9. 1. A solid copper sphere, 10 cm in diameter is deprived of
1020 electrons by a charging scheme. The charge on the sphere is
(A) 160.2 C (B) -160.2 C (C) 16.02 C (D) -16.02 C 2. A lightning
bolt carrying 15,000 A lasts for 100 ms. If the lightning strikes
an airplane flying at 2 km, the charge deposited on the plane is
(A) 13.33 mC (B) 75 C (C) 1500 mC (D) 1.5 C 3. If 120 C of charge
passes through an electric conductor in 60 sec, the current in the
conductor is (A) 0.5 A (B) 2 A (C) 3.33 mA (D) 0.3 mA 4. The energy
required to move 120 coulomb through 3 V is (A) 25 mJ (B) 360 J (C)
40 J (D) 2.78 mJ 5. i = ? (A) 1 A (B) 2 A (C) 3 A (D) 4 A 6. In the
circuit of fig P1.1.6 a charge of 600 C is delivered to the 100 V
source in a 1 minute. The value of v1 must be (A) 240 V (B) 120 V
(C) 60 V (D) 30 V 7. In the circuit of the fig P1.1.7, the value of
the voltage source E is (A) -16 V (b) 4 V (C) -6 V (D) 16 V CHAPTER
1.1 BASIC CONCEPTS Page 3 100 V 60 V v1 20 W Fig. P.1.1.6 1 A 2 A 5
A 4 A3 A i Fig. P.1.1.5 2 V 1 V 5 V 10 V 0 V 4 V E + + + + + Fig.
P.1.1.7 GATE EC BY RK Kanodia www.gatehelp.com
10. 8. Consider the circuit graph shown in fig. P1.1.8. Each
branch of circuit graph represent a circuit element. The value of
voltage v1 is (A) -30 V (B) 25 V (C) -20 V (D) 15 V 9. For the
circuit shown in fig P.1.1.9 the value of voltage vo is (A) 10 V
(B) 15 V (C) 20 V (D) None of the above 10. R1 = ? (A) 25 W (B) 50
W (C) 100 W (D) 2000 W 11. Twelve 6 W resistor are used as edge to
form a cube. The resistance between two diagonally opposite corner
of the cube is (A) 5 6 W (B) 6 5 W (C) 5 W (D) 6 W 12. v1 = ? (A)
-11 V (B) 5 V (C) 8 V (D) 18 V 13. The voltage vo in fig. P1.1.11
is always equal to (A) 1 V (B) 5 V (C) 9 V (D) None of the above
14. Req = ? (A) 11.86 W (B) 10 W (C) 25 W (D) 11.18 W 15. vs = ?
(A) 320 V (B) 280 V (C) 240 V (D) 200 V Page 4 UNIT 1 Networks 5 W
15 V 1 Avo + Fig. P.1.1.9 + 105 V 15 V + 10 V + + 35 V + 55 V + 65
V v1+ + 30 V + 30 V 100 V Fig. P.1.1.8 1 kW 2 kW 3 kW 4 kW 8 V 5 V
6 V 7 V + v1 Fig. P1.1.12 + vo 5 V 1 A 4 W Fig. P1.1.11 5 W 10 W 10
W up to Req 10 W 10 W 10 W 10 W Fig. P1.1.14 180 W 40 W 20 V + 60 W
90 W 180 W vs Fig. P.1.1.15 100 V R1 + 20 V + 70 V R2 60 W Fig.
P.1.1.10 GATE EC BY RK Kanodia www.gatehelp.com
11. 24. Let i t te t ( ) = - 3 100 A and v t t e t ( ) . ( . )=
- - 0 6 0 01 100 V for the network of fig. P.1.1.24. The power
being absorbed by the network element at t = 5 ms is (A) 18.4 mW
(B) 9.2 mW (C) 16.6 mW (D) 8.3 mW 25. In the circuit of fig.
P.1.1.25 bulb A uses 36 W when lit, bulb B uses 24 W when lit, and
bulb C uses 14.4 W when lit. The additional A bulbs in parallel to
this circuit, that would be required to blow the fuse is (A) 4 (B)
5 (C) 6 (D) 7 26. In the circuit of fig. P.1.1.26, the power
absorbed by the load RL is (A) 2 W (B) 4 W (C) 6 W (D) 8 W 27. vo =
? (A) 6 V (B) -6 V (C) -12 V (D) 12 V 28. vab = ? (A) 15.4 V (B)
2.6 V (C) -2.6 V (D) 15.4 V 29. In the circuit of fig. P.1.1.29
power is delivered by (A) dependent source of 192 W (B) dependent
source of 368 W (C) independent source of 16 W (D) independent
source of 40 W 30. The dependent source in fig. P.1.1.30 (A)
delivers 80 W (B) delivers 40 W (C) absorbs 40 W (D) absorbs 80 W
31. In the circuit of fig. P.1.1.31 dependent source (A) supplies
16 W (B) absorbs 16 W (C) supplies 32 W (D) absorbs 32 W Page 6
UNIT 1 Networks 20 A A B C 12 V Fig. P.1.1.25 i v + N Fig. P.1.1.24
i1 RL = 2 W2i11 V 1 W Fig. P.1.1.26 0.2 A 5 W v1 + v2 + 8 W0.3v1 18
W vo + 5v2 Fig. P.1.1.27 a b 2 A 2 W 6 W 2 W 0.2i1i1 0.3i1 R 8 A
Fig. P.1.1.28 500 W ix 400 W 200 W40 V 2ix Fig. P.1.1.29 5 W 20 V 5
W v1 v1 5 Fig. P.1.1.30 ix 8 V+ 2ix4 A Fig. P.1.1.31 GATE EC BY RK
Kanodia www.gatehelp.com
12. 32. A capacitor is charged by a constant current of 2 mA
and results in a voltage increase of 12 V in a 10 sec interval. The
value of capacitance is (A) 0.75 mF (B) 1.33 mF (C) 0.6 mF (D) 1.67
mF 33. The energy required to charge a 10 mF capacitor to 100 V is
(A) 0.10 J (B) 0.05 J (C) 5 10 9 - J (D) 10 10 9 - J 34. The
current in a 100 mF capacitor is shown in fig. P.1.1.34. If
capacitor is initially uncharged, then the waveform for the voltage
across it is 35. The voltage across a 100 mF capacitor is shown in
fig. P.1.1.35. The waveform for the current in the capacitor is 36.
The waveform for the current in a 200 mF capacitor is shown in fig.
P.1.1.36 The waveform for the capacitor voltage is 37. Ceq = ? (A)
3.5 mF (B) 1.2 mF (C) 2.4 mF (D) 2.6 mF 38. In the circuit shown in
fig. P.1.1.38 i t tin ( ) sin= 300 20 mA, for t 0. Let C1 40= mF
and C2 30= mF. All capacitors are initially uncharged. The v tin (
) would be (A) -0.25cos 20t V (B) 0.25cos 20t V (C) -36cos 20t mV
(D) 36cos 20t mV Chap 1.1Basic Concepts Page 7 250m 50m v t(ms) 5 4
v t(ms) 4 v t(ms) 4 v t(ms) 4 50m (C) (D) i(mA) t(ms) 5 4 Fig. P.
