ large) many of them.

2. 343 = 7
3

1331 = 11
3

4913 = 17
3

All numbers given are cube of prime numbers so 13
3
= 2917 satisfy the missing number.

3. The passengers were angry with the airline staff about the delay.

4. Time taken by X to now the lawn = 2 hrs.
? Work done by X in
1
1
2
hr =
Similarly,
Work done by 4 in hr = ?
Work done by x + 4 in 1 hr =
1 1 3
2 4 4
+=
?Total time taken by X & 4 together =
4
3
hours
4
60 min
3
utes =?
= 80 Minutes

5. I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.

6. Given that X = {1, 2, 3}
4 = {2, 3, 4}
1 1 1 2 2 2 3 3 3
, , , ' , , , ,
2 3 4 2 3 4 2 4 3
Z
?
?
=
??
?
?

Minimum value in
1
4
z =
Maximum value in
3
2
z =



3

3
Pr
8
oduct =
7. Let number of boys participated = 4x
Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates =
80 28
7
100 5
xx ?=
Girls candidate who passed =
90 27
3
100 10
xx ?=
Boys candidate who passed = Total passed candidate ? Girls candidate who passed
28 27
5 10
29
10
29
100 72.5%
10 4
xx
x
x
x
=?
=
= ? =
?


8. The correct statement can be concluded from Venn diagram or using the Syllogism.

9.For all digits of a number which lie between 100 and 1000 are even,
Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred?s digit does not include 0 as it will not remain a number which lie between 100 and 1000
? Hundreds digit set is {2, 4, 6, 8}
Total integer be = 5 ? 5 ? 4
Totalchoies
Units Tens Hundreds
for
digit digit digit
? ? ?
?
?
?
?
?

Total integer = 100 numbers

10. Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana
Mita > Ganga
? Mita > Ganga > Rekha, Lakshmi > Sana
? 2 and statement 4 are correct




4

Electrical Engineering
1. Given that
Mean square of random process
2
()
kt
Ex
C
==
Mean is given zero ? E (x) = 0
We know that E(x
2
) ? [E(x)]
2
= variance
KT
Variance
C
=
Standard deviation = variance
KT
C
=

2. Applying R.H criteria for stability
?(S) = S
4
+ 3S
3
+ 3S
2
+ S + K = 0
S
4

S
3



S
2


S
1



S
0

1 3 K
3 1 0
8
3
K 0
8
3
3
8 / 3
K ?
0 0
K

For stability, first column should be greater than zero
8
3
3
00
8 / 3
8
0
9
K
and k
K
?
??
? ? ?


3.
2
3
()
21
S
HS
SS
+
=
++

H (t) = L
?1
[H(S)]



5

11
22
1 1 1
22
33
2 1 ( 1)
1 2 1 2
1
( 1) ( 1)
SS
LL
S S S
S
L L L
S
SS
??
? ? ?
??
++ ??
==
??
??
+ + + ??
??
? ? ? ?
++
??
= = +
? ? ? ?
??
+
?? ++
? ? ? ?


H (t) = e
?t
+2te
?t


4. We know that
Voltage Regulation = 100
NL FL
NL
VV
V
?
?
Given that V FL = 95V
V NL = 100 V
% VR =
100 95
100 5%
100
?
?=

5. We know that P = VI cos ?, as load and voltage are same
? I cos ? = constant
I 1 cos ? 1 = I 2 cos ? 2
I 1 = 200A
Cos ? 1 = 1
Cos ? 2 = 0.5
11
2
2
cos
400
cos
I
IA
?
?
==


6. We know that




6

1
1
2 (2 ) 2
ln ln
4
ln
o r
o
C
bR
ar
C
R
r
? ?
?
? ?
==
? ? ? ?
? ? ? ?
? ? ? ?
?
=
??
??
??



Total portion cover 2 ?
1
12
cov
4 4 2
2
r
portion ers
length for
??
?
? = =
?