1.1.36 (A) (B) v 2 t(ms) 10 4 4 v 2 t(ms) 10 v 2 t(ms) 0.2 4 4 v 2
t(ms) 0.2 (C) (D) (A) (B) i(mA) 2 t(ms) Fig. P. 1.1.34 2.5 Fm 2
Fm1.5 Fm 1 FmCeq Fig. P.1.1.37 6 21 3 v t(ms) Fig. P.1.1.35 6 21 3
t(ms) i(mA) 6 2 1 3 t(ms) i(mA) 600 600 21 3 t(ms) i(mA) 2 1 3
t(ms) i(mA) (C) (D) (A) (B) iin vin + C2 60 mFC1 C2 C2 C2 C1 C1 C1
Fig. P. 1.1.38 GATE EC BY RK Kanodia www.gatehelp.com
13. SOLUTIONS 1. (C) n = 1020 , Q ne e= = =10 16 0220 . C
Charge on sphere will be positive. 2. (D) D DQ i t= = 15000 100m =
15. C 3. (B) i dQ dt = = = 120 60 2 A 4. (B) W Qv= = 360 J 6. (A)
6. (A) In order for 600 C charge to be delivered to the 100 V
source, the current must be anticlockwise. i dQ dt = = = 600 60 10
A Applying KVL we get v1 60 100+ - = 10 20 or v1 = 240 V 7. (A)
Going from 10 V to 0 V 10 5 1+ + +E = 0 or E = -16 V 8. (D) 100 65
352 2= + =v v V v v v3 2 330 65- = = V 105 65 04 3+ - - =v v v4 25=
V v v4 115 55 0+ - + = v1 15= V 9. (B) Voltage is constant because
of 15 V source. 10. (C) Voltage across 60 W resistor = 30 V Current
= = 30 60 0 5. A Voltage across R1 is = - =70 20 50 V R1 50 0 5
100= = . W 11. (C) The current i will be distributed in the cube
branches symmetrically v i i i iab = + + = 6 3 6 6 6 3 5 , R v i eq
ab = = 5 W 12. (C) If we go from +side of 1 kW through 7 V, 6 V and
5 V, we get v1 7 6 5 8= + - = V 13. (D) It is not possible to
determine the voltage across 1 A source. 14. (D) R R R eq eq eq = +
+ + + 5 10 5 10 5 ( ) + = + + +R R R Req eq eq eq 2 15 5 75 10 50 =
=Req 125 1118. W Chap 1.1Basic Concepts Page 9 i a b i 3 i 3 i 6 i
6 i 3 i 3 i Fig. S. 1.1.11 1 A 2 A 5 A 4 A3 A i = 1 A 2 A 1 A 6 A
Fig. S 1.1.5 + v2 + v4 + v3 + 105 V 15 V + 10 V + + 55 V + 65 V v1+
+ 30 V + 30 V 100 V Fig. S 1.1.8 2 V 1 V 5 V 10 V 0 V 4 V E + + + +
+ Fig. S 1.1.7 5 W 10 W 5 W Req Req Fig. S 1.1.14 GATE EC BY RK
Kanodia www.gatehelp.com
14. v v vo o o - + = = 20 5 5 20 5 20 V Power is P v v o= = =1
5 20 20 5 80 W 31. (D) Power P vi i i ix x x= = =2 2 2 ix = 4 A, P
= 32 W (absorb) 32. (D) v v C idtt t t t 2 1 1 2 1 - = = -12 1 2 2
1 C t tm ( ) = 12 2 10C m =C 1 67. mF 33. (B) E Cv= = =-1 2 5 10
100 0 052 6 2 . J 34. (D) v c idtc = = = - - -1 10 10 100 10 2 10 0
2 0 2 3 6 3 m ( ) . V This 0.2 V increases linearly from 0 to 0.2
V. Then current is zero. So capacitor hold this voltage. 35. (D) i
C dv dt = For 0 1<
15. 1. Consider the following circuits : The planner circuits
are (A) 1 and 2 (B) 2 and 3 (C) 3 and 4 (D) 4 and 1 2. Consider the
following graphs Non-planner graphs are (A) 1 and 3 (B) 4 only (C)
3 only (D) 3 and 4 3. A graph of an electrical network has 4 nodes
and 7 branches. The number of links l, with respect to the chosen
tree, would be (A) 2 (B) 3 (C) 4 (D) 5 4. For the graph shown in
fig. P.1.1.4 correct set is Node Branch Twigs Link (A) 4 6 4 2 (B)
4 6 3 3 (C) 5 6 4 2 (D) 5 5 4 1 5. A tree of the graph shown in
fig. P.1.2.5 is (A) a d e h (B) a c f h (C) a f h g (D) a e f g
CHAPTER 1.2 GRAPH THEORY Page 12 Fig. P.1.1.4 (1) (2) (3) (4) (1)
(2) (3) (4) cc 1 2 3 g 5 ed h4 b a f Fig. P.1.2.5 GATE EC BY RK
Kanodia www.gatehelp.com
16. (A) 1 1 0 1 0 1 0 1 1 - - - (B) 1 0 1 1 1 0 0 1 1 - - - (C)
- - - 1 1 0 0 1 1 1 0 1 (D) - - - 1 0 1 0 1 1 1 1 0 13. The
incidence matrix of a graph is as given below A = - - - - - - - 1 1
1 0 0 0 0 0 1 1 1 0 0 1 0 1 0 1 1 0 0 0 1 1 The graph is 14. The
incidence matrix of a graph is as given below A = - - - - - - - 1 0
0 1 1 0 0 0 0 0 1 0 1 1 0 0 1 0 0 0 1 0 1 0 0 1 1 0 1 1 1 0 0 0 0
The graph is 15. The incidence matrix of a graph is as given below
A = - - - - - - 1 1 1 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 0 0 0 1 1 The
graph is 16. The graph of a network is shown in fig. P.1.1.16. The
number of possible tree are (A) 8 (B) 12 (C) 16 (D) 20 Page 14
NetworksUNIT 1 1 2 3 4 4 4 4 1 2 3 1 2 3 1 2 3 (C) (D) (A) (B) 1 2
4 3 1 2 4 3 1 2 4 3 1 2 4 3 (C) (D) (A) (B) 1 2 4 3 5 1 2 4 3 5 (A)
(B) 1 2 4 3 5 1 2 4 3 5 (C) (D) Fig. P.1.1.16 GATE EC BY RK Kanodia
www.gatehelp.com
17. 22. The fundamental cut-set matrix of a graph is QF = - - -
- 1 1 0 0 0 0 0 1 1 0 1 0 0 0 0 0 1 1 0 0 0 1 1 0 The oriented
graph of the network is 23. A graph is shown in fig. P.1.2.23 in
which twigs are solid line and links are dotted line. For this
chosen tree fundamental set matrix is given below. BF = - 1 1 0 0 1
0 0 1 1 1 0 0 0 0 0 1 1 1 The oriented graph will be 24. A graph is
shown in fig. P.1.2.24 in which twigs are solid line and links are
dotted line. For this tree fundamental loop matrix is given as
below BF = 1 1 1 0 1 0 1 1 The oriented graph will be 25. Consider
the graph shown in fig. P.1.2.25 in which twigs are solid line and
links are dotted line. Page 16 NetworksUNIT 1 2 3 1 5 6 4 2 3 1 5 6
4 2 3 1 5 6 4 2 3 1 5 6 4 (C) (D) (A) (B) (A) (B) 1 2 4 5 6 3 Fig.
P. 1.2.23 (C) (D) 4 1 23 Fig. P.1.2.24 41 2 56 3 Fig. P. 1.2.25 (C)
(D) (A) (B) GATE EC BY RK Kanodia www.gatehelp.com
18. A fundamental loop matrix for this tree is given as below
BF = - - - 1 0 0 1 0 1 0 1 0 1 1 0 0 0 1 0 1 1 The oriented graph
will be 26. In the graph shown in fig. P.1.2.26 solid lines are
twigs and dotted line are link. The fundamental loop matrix is (A)
1 1 0 0 0 0 0 1 0 0 1 1 1 0 0 1 0 0 0 0 0 1 1 1 0 0 0 1 0 1 0 1 0 0
0 1 - - - - - - (B) - - - - - - 1 1 0 0 0 0 0 1 0 0 1 1 1 0 0 1 0 0
0 0 0 1 1 1 0 0 0 1 0 1 0 1 0 0 0 1 (C) 1 1 0 0 0 0 0 1 0 0 1 1 1 0
0 1 0 0 0 0 0 1 1 1 0 0 0 1 0 1 0 1 0 0 0 1 - - - (D) 1 1 0 0 0 0 0
1 0 0 1 1 1 0 0 1 0 0 0 0 0 1 1 1 0 0 0 1 0 1 0 1 0 0 0 1 - - 27.
Branch current and loop current relation are expressed in matrix
form as i i i i i i i i 1 2 3 4 5 6 7 8 0 1 1 0 0 0 1 = - - 1 1 0 0
1 1 1 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 - - I I I I 1 2 3 4 where
ij represent branch current and Ik loop current. The number of
independent node equation are (A) 4 (B) 5 (C) 6 (D) 7 28. If the
number of branch in a network is b, the number of nodes is n and
the number of dependent loop is l, then the number of independent
node equations will be (A) n l+ - 1 (B) b - 1 (C) b n- + 1 (D) n -
1 Statement for Q.2930: Branch current and loop current relation
are expressed in matrix form as i i i i i i i i 1 2 3 4 5 6 7 8 0 0
1 0 1 1 = - - -1 0 0 1 0 0 1 0 0 0 0 0 1 1 1 1 0 1 1 0 0 0 0 0 0 1
- - - I I I I 1 2 3 4 where ij represent branch current and Ik loop
current. 29. The rank of incidence matrix is (A) 4 (B) 5 (C) 6 (D)
8 Chap 1.2 Page 17 Graph Theory g bh a c i e d f Fig. P.1.2.26 (C)
(D) (A) (B) GATE EC BY RK Kanodia www.gatehelp.com
19. 30. The directed graph will be 31. A network has 8 nodes
and 5 independent loops. The number of branches in the network is
(A) 11 (B) 12 (C) 8 (D) 6 32. A branch has 6 node and 9 branch. The
independent loops are (A) 3 (B) 4 (C) 5 (D) 6 Statement for Q.3334:
For a network branch voltage and node voltage relation are
expressed in matrix form as follows: v v v v v v v v 1 2 3 4 5 6 7
8 1 0 0 1 0 1 0 0 0 = 0 1 0 0 0 0 1 1 1 0 0 0 1 1 0 0 0 1 1 1 0 1 0
- - - - V V V V 1 2 3 4 where vi is the branch voltage and Vk is
the node voltage with respect to datum node. 33. The independent
mesh equation for this network are (A) 4 (B) 5 (C) 6 (D 7 33. The
oriented graph for this network is ************ Page 18
NetworksUNIT 1 5 3 421 5 3 421 5 3 421 5 3 421 (C) (D) 5 3 4 2 1 7
6 8 5 3 4 2 1 7 6 8 5 3 4 2 1 7 6 8 5 3 4 2 1 7 6 8 (C) (D) GATE EC
BY RK Kanodia www.gatehelp.com
20. SOLUTIONS 1. (A) The circuit 1 and 2 are redrawn as below.
3 and 4 can not be redrawn on a plane without crossing other
branch. Fig. S1.2.1 2. (B) Other three circuits can be drawn on
plane without crossing Fig. S1.2.1 3. (C) l b n= - - =( )1 4. 4.