1
3
2
r
and lengthfor
?
?
Both are connected in parallel

C 2 = C r1 + C r2



7

2
2
2 ( )
2 (2 ) 3
22
ln ln
[3 ]
2
ln
ro
o
r
o
RR
rr
R
r
?
? ??
?
??
?
= ? + ?
? ? ? ?
? ? ? ?
? ? ? ?
?
?
=+
??
??
??

21
2
2
2
4
[3 ] 2( )
2
ln ln
10
r
oo
r
GivenC C
E
RR
rr
??
=
?
?
+=
? ? ? ?
? ? ? ?
? ? ? ?
?=


7.


1
( .)
mm
m
V V
II
f Xm
V
f constt
? ? ? ?
??
?
?
=

By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which
changes the magnetizing reactance

8.
2
10
H( )
s(s 100 2)
s
s
=
++

For finding steady state value, we will apply final value theorem



8

( ) ( )
( )
( )
( )
0
2
0
lim lim
10
lim
100 2
1
10 2
ts
s
y t sY s
ys
s s s
y
? ? ?
?
=
? = ?
++
?=


9.
0.25
(s)
s
e
G
s
?
?
=
Nyquist plot cut the negative real
Axis at w = phase cross over frequency
0.25
()
180
90 0.25
( )| 180
180
90 0.25 180
45
90
2
j
pc
pc
pc
pc
pc
e
Gj
j
Gj
?
??
??
?
?
?
??
?
?
??
?
?
?
??
?
=
=
=
?
= ? ? ? ?
? = ?
= ? ? ? = ?
??
=
??
??
??
=

Magnitude at cutting point
( )
pc
X G j
?
? =
2
1
2
pc
x
??
??
==
=

Then, the co-ordinates becomes (-0.5, j0).




9


10. Given Z in = 10?, Z o/p = 100?
For CCCS

Series connection is output
Z o/p = Z o/p (1+A?) = 100 (1 + 9)
= 100 K?

11. We know that,
For 6-pulse converter harmonic present in AC current are 6K ? 1
General expression NK ? 1 [k = 0, 1, 2, 3]
For 6 pulse n = 6
Lowest order harmonic = 5
Lower harmonic frequency = 5 ? 50 = 250 Hz

12.


Applying nodal analysis at point 1 whose voltage is assumed as V 1.
11
1
20 5
2 0.......(1)
23
20
...........................(2)
2
V V I
V
I
??
? + =
?
=

Solving (1) and (2)
1
5
20
3
VI
I
?
? ? + =
8I = V 1 ? 6



10

8I = 20 ? 2I ? 6
10I = 14
I = 1.4 A

13.
2 2 2
2
2 2 2
22
2
22
0
d u d u d u
Waveequation c
dt dx dy
d u d u
Laplaceequation U
dx dy
??
=+
??
??
??
? = + =

2
222
2 2 2
0
Poissionequation U f
du d u d u d u
Heat equation
dt
dy dy dz
?
?=
??
? + + =
??
??
??


14.
2
1
2
z
For
z
?
+
, the singularity z = ?2 lies outside the |Z| < |
? By Cauchy?s integral theorem
2
1
0 | | |
2
z
dz for z
z
?
=?
+
?


15.
Given that
y = 2x
3
+ 3y
2
+ 4z
2
2
.?
? ??
? ??
? ??
6 6 4
. 6 6 4
C
grad f dr
dr dxi dyj dzk
df df df
grad f i j k
dx dy dz
x i yj k
grad f dr x dx ydy zdz
=
= + +
= + +
= + +
= + +
?
? ? ? ?

Applying the limits



11

2 3 2
2
3 3 2
. [ 6 6 4 ]
C
grad f dr x dx ydy dz
?
??
= + +
? ? ? ?

2 6 2 2 6 1
22
2 3 2 2 6 2
[ 6 6 4 ] [ 6 6 4 ] x dx ydy dz x ydy dz
?
?
= + + + + +
? ? ? ? ? ?

3 2 2 2 1
3 3 2
[2 ] [3 ] [4 ]
70 81 12 139
x y z
?
??
= + +
= + ? =


16.