(B) There are 4 node and 6 branches. t n= - =1 3, l b n= - + =1 3
5. (C) From fig. it can be seen that a f h g is a tree of given
graph 6. (B) From fig. it can be seen that a d f is a tree. 7. (D)
D is not a tree Fig. S .1.2.7 8. (D) it is obvious from the
following figure that 1, 3, and 4 are tree Fig. S. 1.2.8 Chap 1.2
Page 19 Graph Theory (1) (2) (3) cc ged h b a f Fig. S 1.2.5 (1)
(2) c e d fa b Fig. S. 1.2.6 2 2 a a 1 1 3 3 c c d d e e b b 4 4 f
f 2 a 1 3 c d e b 4 f 2 a 1 3 c d e b 4 f 2 a 1 3 c d e b 4 f (5)
(1) (2) (3) (4) (C) (D) (A) (B) GATE EC BY RK Kanodia
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21. This in similar to matrix in (A). Only place of rows has
been changed. 27. (A) Number of branch =8 Number of link =4 Number
of twigs =8 4 4- = Number of twigs =number of independent node
equation. 28. (D) The number of independent node equation are n -
1. 29. (A) Number of branch b = 8 Number of link l = 4 Number of
twigs t b l= - = 4 rank of matrix = - = =n t1 4 30. (B) We know the
branch current and loop current are related as [ ] [ ][ ]i B Ib T
L= So fundamental loop matrix is Bf = - - - - - - 0 1 0 1 0 1 1 0 0
1 1 0 0 1 0 0 1 1 0 0 1 0 0 0 0 0 0 0 1 1 0 1 f-loop 1 include
branch (2, 4, 6, 7) and direction of branch2 is opposite to other
(B only). 31. (B) Independent loops =link l b n= - -( )1 = -5 7b ,
b = 12 32. (B) Independent loop =link l b n= - - =( )1 4 33. (A)
There are 8 branches and 4 1 5+ = node Number of link = - + =8 5 1
4 So independent mesh equation =Number of link. 34. (D) We know
that [ ] [ ]v A Vb r T n= So reduced incidence matrix is Ar = - - -
- 1 0 0 0 1 0 0 1 0 1 0 0 1 1 0 0 0 0 1 0 0 1 1 1 0 0 0 1 0 0 1 0
At node-1, three branch leaves so the only option is (D).
*********** Page 22 NetworksUNIT 1 g bh a c i e d fl3 l4 l1 l2 Fig.
S 1.2.26 GATE EC BY RK Kanodia www.gatehelp.com
22. 1. v1 = ? (A) 0.4vs (B) 1.5vs (C) 0.67vs (D) 2.5vs 2. va =
? (A) -11 V (B) 11 V (C) 3 V (D) -3 V 3. v1 = ? (A) 120 V (B) -120
V (C) 90 V (D) -90 V 4. va = ? (A) 4.33 V (B) 4.09 V (C) 8.67 V (D)
8.18 V 5. v2 = ? (A) 0.5 V (B) 1.0 V (C) 1.5 V (D) 2.0 V 6. ib = ?
(A) 0.6 A (B) 0.5 A (C) 0.4 A (D) 0.3 A CHAPTER 1.3 Page 23 METHODS
OF ANALYSIS 6R 3R 4vs vsv16R + Fig. P1.3.1 3 A 2 W va 3 W 1 A Fig.
P1.3.2 12 V 10 W 4 W 4 A1 W 2 W 10 V va Fig. P1.3.4 v2 60 W 30 W
0.5 A 10 V 30 W 20 W + Fig. P1.3.5 10 W 30 V 3 A 20 W 30 W v1 + 6 A
60 W9 A60 W Fig. P1.3.3 10 V 36 W 69 W 0.5 A 37 W64 W ib Fig.
P1.3.6 GATE EC BY RK Kanodia www.gatehelp.com
23. 7. i1 = ? (A) 3.3 A (B) 2.1 A (C) 1.7 A (D) 1.1 A 8. i1 = ?
(A) 1 mA (B) 1.5 mA (C) 2 mA (D) 2.5 mA 9. i1 = ? (A) 4 A (B) 3 A
(C) 6 A (D) 5 A 10. i1 = ? (A) 20 mA (B) 15 mA (C) 10 mA (D) 5 mA
11. i1 = ? (A) 0.01 A (B) -0.01 A (C) 0.03 A (D) 0.02 A 12. The
value of the current measured by the ammeter in Fig. P1.3.12 is (A)
2 3 A (B) 5 3 A (C) - 5 6 A (D) 2 9 A 13. i1 = ? (A) 10 mA (B) -10
mA (C) 0.4 mA (D) -0.4 mA Page 24 UNIT 1 Networks 2 A 4 A 3 W 2 W
i1 4 W 3 V Fig. P1.3.9 i145 V 500 W 2 kW 15 mA Fig. P1.3.10 7.5mA
90 kW 10 kW 90 kW 10 kW i2 i1 75 V Fig. P1.3.8 6.6 V i1 100 W 40 W
0.1 A 0.06 A 50 W 60W 0.1A Fig. P1.3.11 75 V 4 W 3 W 5 W 10 W 2 W8
W 6 A i1 Fig. P1.3.7 2 A 3 A 6 W 4 W7 W 5 W 2 W Ammeter Fig.
P1.3.12 40 mA 100 W i1 200 W 50 W 10 mA Fig. 1.3.13 GATE EC BY RK
Kanodia www.gatehelp.com
24. 14. The values of node voltage are va = 12 V, vb = 9.88 V
and vc = 5.29 V. The power supplied by the voltage source is (A)
19.8 W (B) 27.3 W (C) 46.9 W (D) 54.6 W 15. i i i1 2 3, , = ? (A) 3
A, 2 A, and 4 A (B) 3 A, 3 A, and 8 A (C) 1 A, 3 A, and 4 A (D) 1
A, 2 A, and 8 A 16. vo = ? (A) 6 5 V (B) 8 5 V (C) 6 7 V (D) 5 7 V
17. The mesh current equation for the circuit in Fig. P1.3.17 are
(A) 4 2 0 2 8 2 0 2 5 12 8 20 1 2 3 - - - - = - i i i (B) 6 2 0 2
12 2 0 2 7 12 8 20 1 2 3 - - - = i i i (C) 6 2 0 2 12 2 0 2 7 12 8
20 1 2 3 - - - - = i i i (D) 4 2 0 2 8 2 0 2 5 12 8 20 1 2 3 - - -
= i i i 18. For the circuit shown in Fig. P1.3.18 the mesh equation
are (A) 6 12 12 6 6 18 1 1 0 1 2 3 k k k k k k k k k - - - - - - i
i i = - 6 0 5 (B) 6 12 12 6 6 18 1 1 0 1 2 3 k k k k k k k k k - -
- - i i i = - 6 0 5 (C) - - - 6 12 12 6 6 18 1 1 0 1 2 3 k k k k k
k k k k i i i = - 6 0 5 (D) - - - - - 6 12 12 6 6 18 1 1 0 1 2 3 k
k k k k k k k k i i i = - 6 0 5 Chap 1.3Methods of Analysis Page 25
15 V 3 W 9 W 2 W i2 i3 21 V6 W i1 Fig. P1.3.15 i2 i3i1 4 W 8 W 5 W
12 V 2 W 2 W 8 V 20 V Fig. 1.3.17 4 mA vo2 kW1 mA 2 kW 1 kW 2 mA 1
kW + 1 kW Fig. P1.3.16 va 6 W 4 W 3 W 2 W1 A12 V vb vc Fig. 1.3.14
6 kW 5 mA6 V 6 kW 6 kW 6 kW i1 i2 i3 Fig. 1.3.18 GATE EC BY RK
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25. The value of R4 is (A) 40 (B) 15 (C) 5 (D) 20 26. va = ?
(A) 26 V (B) 19 V (C) 13 V (D) 18 V 27. v = ? (A) 60 V (B) -60 V
(C) 30 V (D) -30 V 28. i1 = ? (A) 66.67 mA (B) 46.24 mA (C) 23.12
mA (D) 33.33 mA 29. va = ? (A) 342 V (B) 171 V (C) 198 V (D) 396 V
30. ia = ? (A) 14 mA (B) -6 5. mA (C) 7 mA (D) -21 mA 31. v2 = ?
(A) 5 V (B) 75 V (C) 3 V (D) 10 V 32. i1 = ? Chap 1.3Methods of
Analysis Page 27 2.5 Wk 10 kW 10 Wk va 10 Wk 5 Wk 4 mA20 V Fig.
P1.3.26 2 A 10 W 20 Wv 15 W 5 W 4 A Fig. P.3.1.27 40 V 300 W 0.4i1
i1 500 W Fig. P1.3.28 10 A 10 W 4 A 200 W 5 A 100 W 20 W 20 A 50 W
40 Wva Fig. P1.3.29 50 W 150 W 4 V 8 V2 V 100 W 200 W225 W 75 W 50
W ia Fig. P1.3.30 v2 + 50 W 100 W10 V 0.04v2 Fig. P1.3.31 2 W 4 A8
V i1 0.5i1 4 W 6 V Fig. P1.3.32 v1 R4 v2 25i2 R2 R3 R1 i1 i2 i3
Fig. P1.3.25 GATE EC BY RK Kanodia www.gatehelp.com
26. (A) -1.636 A (B) -3 273. A (C) -2.314 A (D) -4 628. A 33.
vx = ? (A) 32 V (B) -32 V (C) 12 V (D) -12 V 34. ib = ? (A) 4 mA
(B) -4 mA (C) 12 mA (D) -12 mA 35. vb = ? (A) 1 V (B) 1.5 V (C) 4 V
(D) 6 V 36. vx = ? (A) -3 V (B) 3 V (C) 10 V (D) -10 V 37. va = ?
(A) 25.91 V (B) -25.91 V (C) 51.82 V (D) -51.82 V 38. For the
circuit of Fig. P1.3.38 the value of vs , that will result in v1 =
0, is (A) 28 V (B) -28 V (C) 14 V (D) -14 V 39. i i1 2, = ? (A) 2.6
A, 1.4 A (B) 2.6 A, -1.4 A (C) 1.6 A, 1.35 A (D) 1.2 A, -1.35 A 40.
v1 = ? Page 28 UNIT 1 Networks 1 kW 4va6 V 3 kW ib 2 kW va Fig.
P1.3.34 ia 4 kW 2 V 2 kW 5ia vb Fig. P1.3.35 iy 2 A 100 W vx50 W +
25iy 0.2vx 50 W Fig. P1.3.36 1.6 A 100 W 0.02vx 50 W vx + Fig.
P1.3.33 0.8va 10 A5 W2.5 W va 16 A 2 W Fig. P1.3.37 3 A 2 A 0.1v1
10 W 20 W 40 Wvs 48 V + v1 Fig. P1.3.38 4 W 2 W 18 Vi115 V i2 2ix 6
W ix Fig. P1.3.39 3 W 14 V 2vy + v1 7 A 6 W 2 A + vy 3 W 2 W Fig.
P1.3.40 GATE EC BY RK Kanodia www.gatehelp.com
27. (A) 10 V (B) -10 V (C) 7 V (D) -7 V 41. vx = ? (A) 9 V (B)
-9 V (C) 10 V (D) -10 V 42. The power being dissipated in the 2 W
resistor in the circuit of Fig. P1.3.42 is (A) 76.4 W (B) 305.6 W
(C) 52.5 W (D) 210.0 W 43. i1 = ? (A) 0.12 A (B) 0.24 A (C) 0.36 A
(D) 0.48 A ***************** SOLUTIONS 1. (B) Applying the nodal
analysis v v R v R R R R v s s s1 4 6 3 1 6 1 3 1 6 15= + + + = .
2. (C) va = + + =2 3 1 3 1 11( ) ( ) V 3. (D) - + - + = v v1 1 60
60 6 9 = -v1 90 V 4. (C) v va a- + = 10 4 2 4 =va 8 67. V 5. (D) v
v2 2 20 10 30 0 5+ + = . =v2 2 V 6. (B) Using Thevenin equivalent
and source transform va = + + + + = 25 2 60 15 3 14 1 3 1 15 15 23
8 3 . V i1 25 15 23 14 3 2 09= - = . . A 7. (A) ib = + + = 10 64 36
0 5 0 6. . A 8. (B) 75 90 10 7 51 1= + -k k mi i( . ) 150 100 151
1= =ki i . mA 9. (B) 3 2 3 4 31 1 1= + - =i i i( ) A 10. (B) 45 2
500 151 1= + +k mi i( ) =i1 15 mA 11. (D) 6 6 50 100 0 1 40 0 06 60
0 11 1 1 1. ( . ) ( . ) ( . )= + + + - + -i i i i i1 0 02= . A Chap
1.3Methods of Analysis Page 29 +vx 0.6 A 600 W 900 W 0.3 A 500 W
0.5vx 500 W Fig. P1.3.41 2.5 A 2 W3 W 30 V 5 W 6ia4 W 2 A ia Fig.
P1.3.42 500 W 100 W 100 W 400 W + + vy vx 0.005vy 0.6 A 0.001vy 180
V Fig. P1.3.43 25 V W 2 W 10 W 60 V 5 W3 W i1 va 8 3 Fig. S.1.3.6
GATE EC BY RK Kanodia www.gatehelp.com
28. 38. (D) If v1 0= , the dependent source is a short circuit
v v v vs1 1 1 40 10 48 20 2 3+ - + - = - v1 0= - - = - vs 10 48 20
1 vs = - 14 V 39. (D) i i ix = -1 2 15 4 2 61 1 2 1 2= - - + -i i i
i i( ) ( ) 8 4 151 2i i- = K(i) - = + -18 2 62 2 1i i i( ) 3 4 91
2i i- = K(ii) i1 12= . A, i2 1 35= - . A 40. (B) 14 3 6 2 7 2 2 71
1 1= + + - - + + -i v i v iy y( ) ( ) v iy = -3 21( ) 14 3 9 2 6 9
2 71 1 1 1= + - + - + -i i i i( ) ( ) ( ) 14 20 18 54 141= - - -i
i1 5= A v1 6 5 2 7 2 3 5 2 2 5 7 10= - - + - + - = -( ) ( ) ( ) V
41. (D) Let i1 and i2 be two loop current 0 5 500 5001 1 2. ( )v i
i ix = + - , v ix = -500 1 5 2 01 2i i- = K(i) 500 900 0 3 600 0 6
02 1 2 2( ) ( . ) ( . )i i i i- + + + - = - + =5 20 0 91 2i i .
K(ii) i1 20= mA, vx = - = -500 20 10m V 42. (C) 30 5 3 2 5 4 2 5 2=
+ - + - +i i ia a a( . ) ( . ) ia = + + = 30 7 5 2 12 329 . . A 6
19 75ia = . V voltage across 2 W resistor 30 19 75 10 25- =. . V, P
= = ( . ) . 10 25 2 52 53 2 W 43. (A) v ix = 500 1 v i vy x= -400 0
0011( . ) = - =400 0 5 2001 1 1( . )i i i 180 500 100 0 6 200 100 0
0051 1 1 1= + - + + +i i i i vy( . ) ( . ) 180 900 60 100 0 005
2001 1= - + i i. i1 0 12= . A ************ Page 32 UNIT 1 Networks
3 A 2 A 10 W 20 W 40 Wvs 48 V + v1 Fig. S1.3.38 3 W 14 V 2vy + v1
i1 7 A 6 W 2 A + vy 3 W 2 W Fig. S1.3.40 GATE EC BY RK Kanodia
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29. 1. v RTH TH, = ? (A) 2 V, 4 W (B) 4 V, 4 W (C) 4 V, 5 W (D)
2 V, 5 W 2. i RN N, = ? (A) 3 A, 10 3 W (B) 10 A, 4 W (C) 1,5 A, 6
W (D) 1.5 A, 4 W 3. v RTH TH, = ? (A) -2 V, 6 5 W (B) 2 V, 5 6 W
(C) 1 V, 5 6 W (D) -1 V, 6 5 W 4. A simple equivalent circuit of
the 2 terminal network shown in fig. P1.4.4 is 5. i RN N, = ? (A) 4
A, 3 W (B) 2 A, 6 W (C) 2 A, 9 W (D) 4 A, 2 W CHAPTER 1.4 Page 33
NETWORKS THEOREM 3 W 2 W 6 V 6 W RTHvTH, Fig. P.1.4.1 2 W 2 W 4 W15
V RN iN, Fig. P.1.4.2 2 W 3 W 1 W2 A RTHvTH, Fig. P.1.4.3 R R R R i
i v (C) (D) (A) (B) vi R Fig. P.1.4.4 2 W 4 W 3 W6 A iN RN Fig.
P.1.4.5 GATE EC BY RK Kanodia www.gatehelp.com
30. 6. v RTH TH, = ? (A) -100 V, 75 W (B) 155 V, 55 W (C) 155
V, 37 W (D) 145 V, 75 W 7. RTH = ? (A) 3 W (B) 12 W (C) 6 W (D) 8.
The Thevenin impedance across the terminals ab of the network shown
in fig. P.1.4.8 is (A) 2 W (B) 6 W (C) 6 16. W (D) 4 3 W 9. For In
the the circuit shown in fig. P.1.4.9 a network and its Thevenin
and Norton equivalent are given The value of the parameter are vTH
RTH iN RN (A) 4 V 2 W 2 A 2 W (B) 4 V 2 W 2 A 3 W (C) 8 V 1.2 W 30
3 A 1.2 W (D) 8 V 5 W 8 5 A 5 W 10. v1 = ? (A) 6 V (B) 7 V (C) 8 V
(D) 10 V 11. i1 = ? (A) 3 A (B) 0.75 mA (C) 2 mA (D) 1.75 mA
Statement for Q.1213: A circuit is given in fig. P.1.4.1213. Find
the Thevenin equivalent as given in question.. 12. As viewed from
terminal x and x is (A) 8 V, 6 W (B) 5 V, 6 W (C) 5 V, 32 W (D) 8
V, 32 W Page 34 UNIT 1 Networks 30 W 25 W 20 W 5 A 5 V RTHvTH, Fig.
P.1.4.6 2 A 5 V 6 W 6 W RTH Fig. P.1.4.7 3 W 6 W2 A 8 W 8 W 2 V b a
Fig. P.1.4.8 2 W 3 W 2 A4 V RTH vTH iN RN Fig. P.1.4.9 1 W 1 W 3 W
2 W 6 W8 V v1 + 2 W 6 W 18 V Fig. P.1.4.10 4 kW 4 kWi1 6 kW 20 V 24
V4 kW 3 kW12 V Fig. P.1.4.11 10 W x 16 Wx 5 V 1 A40 W 8 W y y Fig.