Net reactance of generator
0.25
0.05 . .
5
Pr 1
20 . .
0.05
SC
X pu
e faultvoltage
I pu
X
==
?
= = =

Short Circuit MVA = I SC ?Base MVA
= 20 ? 5 = 100 MVA
17. For NMOS transistor to be in saturation the condition will be
V GS > V th
And V DS > V GS ? V Th
18.
I sec = 5 ? 20 = 100 A
V = I sec R = 100 ? 0.01 = 1V
VA output of CT = VI sec = 100 ? 1 100 VA

19.
Y 12 = ? (y 12) = ? j20
Series admittance of each line =
12
20
10
22
Y j
j
?
= = ?



12

Series reactance of each line =
1
0.1 . .
10
j pu
j
=
?

20.
011
101
110
M
??
??
=
??
??
??

Determinant of M = |M|
|M| = 0 [0 ? 1] ? 1 [0 ? 1] + 1 [1 ? 0]
|M| = 2
|M| ? 0
? Rank of M = number of columns
P (M) = 3

21. H (t) = 1 + e
?at
u (t)
?1? is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy
h (t) = 0 t < 0
?0 t > 0
Which it is not satisfying due to presence of constant
? It is not causal
22.
2
1 1 1
22
22
11
1
2
2 2 2
()
0
()
a s b s c
Hs
a s b s C
ab
C
Hs
a S b S C
++
=
++
==
=
++

At s = 0
H (0) = constant
At s = ?

? It is a low par filter
23. Waveform for output voltage of single phase full bridge PWM inverter



13


61
4
sin sin
2
o
nk
Vdc n
V nd n t
n
?
?
?
=?
=
?

V o1rms = fundamental r ms output voltage
1
22
sin sin
2
o
V Vdc d
?
?
=
Given, V o1 = 0.754 V dc
1
22
0.75 sin
0.75
sin 56.44
0.9
dc dc
V V d
d
?
?
=
??
==
??
??

Pulse width = 2d = 112.88

24. For series R ? L circuit, I (t) expression is


A. 60 B. 90
C. -30 D. -45
/
2 2 2 2
( ) sin( sin( )
t mm
LL
VV
i t e t
R X R X
?
? ? ? ?
?
?
?
???
= ? + ?
??
++ ??
?
?





14

/
( ) sin( )
sin( )
t M
m
V
i t Ae t
Z
V
DC offset A
z
?
??
??
?
= + ?
?
= = ?

For Maximum value of DC offset A
? ? ? = ? 90
1
3
1
tan 90
377 10 10
tan 90
3.77
45 90
45
L
R
?
?
?
?
?
?
?
?
??
? = ?
??
??
??
??
? = ?
??
??
??
? ? = ? ?
= ? ?


25. M is a 2 ? 2 Matrix with Eigen value 4 and 9 If has ? 1, ? 2 _ _ _ _ _ ? n Eigen values
M
n
? ? 1
n
, ? 2
n
_ _ _ ? n
n
Eigen values
M
2
? 4
2
, 9
2

? M
2
has Eigen values as 16 and 81

26. V S = 400 KV
l = 300 km
L 1 = 1 mH / km / phase
C 1 = 0.01 ?F / km / phase
5
36
11
5
22
11
3.16 10 /
1 10 0.01 10
2 2 50 300
' 0.29
3.16 10
(0.29)
1 1 0.955
22
400
418.85KV
0.955
s
R
v km s
LC
fl
v
A
V
V
A
??
?
?
??
= = = ?
? ? ?
??
= = =
?
= ? = ? =
= = =


27. According to Mill man?s Theorem, the equivalent circuit of the given circuit is




15




3 1 2 4
1 2 3 4
1 2 3 4
1 1 1 1
eq
E E E E
R R R R
E
R R R R
+++
=
+++

200 160 100 80
5 40 25 20
1 1 1 1
50 40 25 20
? ? ? ?
+ + ? + ?
? ? ? ?
? ? ? ?
=
+ + +