P.1.4.1213 GATE EC BY RK Kanodia www.gatehelp.com
31. 13. As viewed from terminal y and y is (A) 8 V, 32 W (B) 4
V, 32 W (C) 5 V, 6 W (D) 7 V, 6 W 14. A practical DC current source
provide 20 kW to a 50 W load and 20 kW to a 200 W load. The maximum
power, that can drawn from it, is (A) 22.5 kW (B) 45 kW (C) 30.3 kW
(D) 40 kW Statement for Q.1516: In the circuit of fig. P.1.4.1516
when R = 0 W , the current iR equals 10 A. 15. The value of R, for
which it absorbs maximum power, is (A) 4 W (B) 3 W (C) 2 W (D) None
of the above 16. The maximum power will be (A) 50 W (B) 100 W (C)
200 W (D) value of E is required 17. Consider a 24 V battery of
internal resistance r = 4 W connected to a variable resistance RL .
The rate of heat dissipated in the resistor is maximum when the
current drawn from the battery is i . The current drawn form the
battery will be i 2 when RL is equal to (A) 2 W (B) 4 W (C) 8 W (D)
12 W 18. i RN N, = ? (A) 2 A, 20 W (B) 2 A, -20 W (C) 0 A, 20 W (D)
0 A, -20 W 19. v RTH TH, = ? (A) 0 W (B) 1.2 W (C) 2.4 W (D) 3.6 W
20. v RTH TH, = ? (A) 8 V, 5 W (B) 8 V, 10 W (C) 4 V, 5 W (D) 4 V,
10 W 21. RTH = ? (A) 3 W (B) 1.2 W (C) 5 W (D) 10 W 22. In the
circuit shown in fig. P.1.4.22 the effective resistance faced by
the voltage source is (A) 4 W (B) 3 W (C) 2 W (D) 1 W Chap
1.4Network Theorems Page 35 4 W 2 W 2 W 4 W R 2 W 4 AE iR Fig.
P.1.4.1516. 10 W i1 5 W 30 W20i1 RNiN, Fig. P.1.4.18 6 W i1 4 W3i1
RNiN, Fig. P1.4.19 v1 + 5 W0.1v1 4 V vTH RTH Fig. P.1.4.20 vx + 2 W
3 W 4 V vx 4 RTH Fig. P.1.4.21 4 W i i vs 4 Fig. P.1.4.22 GATE EC
BY RK Kanodia www.gatehelp.com
32. 23. In the circuit of fig. P1.4.23 the value of RTH at
terminal ab is (A) -3 W (B) 9 8 W (C) - 8 3 W (D) None of the above
24. RTH = ? (A) (B) 0 (C) 3 125 W (D) 125 3 W 25. In the circuit of
fig. P.1.4.25, the RL will absorb maximum power if RL is equal to
(A) 400 3 W (B) 2 9 kW (C) 800 3 W (D) 4 9 kW Statement for Q.2627:
In the circuit shown in fig. P1.4.2627 the maximum power transfer
condition is met for the load RL . 26. The value of RL will be (A)
2 W (B) 3 W (C) 1 W (D) None of the above 27. The maximum power is
(A) 0.75 W (B) 1.5 W (C) 2.25 W (D) 1.125 W 28. RTH = ? (A) 100 W
(B) 136.4 W (C) 200 W (D) 272.8 W 29. Consider the circuits shown
in fig. P.1.4.29 Page 36 UNIT 1 Networks 0.75va va + 4 W a 8 W 9 V
b Fig. P.1.4.23 + va 200 W 50 W100 W va 100 RTH Fig. P.1.4.24 40 W
100 W 200 W 3i RL 6 V i Fig. P.1.4.25 ix 3 W RL0.9 A 2 W 16 V Fig.
P.1.4.2627 100 W -2ix 0.01vx ix 800 W 100 W 300 W + vx RTH Fig.
P.1.4.28 2 W 6 W 6 W 6 W 2 W 2 W 12 V18 V 3 A ib ia 2 W 6 W 6 W 6 W
2 W 2 W 8 V 12 V 12 V Fig. P.1.4.29a & b GATE EC BY RK Kanodia
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33. The relation between ia and ib is (A) i ib a= + 6 (B) i ib
a= + 2 (C) i ib a= 15. (D) i ib a= 30. Req = ? (A) 18 W (B) 72 13 W
(C) 36 13 W (D) 9 W 31. In the lattice network the value of RL for
the maximum power transfer to it is (A) 6.67 W (B) 9 W (C) 6.52 W
(D) 8 W Statement for Q.3233: A circuit is shown in fig.
P.1.4.3233. 32. If v vs s1 2 6= = V then the value of va is (A) 3 V
(B) 4 V (C) 6 V (D) 5 V 33. If vs1 6= V and vs2 6= - V then the
value of va is (A) 4 V (B) -4 V (C) 6 V (D) -6 V 34. A network N
feeds a resistance R as shown in fig. P1.4.34. Let the power
consumed by R be P. If an identical network is added as shown in
figure, the power consumed by R will be (A) equal to P (B) less
than P (C) between P and 4P (D) more than 4P 35. A certain network
consists of a large number of ideal linear resistors, one of which
is R and two constant ideal source. The power consumed by R is P1
when only the first source is active, and P2 when only the second
source is active. If both sources are active simultaneously, then
the power consumed by R is (A) P P1 2 (B) P P1 2 (C) ( )P P1 2 2
(D) ( )P P1 2 2 36. A battery has a short-circuit current of 30 A
and an open circuit voltage of 24 V. If the battery is connected to
an electric bulb of resistance 2 W, the power dissipated by the
bulb is (A) 80 W (B) 1800 W (C) 112.5 W (D) 228 W 37. The following
results were obtained from measurements taken between the two
terminal of a resistive network Terminal voltage 12 V 0 V Terminal
current 0 A 1.5 A The Thevenin resistance of the network is (A) 16
W (B) 8 W (C) 0 (D) Chap 1.4Network Theorems Page 37 12 W 4 W 6 W 2
W 6 W 9 W 18 W Req Fig. P.1.4.30 RL 7 W 6 W 5 W 9 W Fig. P.1.4.31
12 W 1 W 3 W 3 W 1W 1 W vs2vs1 + va Fig. P.1.4.3233 N R N R N Fig.
P.1.4.34 GATE EC BY RK Kanodia www.gatehelp.com
34. 38. A DC voltmeter with a sensitivity of 20 kW/V is used to
find the Thevenin equivalent of a linear network. Reading on two
scales are as follows (a) 0 10- V scale : 4 V (b) 0 -15 V scale : 5
V The Thevenin voltage and the Thevenin resistance of the network
is (A) 16 3 V, 200 3 kW (B) 32 3 V, 1 15 MW (C) 18 V, 2 15 MW (D)
36 V, 200 3 kW 39. Consider the network shown in fig. P.1.4.39. The
power absorbed by load resistance RL is shown in table : RL 10 kW
30 kW P 3.6 MW 4.8 MW The value of RL , that would absorb maximum
power, is (A) 60 kW (B) 100 W (C) 300 W (D) 30 kW 40. Measurement
made on terminal ab of a circuit of fig.P.1.4.40 yield the
current-voltage characteristics shown in fig. P.1.4.40. The
Thevenin resistance is (A) 300 W (B) -300 W (C) 100 W (D) -100 W
*********** SOLUTIONS 1. (B) vTH = + = ( )( )6 6 3 6 4 V, RTH = +
=( || )3 6 2 4 W 2. (A) RN = + =2 4 2 10 3 || W, v1 15 2 1 2 1 2 1
4 6= + + = V i i v sc N= = =1 2 3 A 3. (C) vTH = + = ( )( )( )2 3 1
3 3 1 V, RTH = =1 5 5 6 || W 4. (B) After killing all source
equivalent resistance is R Open circuit voltage = v1 5. (D) The
short circuit current across the terminal is i isc N= + = = 6 4 4 2
4 A , RN = =6 3 2|| W 6. (B) For the calculation of RTH if we kill
the sources then 20 W resistance is inactive because 5 A source
will be open circuit RTH = + =30 25 55 W, vTH = + =5 5 30 155 V
Page 38 UNIT 1 Networks RL Linear Network + vab Fig. P.1.4.39
21-1-2-3-4 0 i(mA) 30 20 10 v Resistive Network + vab a b Fig.
P.1.4.40 2 W 2 Wv1 4 W15 V isc Fig. S.1.4.2 2 W 3 W4 W6 A isc Fig.
S1.4.5 GATE EC BY RK Kanodia www.gatehelp.com
35. 7. (C) After killing the source, RTH = 6 W 8. (B) After
killing all source, RTH = + =3 6 8 8 6|| || W 9. (D) v voc TH= + =
=2 2 4 8 V R RTH N= + = =2 3 5 W , i v R N TH TH = = 8 5 A 10. (A)
If we solve this circuit direct, we have to deal with three
variable. But by simple manipulation variable can be reduced to
one. By changing the LHS and RHS in Thevenin equivalent v1 4 1 1 12
1 2 1 1 1 1 6 1 1 2 6= + + + + + + + = V 11. (B) If we solve this
circuit direct, we have to deal with three variable. But by simple
manipulation variable can be reduced to one. By changing the LHS
and RHS in Thevenin equivalent i1 20 6 8 2 4 2 0 75= - - + + = k k
k . mA 12. (B) We Thevenized the left side of xx and source
transformed right side of yy v vxx TH = = + + = 4 8 8 24 1 8 1 24 5
V, RTH = + =8 16 8 6||( ) W 13. (D) Thevenin equivalent seen from
terminal yy is v vyy TH = = + + = 4 24 8 8 1 24 1 8 7 V, RTH = + =(
)||8 16 8 6 W 14. (A) ir r + = 50 50 20 2 k, ir r + = 200 200 20 2
k ( ) ( )r r+ = +200 4 502 2 r = 100 W i = 30 A, Pmax = = ( ) . 30
100 4 22 5 2 kW 15. (C) Thevenized the circuit across R, RTH = 2 W
16. (A) isc = 10 A, RTH = 2 W, Pmax = = 10 2 2 50 2 W Chap
1.4Network Theorems Page 39 4 V 12 V 1 W 1 W v1 + 2 W 6 W 1 W Fig.
S1.4.10 2 kW 4 kWi1 2 kW 20 V 6 V 8 V Fig. S1.4.11 8 W x 16 Wx 4 V
8 V y y 8 W Fig. S1.4.12 i RLr Fig. S1.4.14 4 W 2 W 2 W 4 W 2 W
Fig. S1.4.15 6 W 6 W RTH Fig. S.1.4.7 GATE EC BY RK Kanodia
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36. Now in this circuit all straight-through connection have
been cut as shown in fig. S1.4.32b va = + + + = 6 2 3 2 3 1 5 ( ) V
33. (B) Since both source have opposite polarity, hence short
circuit the all straight-through connection as shown in fig.
S.1.4.33 va = - + = - 6 6 3 2 1 4 ( || ) V 34. (C) Let Thevenin
equivalent of both network P V R R RTH TH = + 2 P V R R R V R R RTH
TH TH TH = + = + 2 4 2 2 2 Thus P P P< < 4 35. (C) i P R 1 1
= and i P R 2 2 = using superposition i i i= + =1 2 P R P R 1 2 i R
P P2 1 2 2 = ( ) 36. (C) r v i oc sc = = 1 2. W P = + = 24 1 2 2 2
112 5 2 2 ( . ) . W 37. (B) R v i TH oc sc = = = 12 15 8 . W 38.
(A) Let 1 1 20 50 sensitivity = = k mA For 0 -10 V scale Rm = =10
20 200k kW For 0 -50 V scale Rm = =50 20 1k MW For 4 V reading i =
= 4 10 50 20 mA v R RTH TH TH= + = +20 20 200 4 20m m mk ...(i) For
5 V reading i = = 5 50 50 5m mA v R RTH TH TH= + = +5 5 1 5 5m m mM
...(ii) Solving (i) and (ii) vTH = 16 3 V, RTH = 200 3 kW 39. (D)
v10 10 3 6 6k k m= =. v30 30 4 8 12k k m= =. V 6 10 10 = + R v TH
TH 10 6 60v RTH TH= + 12 30 30 = + v R TH TH 5 2 60v RTH TH= + RTH
= 30 kW 40. (D) At v = 0 , isc = 30 mA At i = 0, voc = - 3 V R v i
TH oc sc = = - = - 3 30 100 m W ************ Page 42 UNIT 1
Networks vTHvTHvTH RTH R R RTH RTH Fig. S1.4.34 6 V + va 2 W 6 W 1
W 3 W Fig. S1.4.33 6 V + va 2 W 6 W 1 W 3 W Fig. S.1.4.32b GATE EC
BY RK Kanodia www.gatehelp.com
37. 1. The natural response of an RLC circuit is described by
the differential equation d v dt dv dt v 2 2 2 0+ + = , v( )0 10= ,
dv dt ( )0 0= . The v t( ) is (A) 10 1( )+ - t e t V (B) 10 1( )- -
t e t V (C) 10e t- V (D) 10te t- V 2. The differential equation for
the circuit shown in fig. P1.6.2. is (A) + + =v t v t v t v ts( ) (
) . ( ) ( )3000 102 10 108 8 (B) + + =v t v t v t v ts( ) ( ) . ( )
( )1000 102 10 108 8 (C) + + = v t v t v t v ts ( ) ( ) . ( ) ( )
10 2 10 1028 5 (D) + + = v t v t v t v ts ( ) ( ) . ( ) ( ) 10 2 10
1988 5 3. The differential equation for the circuit shown in fig.
P1.6.3 is (A) + + =i t i t i t i tL L L s( ) ( ) ( ) ( )1100 11 10
108 8 (B) + + =i t i t i t i tL L L s( ) ( ) ( ) ( )1100 11 10 108
8 (C) + + = i t i t i t i tL L L s ( ) . ( ) . ( ) ( ) 10 11 10 118
4 (D) + + = i t i t i t i tL L L s ( ) ( ) ( ) ( ) 10 11 10 118 4
4. In the circuit of fig. P.1.6.4 vs = 0 for t > 0. The initial
condition are v( )0 6= V and dv dt( )0 3000= - V s. The v t( ) for
t > 0 is (A) - +- - 2 8100 400 e et t V (B) 6 8100 400 e et t- -
+ V (C) 6 8100 400 e et t- - - V (D) None of the above 5. The
circuit shown in fig. P1.6.5 has been open for a long time before
closing at t = 0. The initial condition is v( )0 2= V. The v t( )
for t > is (A) 5 7 3 e et t- - - V (B) 7 5 3 e et t- - - V (C) -
+- - e et t 3 3 V (D) 3 3 e et t- - - V CHAPTER 1.6 THE RLC
CIRCUITS Page 54 2 W vs 10 Fm100 W 1 mH v Fig. P1.6.2 10 Fm 10 W
100 W iL is Fig. P.1.6.3 vs 25 Fm 1 H 80 W vC + Fig. P1.6.4 t=0 W1
H vC + F4 3 3 1 Fig. P.1.6.5 GATE EC BY RK Kanodia
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38. Statement for Q.67: Circuit is shown in fig. P.1.6. Initial
conditions are i i1 20 0 11( ) ( )= = A 6. i1 1( )s = ? (A) 0.78 A
(B) 1.46 A (C) 2.56 A (D) 3.62 A 7. i2 1( )s = ? (A) 0.78 A (B)
1.46 A (C) 2.56 A (D) 3.62 A 8. v tC( ) = ? for t > 0 (A) 4 1000
2000 e et t- - - V (B) ( )3 6000 2000 + - t e t V (C) 2 1000 2000 e
et t- - + V (D) ( )3 6000 2000 - - t e t V 9. The circuit shown in
fig. P1.6.9 is in steady state with switch open. At t = 0 the
switch is closed. The output voltage v tC( ) for t > 0 is (A) -
+- - 9 12400 300 e et t (B) e t tt- +400 3 300 4 300[ cos sin ] (C)
e t tt- +300 3 400 4 300[ cos sin ] (D) e t tt- +300 3 400 2 25
300[ cos . sin ] 10. The switch of the circuit shown in fig.
P1.6.10 is opened at t = 0 after long time. The v t( ) , for t >
0 is (A) 4 22 e tt- sin V (B) - - 4 22 e tt sin V (C) 4 22 e tt-
cos V (D) - - 4 22 e tt cos V 11. In the circuit of fig. P1.6.23
the switch is opened at t = 0 after long time. The current i tL ( )
for t > 0 is (A) e t tt- +2 2 4( cos sin ) A (B) e t tt- -2 3 4(
sin cos ) A (C) e t tt- - +2 4 2( sin cos ) A (D)e t tt- -2 2 4(
sin cos ) A Statement for Q.1214: In the circuit shown in fig.
P1.6.1214 all initial condition are zero. 12. If i ts( ) = 1 A,
then the inductor current i tL ( ) is (A) 1 A (B) t A (C) t + 1 A
(D) 0 A 13. If i t ts( ) .= 0 5 A, then i tL ( ) is (A) 0 5 3 25 10
3 . .t + - A (B) 2 3250t - A (C) 0 5 0 25 10 3 . .t - - A (D) 2
3250t + A 14. If i t es t ( ) = - 2 250 A then i tL ( ) is (A) 4000
3 250 te t- A (B) 4000 3 250 e t- A (C) 200 7 250 e t- A (D) 200 7
250 te t- A Chap 1.6The RLC Circuits Page 55 3 H1 W 2 H 2 W i1 i2
Fig. P1.6.67 30 (- ) mAu t 10 Fm 25 mH 100 W vC + Fig. P1.6.8 0.8 H
500 W 9 V 250 W 5 Fm vC + t=0 Fig. P1.6.9 3 W 1 W H6 V t=0 F2 1 4 1
vC + Fig. P1.6.10 2 W t=0 4 H 7 A4 W 8 W iL F4 1 Fig. P1.6.11 W 100
65 i u ts ( ) A 1 mF iL vL + 10 mH Fig. P1.5.12-14 GATE EC BY RK
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39. 15. The forced response for the capacitor voltage v tf ( )
is (A) 0 2 117 10 3 . .t + - V (B) 0 2 117 10 3 . .t - - V (C) 117
10 0 23 . . -- t V (D) 117 10 0 23 . . +- t V 16. For a RLC series
circuit R L= =20 0 6W , . H, the value of C will be [CD =critically
damped, OD =over damped, UD =under damped]. CD OD UD (A) C = 6 mF C
>6 mF C 6 mF (C) C >6 mF C = 6 mF C < 6 mF (D) C 6 mF 17.