E eq = 0V
So, the current I flowing is 0 A

28. For synchronous motor
E g = V 1 ? IZ
220
()
3
t
V V Phase =
Z = (0.25 + j 2.5)?
I = 10 ? ?36.86 A
220
(0.25 2.5) 10 36.86
3
g
Ej = ? + ? ? ?
E g = 141.658 ??8.7 V (phase)
E g = 245.36 V (line)




16

29.
32
| | 5
32
2
8
2 ( )
2
(8
2 lim (z 2)
2
8 4 8
28
1
z
z
zz
dz j sumof residues
z
zz
j
z
jj
?
?
??
=
?
++
=
+
??
++
= ? +
??
+
??
??
? + +
??
==
??
??
?


30.
V (t) = ? 170 sin (377 )
6
t
?
?
I (t) = 8 cos (377 )
6
t
?
+

V(t) = ? 170 sin (377 )
6
t
?
?
V(t) = 170 cos (377 )
62
t
??
?+
V(t)=170 cos (377 )
3
t
?
+
P = V rms I rms cos?
170 8
cos30
22
P =
P = 588.89 watts

31. Given R 1 = 5.39?, R 2 = 5.72?, X 1 = X 2 = 8.22?
for frequency ? 10 Hz



17

12
10
8.22 1.644
50
XX = = ? = ?
Starting phase current at 10 Hz
22
1 2 1 2
22
( ) ( )
100
(5.39 5.72) (1.644 1.644)
pn
pn
V
I
R R X X
=
+ + +
=
++

I Pn = 8.63A
Starting line current = 3
L Ph
II =
3 8.63
14.95
L
L
I
IA
=?
=


32. Given data L = 50mH, C = 0.05 ?F
Critical resistance to avoid current shopping will be given as
3
6
1 1 50 10
22
0.05 10
L
R
C
?
?
?
==
?

R = 500?

33.

X eq = 0.25 + 0.2 +
0.4
2

X eq = 0.65 PU
P = V PU I PV cos?
0.8 = 1 ? I PV ? 0.8
I PU = 1 PU
1 36.86 I = ? ? [as 0.8 pf lagging]
1 1 36.86 0.65 1.484 20.51
eq
E V jIX
E j Pu
=+
= + ? ? ? = ?
-



18

? = 20.51 degrees

34.

A. 600 mV B. 500 mV
C. 400 mV D. 100 mV

2
12
22
1
11
22
21
1 2 1 1
22
21
1 2 1 1
2 [ ]
1
1
1
out x
L
out
L
out
R
Vx V VoltagedivisionRule
RR
RR
V V V
RR
R R R
V V V
R R R R
R R R
V V V
R R R R
=
+
??
= + ?
??
??
??
= + ?
??
+
??
??
= + ?
??
+
??




19

2 2 2
2 1 2 1
1 1 1
()
100
(50 10)
10
400
out
out
out
R R R
V V V V V
R R R
V
V mV
= ? = ?
=?
=


35.



Output XY XY
XY
=+
=?

The above expression is for XOR gate

36. Discharging of capacitor equation
V C (t) = V oe
?t/?

Where ? = RC = (10
3
) (10
?7
) = 10
?4
sec
V o = 100V
V c(t) = 100 e
?104t

V c(t) = 1V
1 = 100 e
?104t

T = 0.46 msec

37.
( )
0
11
cos sin .
nn
nn
f t a a nt b nt
??
==
= + +
??





20

2
11
2
1
1
2
( )cos ( )
2
| sin cos
2
sin cos
sin2 cos2
2 2 2
2
( )sin ( )
T
n
o
x
o
T
o
o
o
T
n
o
a x t n t d t
T
a A t tdt
x
A
t tdt
A t A t
a
ao
b x t n t d t
T
?
?
?
? ?
??
?
??
??
=
=
=
=
=
?
??
==
??
??
=
=
?
?
?
?
?