The circuit shown in fig. P1.6.17 is critically damped. The value
of R is (A) 40 W (B) 60 W (C) 120 W (D) 180 W 18. The step response
of an RLC series circuit is given by d i t dt di t dt i t 2 2 5 10
( ) ( ) ( )+ + = , i( )0 2+ = , di dt ( )0 4 + = . The i t( ) is
(A) 1 4+ - e tt cos A (B) 4 2 4- - e tt cos A (C) 2 4+ - e tt sin A
(D) 10 4+ - e tt sin A 19. In the circuit shown in fig. P 1.5.19 v
t( ) for t > 0 is (A) 50 46 5 3 62 3 4 - + - ( . sin cos )t t e
t V (B) 50 46 5 3 62 3 4 + + - ( . sin cos )t t e t V (C) 50 62 4
46 5 4 3 + + - ( cos . sin )t t e t V (D) 50 62 4 46 5 4 3 - + - (
cos . sin )t t e t V 20. In the circuit of fig. P1.6.20 the switch
is closed at t = 0 after long time. The current i t( ) for t > 0
is (A) -10 8sin t A (B) 10 8sin t A (C) -10 8cos t A (D) 10 8cos t
A 21. In the circuit of fig. P1.6.21 switch is moved from 8 V to 12
V at t = 0. The voltage v t( ) for t > 0 is (A) 12 4 2 2 2- + -
( cos sin )t t e t V (B) 12 4 2 8 2- + - ( cos sin )t t e t V (C)
12 4 2 8 2+ + - ( cos sin )t t e t V (D) 12 4 2 2 2+ + - ( cos sin
)t t e t V Page 56 UNIT 1 Networks 120 WR 10 mF 4 H Fig. P1.6.17 2
(- ) Au t 1 H 0.04 F 4 W 2 W 50 ( ) Vu t vC + Fig. P1.6.19 5 W iL
20 V t=0 H 1 4 F 1 16 vC + Fig. P1.6.20 2 W 12 V 1 H 8 V F 1 6 vC +
t=0 Fig. P1.6.21 vx + iL 20 mH50 Wavx 100 W Fig. P1.6.15 GATE EC BY
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40. 22. In the circuit of fig. P1.5.22 the voltage v t( ) is
(A) 40 20 0 6 15 0 6 0 8 - + - ( cos . sin . ) . t t e t V (B) 35
15 0 6 20 0 6 0 8 + + - ( cos . sin . ) . t t e t V (C) 35 15 0 6
20 0 6 0 8 - + - ( cos . sin . ) . t t e t V (D) 35 15 0 6 0 8 - -
cos . . t e t V 23. In the circuit of fig. P1.6.23 the switch is
opened at t = 0 after long time. The current i t( ) for t > 0 is
(A) e et t- - +2 306 0 869. . A (B) - +- - e et t2 306 0 869 2. . A
(C) e et t- - +4 431 0 903. . A (D) 2 4 431 0 903 e et t- - -. . A
24. In the circuit of fig. P1.6.24 switch is moved from position a
to b at t = 0. The i tL ( ) for t > 0 is (A) ( )4 6 4 - t e t A
(B) ( )3 6 4 - - t e t A (C) ( )3 9 5 - - t e t A (D) ( )3 8 5 - -
t e t A 25. In the circuit shown in fig. P1.6.25 a steady state has
been established before switch closed. The i t( ) for t > 0 is
(A) 0 73 4 582 . sin .e tt- A (B) 0 89 6 382 . sin .e tt- A (C) 0
73 4 584 . sin .e tt- A (D) 0 89 6 384 . sin .e tt- A 26. The
switch is closed after long time in the circuit of fig. P1.6.26.
The v t( ) for t > 0 is (A) - + - 8 6 43 e tt sin V (B) - + - 12
4 43 e tt cos V (C) - + + - 12 4 4 3 4 3 ( cos sin )t t e t V (D) -
+ + - 12 4 4 6 4 3 ( cos sin )t t e t V 27. i t( ) = ? (A) 6 6 500
6 5000 50 - + - ( cos sin )t t e t mA (B) 8 8 500 0 06 5000 50 - +
- ( cos . sin )t t e t mA (C) 6 6 5000 0 06 5000 50 - + - ( cos .
sin )t t e t mA (D) 6 500050 e tt- sin mA Chap 1.6The RLC Circuits
Page 57 10 W 5 W 10 W 2 A H 3 4 F 1 3 t=0 Fig. P1.6.23 14 W 0.02 F
t=0 b a 2 W 12 V 4 A 6 W iL 2 H Fig. P1.6.24 5 W 20 W 1 H 5 W i 100
V F 1 25 t=0 Fig. P1.6.25 1 H 6 W 1 W 4 V t=0 2 A F 1 25 vC + Fig.
P1.6.26 2 kW 5 Fm i 12 ( ) Vu t 8 mH Fig. P1.6.27 0.2 F 1 W5 H 5 W3
( ) Au t 2 W 20 V vC + Fig. P1.6.22 GATE EC BY RK Kanodia
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41. 28. In the circuit of fig. P1.6.28 i( )0 1= A and v( )0 0=
. The current i t( ) for t > 0 is (A) 4 6 38 0 5 + - . . e At
(B) 4 6 38 0 5 - - . . e t A (C) 4 3 1 32 113 1 32 0 5 + + - ( cos
. . sin . ) . t t e t A (D) 4 3 1 32 113 1 32 0 5 - + - ( cos . .
sin . ) . t t e t A 29. In the circuit of fig. P1.6.29 a steady
state has been established before switch closed. The v to( ) for t
> 0 is (A) 100 10 te t- V (B) 200 10 te t- V (C) 400 50 te t- V
(D) 800 50 te t- V 30. In the circuit of fig. P1.6.30 a steady
state has been established before switch closed. The i t( ) for t
> 0 is (A) 2 22 e tt- sin A (B) - - e tt2 2sin A (C) - - - 2 1 2
( )t e t A (D) 2 1 2 ( )- - t e t A 31. In the circuit of fig.
P1.6.31 a steady state has been established. The i t( ) for t >
0 is (A) 9 2 810 2 5 + -- - e et t. A (B) 9 8 210 2 5 - + - e et t.
A (C) 9 2 10 10 2 5 + + - ( cos sin ) . t t e t A (D) 9 10 2 10 2 5
+ + - (cos sin ) . t t e t A *************** SOLUTIONS 1. (A) s s2
2 1 0+ + = s = - -1 1, , v t A A t e t ( ) ( )= + - 1 2 v( )0 10=
V, dv dt A A ( )0 0 1 1 2= = - + A A1 2 10= = 2. (A) i v dv dt L =
+ - 100 10 10 6 v i di dt vs L L = + +- 2 10 3 = + + + - - - - 2
100 10 10 10 1 100 10 106 3 6 2 v dv dt dv dt d vt dt v2 + 10 3000
1028 v t v t v t v ts( ) ( ) ( ) . ( )= + + 3. (C) i v i dv dt s C
L C = + + 100 10m v i di dt C L L = + - 10 10 3 i i di dt i d dt i
di dt s L L L L L = + + + +- - - 0 1 10 10 10 105 5 3 . ( ) = + + +
+- - - 0 1 10 10 105 4 8 2 2 . i di dt i di dt d i dt L L L L L + +
= i t i t i t i tL L L s ( ) . ( ) . ( ) ( ) 10 11 10 118 4 4. (A)
v dv dt v v dts 80 25 0+ + - =m ( ) d v dt dv dt 2 2 500 40000 0+ +
= s s2 500 40000 0+ + = s = - -100 400, , v t Ae Bet t ( ) = +-
-100 400 A B+ = 6, - - = -100 400 3000A B B = 8, A = -2 5. (C) The
characteristic equation is s s RC LC 2 1 0+ + = After putting the
values, s s2 4 3 0+ + = v t Ae Bet t ( ) = +- -3 , Page 58 UNIT 1
Networks 10 W 10 mF5 W3 A vo + 1 H t=0 Fig. P1.6.29 i 1 H 2 W 6 V 1
W F 1 4 t=0 Fig. P1.6.30 40 W 4 H10 W 6 ( ) Au t3 A i 10 mF Fig.