1
2
1
1
1
2
sin sin
2
sin
1 cos2
()
22
2
o
o
o
b A t t dt
A
b t dt
At
b dt
A
b
?
?
?
?
?
?
=
=
=?
=
?
?
?


38.



21

2 2 2
3 2 2 2 3 2
2 2 3
3 2 2 2 3 2
2 2 3
? ??
2 3 4 ,
??
(2 2 2 ) (3 3 3 )
?
(4 4 4 )
( ) (2 2 2 ) (3 3 3 )
4
(4 4 4 )
A xi yj zk U x y z
UA x xy xz i x y y yz j
x z y z z k
dd
div UA x xy xz x y y yz
dx d
d
x z y z z
dz
= + + = + +
= + + + + +
+ + +
= + + + + +
+ + +

div (UA) = (6x
2
+ 2y
2
+ 2z
2
) + (3x
2
+ 9y
2
+ 3z
2
) + (4x
2
+ 4y
2
+ 12z
2
)
at (1, 1, 1) ? x = 1, y = 1, z = 1
div (UA) = 45

39. PMMC Instrument
I fs = 10 mA
R m = 10?

100 = I fs (R m + R se)
100 = 10 ? 10
?3
(10 + R sc)
R se = 10000 ? 10 = 9990?

40.

11
22
1
2
0 1 0
2
10
xx
r
xx
x
y
x
? ? ?
? ? ? ? ? ? ? ?
=+
? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ? ?
??
=??
??
??
??


We know
X AX Bu
Y CX Du
=+
=+

Comparing the above equation with the given problem
0 1 0
2
AB
? ? ?
? ? ? ?
==
? ? ? ?
??
? ? ? ?

C = (1 0)
Characteristic equation is
|SI ? A| = 0



22

0 0 1
0
02
1
2
S
S
S
S
??
??
? ? ? ?
?=
? ? ? ?
??
? ? ? ?
?
=
+

s
2
+ 2S? + ? = 0 (1)
s
2
+ 2?? ns + ? n
2
= 0 (2)
Comparing (1) and (2)
? n
2
= ?
n
?? =
2?? n = 2?
n
??
?
?
?
==

41.


( )
2
6 6 2
2
1
( C ) ( )
2
1
5 10 (100 10 ) (10) ( 0.5)
2
0.5 0.01
0.6V 600mV
D n ox gs t
out
out
out
w
I V V
L
V
V
V
?
??
??
=?
??
??
? = ? ? ? ?
?=
==


42.



23



From the given Bode plot,
T(S) = Transfer function =
s 1 1
1 20
K
ss
? ? ? ?
++
? ? ? ?
? ? ? ?

It has three poles and no zero
So, statement 1 is false
?T(s) = ? 90 ? tan
?1
w ? tan
?1

20
w

?T(jw) |w ? ? = ? 270
o

So, statement 2 is true


43. Load supplied previously before adding extra load
12 KW at pf of 0.6
S Load = 12 + j16
Now, Let P be extra load added (Q extra = as unity p.f)
S Load = 12 + P + j16
|S Load| =
22
(12 ) 16 P ++
Rated KVA |S rated| = 25
22
2 2 2
25 (12 ) 16
25 (12 ) 16
7.5, 31.2
P
P
P
= + +
= + +
=?

So, 7.20 KW is extra load which is added

44.
M
?1
M = I



24

1
12
2
1 1 1 2
2 1 2 2
10
[]
01
10
01
T
T
VV
Vv
U
VV
U
U T U T
U T U T
??
??
= ??
??
?? ??
??
? ? ? ?
=
? ? ? ?
? ? ? ?

1 1 1 2
2 1 2 2
10
01
TT
TT
U V U V
U V U V
==
==

Statement 1 and 2 are both correct

45.
V sr I sr cos ? = V oI o
For single phase fully ? controlled converter
I o = I sr = 10A
180
cos 0.78
230
o
sr
V
V
? = = =

46. Given that
Switch frequency, f s = 250Hz
Load resistance R L = 24?
Supply voltage V s = 48V
T ON = 1 msec
22 2
1
4
0.25
() (0.25 48)
24
s
ON
os
T ms
f
T
T
VV
Loadpower
RR
?
?
==
==
?
= = =

P = 6 watts

47. P o = 120w, Vs = 24V, V o = 48V
1
24
1
48
s
o
V
V
?
?
=
?
?=

? = 0.5 [Duty cycle]



25

P o = V oI o = 120
120
2.54
48
o
IA ==
V SI S = V oI o

120
5
24
s
IA ==
At boundary of continuous & discontinuous
3
2
25
0.5 24
24
50 10 10
L
LS
S
L
Lc
C
I
II
V
I
f
LH
?
?
?
==
? = = ?
?
==
??


48.


No load
I NL = 3A
220 220
1
220
C
IA
Rf
===
I a = I L ? I f = 2A
Back cmf = Eb N = V ? I aR a
= 220 ? 2 ? 0.5 = 219V
Full load
I FL = 25A N f = 1500 rpm
I f = 1A
I a = I FL ? I f = 24A
EbF = V ? I aR a = 220 ? 24 ? 0.5 = 208 V
We know E ? speed (N)
f
bF
bN N
N
E
EN
= (N N = speed at no load)



26

208 1500
219
N
N
=
N N = 1579.33 rpm
49.
Ac line current rms = (I s) rms =
22
100 81.65
33
o
IA ==
50.




F (P, Q, R, S) = S QR +

51.
P = 0.02
n = 50
? = np = 50 (0.02) = 1
P (x > 2) = 1 ? P (x < 2)
= 1 ? [P(x =0) + P (x = 1)]
1
1 1 (1 )
01 11
o
ee
e
??
?
??
??
?
??
= ? + = ? +
??
??
??




27

P (x > 2) = 1 ? e
?1
(1 + 1) = 0.26
52.

L air = 0.2 cm
L m = 40 cm
Given B o = 1 Tesla at
r
? ? ?
L core = 40 ? 0.2 = 39.8cm
Let a = uniform cross ? sectional area
We know that
Re tan
MMF NI
flux
Total luc ce S
? = = =
S T = S airgap + S core
(1)A
1
air core
o o r
core
air
or
LL
A
L
SL
A
? ? ?
??
=+
??
=+
??
??

Case 1: when ? r ? ?, B = 1T
MMF = NI 1 = B 1 A
1
core
air
ro
L
L
A ??
??
+ ? ?
??
??

NI 1 = 1 (a) [l air]
1
air
oo
L
A ??
?=
1
(1)
air
o
l
NI
?
=
Case 2:
? r = 1000
MMF = Same



28

NI 1 = B 2 A
1
core
air
ro
L
L
A ??
??
+
??
??

Put NI 1 from (1)
2
2
1
1000
39.8
0.2 0.2
1000
air core
air
o o
LL
BL
B
??
??
=+
??
??
??
=+
??
??

B L = 0.834 Tesla

53. Fault current for SLG fault
l
1 2 0
3
3
FG
V
I
X X X Xn
=
+ + +

Fault current for 3? fault
3
1
1 2 0 1
1 0 2
3
3
2
3
2(0.25) 0.05 0.15
3
f
n
V
I
X
VV
X X X X X
XX
Xn
X
?
=
=
+ + +
? ? ?
=
??
=

X n = 0.1 Pu
X n (in?) =
2
30
0.1
50
?
bass
Z MVA
Zpu
KVL
?
??
=
??
??

X n (in?) = 1.8?

54.



29


( ) 100A
N a b c
N
rms
I I I I
I
= + +
=


55.
( )
1
3
3
.
1
??
+
??
??
=
??
+
??
??
s
T
Ds
s
T


T (s) =
13
1
TS
TS
+
+

Frequency at which ?T (jw) is maximum (i)
1
m
W
T ?
=

1
()
1
TS
T S is
TS
? +
=
+
The general phase lead compensator
? ? = 3
2
11
3
3
m
w
T
T
==
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