P1.6.31 0.5 F2 W1 H i 4 ( ) Au t Fig. P1.6.28 iL 2 W vs 10 Fm100 W
1 mH v Fig. S1.6.2 GATE EC BY RK Kanodia www.gatehelp.com
42. v( )0 2+ = V A B+ = 2 iL ( )0 0+ = iR( )0 2 3 4 8 3 = = , -
= + C dv dt ( )0 8 3 dv dt ( )0 8 + = - , - - = -A B3 8, B A= = -3
1, 6. (D) i di dt di dt 1 1 2 5 3 0+ - = , 2 3 3 02 2 1 i di dt di
dt + - = ( )1 5 3 01 2+ - =s i si , - + + =3 2 3 01 2si s i( ) ( )
( )( ) 1 5 3 3 2 3 01 1 + - + =s i s s i s 6 13 2 02 s s+ + = s = -
- 1 6 2, i A e Be t t 1 1 6 2 = + - - , i A B( )0 11= + = In
differential equation putting t = 0 and solving di dt di dt 1 20 33
2 0 143 6 ( ) , ( )+ + = - = - - - = - A B 6 2 33 2 , A B= =3 8, ,
i e e t t 1 6 2 3 8= + - - , i e e1 1 6 2 1 3 8 3 62( ) .s = + = -
- A 7. (A) i Ce De t t 2 6 2 = + - - i C D2 0 11( ) = = + , di dt C
D2 0 143 6 6 2 ( ) = - = - - C = -1 and D = 12 i e e t t 2 6 2 12=
- + - - A, i e e2 1 6 2 1 12 0 78( ) .s = + = - - A 8. (B) vC( )0
30 100 3+ = =m V C dv dt i i C dv dt C L L C( ) ( ) ( ) ( )0 0 0 0
0- - + + = = = = s s2 3 3 6 100 25 10 1 25 10 10 10 + + - - - s =
-2000, -2000 v t A A t eC t ( ) ( )= + - 1 2 2000 dv t dt A e A A t
eC t t( ) ( ) ( )= + + -- - 2 2000 1 2 2000 2000 v AC( )0 31 + = =
, dv dt AC( )0 2000 3 02= - = A2 6000= 9. (B) v iC L( ) , ( )0 3 0
12+ + = = -V mA v i dv dt C L C 250 5 10 06 + + =- 3 250 12 5 10 0
06 - + =- + m dv dt C( ) = + dv dt C( )0 0 s s2 6 6 250 5 10 1 0 8
5 10 0+ + =- - . + + =s s2 4 800 25 10 0 = - s j400 300 v t e A t A
tC t ( ) ( cos sin )= +-400 1 2300 300 A1 3= , dv dt A A AC( ) , 0
400 300 41 2 2= - + = 10. (B) v( )0 0+ = , iL ( )0 2+ = A, 1 4 0 2
dv xdt C( )+ = - s s2 4 8 0+ + = s j= - 2 2 v t e A t A tC t ( ) (
cos sin )= +-2 1 22 2 A1 0= , dv dt AC ( ) ( ) ( ) 0 8 2 0 0 0 2 2
+ = - = - + + + , A2 4= - 11. (D) iL ( )0 4+ = - , vC( )0 8+ = V 4
0 8 4 8 di dt L ( ) ( ) + = - - di dt L ( )0 10 + = s v v iC C L 4
1 2 0+ + = , v si iC L L= +4 8 s i si s jL L 2 4 5 0 2+ + = = - , i
t e A t A tL t ( ) ( cos sin )= +-2 1 2 A1 4= - , di dt A AL ( ) (
) 0 10 2 01 2 + = = - + + , A2 2= 12. (A) i v dv dt is L= + +- 100
65 10 3 , v di dt L = - 10 10 3 i di dt d i dt is L L L= + + =- -
-65 100 10 10 10 10 10 03 3 3 2 ( ) ( ) d i dt di dt i iL L L s 2 5
5 650 10 10+ + = Trying i t BL ( ) = 0 0 10 105 5 + + =B , B = 1,
iL = 1 A 13. (A) Trying i t At BL ( ) = + , 0 650 10 10 0 55 5 + +
+ =A At B t( ) ( . ), A = 0 5. Chap 1.6The RLC Circuits Page 59 8
W2 W iL 4 H F4 1 vC + Fig. S1.6.11 GATE EC BY RK Kanodia
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43. di dt A B ( ) . . 0 16 3 4 431 0 903 + = - = - - A B= =1 1,
24. (C) vC( )0 0= , iL ( )0 4 6 6 2 3= + = 0 02 0 0 3. ( ) ( ) dv
dt iC L= = dv dt C( )0 150= a = + = 6 14 2 2 5, wo = = 1 2 0 02 5 .
a wo= critically damped v t A Bt e t ( ) ( )= + + - 12 5 0 12 150
5= + = - +A A B, A B= - =12 90, v t t e t ( ) ( )= + - - 12 90 12 5
i t e t e t eL t t t ( ) . ( ) ( ) . ( ) ( )= - - + = -- - - 0 02 5
90 12 0 02 90 3 95 5 5 25. (A) v( )0 100 5 5 5 20 50 3 + = + + = ,
iL ( )0 0+ = if = 0 A di dt L ( )0 20 50 3 10 3 + = - = a = = 4 2 1
2, wo = = 1 1 1 25 5 s j= - - = - 2 4 25 2 4 58. i t A t B t e t (
) ( cos . sin . )= + - 4 58 4 58 2 26.(A) iL ( )0 0+ = , vL ( )0 4
12 8+ = - = - 1 25 0 0 0 dv dt iL L ( ) ( ) + + = = a = = 6 2 3, Wo
= = 1 1 1 25 5 / b = - - = - 3 9 25 3 4j v t A t B t e t 1 3 12 4
4( ) ( cos sin )= - + + - v AL ( )0 8 12= - = + , =A 4 dv dt A BL (
)0 0 3 4= = - + , =B 3 27. (C) a = = = 1 2 1 2 2 54 50 RC k W LC m
o = = = 1 1 8 5 5000 m a < Wo, underdamped response. s j= - - =
- 50 50 5000 50 50002 2 i t A t B t e t ( ) ( cos sin )= + + - 6
5000 5000 50 mA i A( )0 6 6= = + , = -A 6 di dt A B ( )0 50 5000 0=
- + = , B = -0 06. 28.(D) i( )0 1+ = A, v Ldi dt ( ) ( ) 0 0+ + = a
= = 1 2 2 0 5 0 5 . . , Wo = - = 1 1 0 5 2 . s j= - - = 0 5 0 5 2 0
5 1 3232 . . . . i t A t B t e t( ) ( cos . sin . ) . = + + - 4 1
32 1 32 0 5 1 4= + A, = -A 3 di dt A B ( ) . . 0 0 0 5 1 32= = + ,
B = -113. 29. (B) Vo( )0 0+ = ,iL ( )0 1+ = A di dt vL ( ) ( ) 0 0
01 + = = a = = 1 2 5 0 01 10 . , Wo = = 1 1 0 01 10 . a = Wo, so
critically damped response s = - -10 10, i t A Bt e t ( ) ( )= + -
3 10 , i A( )0 1 3= = + di dt A B ( )0 10 + = - + i t t eL t ( ) (
)= - + - 3 2 20 10 , v Ldi t dt teo L t = = -( ) 200 10 30. (C) i(
)0 6 1 2 2+ = - + = - A, v di dt c( ) ( ) 0 2 1 2 0+ + = = = a = =
= 1 2 1 2 1 0 25 2 RC . , W LC o = = 1 2 a = Wo, critically damped
response s = - -2 2, i t A Bt e t ( ) ( )= + -2 , A = -2 di t dt Bt
e B et t( ) ( ) ( ) ( )= - + - + + - 2 2 02 2 At t = 0, = -B 2 31.
(A) i( )0 3+ = A, vC( )0 0+ = V = + 4 0di dt ( ) is = 9 A, R = =10
40 8|| W a = = = 1 2 1 2 8 0 01 6 25 RC . . W LC o = = = 1 1 4 0 01
5 . a > Wo, so overdamped response s = - - = - -6 25 6 25 25 10
2 52 . . , . i t Ae Bet t ( ) . = + +- - 9 10 2 5 3 9= + +A B, 0 10
2 5= - -A B. On solving, A = 2, B = -8 ************ Chap 1.6The RLC
Circuits Page 61 GATE EC BY RK Kanodia www.gatehelp.com
44. 1. i t( ) = ? (A) 20 300 68 2cos ( . )t + A (B) 20 300 68
2cos( . )t - A (C) 2 48 300 68 2. cos( . )t + A (D) 2 48 300 68 2.
cos( . )t - A 2. v tC( ) ?= (A) 0 89 10 63 433 . cos ( . )t - V (B)
0 89 10 63 433 . cos ( . )t + V (C) 0 45 10 26 573 . cos ( . )t + V
(D) 0 45 10 26 573 . cos ( . )t - V 3. v tC( ) = ? (A) 1 2 2 45cos
( )t - V (B) 1 2 2 45cos ( )t + V (C) 1 2 2 45sin ( )t - V (D) 1 2
2 45sin ( )t + V 4. v tC( ) ?= (A) 2 25 5 150. cos ( )t + V (B) 2
25 5 150. cos ( )t - V (C) 2 25 5 140 71. cos ( . )t + V (D) 2 25 5
140 71. cos ( . )t - V 5. i t( ) = ? (A) 2 2 5 77sin ( . )t + A (B)
cos ( . )2 84 23t - A (C) 2 2 5 77sin ( . )t - A (D) cos ( . )2 84
23t + A CHAPTER 1.7 SINUSOIDAL STEADY STATE ANALYSIS Page 62 i 25
mH20cos 300 Vt 3 W ~ Fig. P1.7.1 2 Wcos 10 A 3 t 1 mF~ vC + Fig.
P1.7.2 5 W 0.1 F vC + cos 2 Vt ~ Fig. P1.7.3 3 H 50 mF 9 W vC +
8cos 5 Vt ~ Fig. P1.7.4 1 W 4 W 0.25 F 4 H i 10cos 2 Vt ~ Fig.
P1.7.5 GATE EC BY RK Kanodia www.gatehelp.com
45. 13. In the bridge shown in fig. P1.7.13, Z1 300= W, Z j2
300 600= -( ) W, Z j3 200 100= +( )W. The Z4 at balance is (A) 400
300+ j W (B) 400 300- j W (C) j100 W (D) - j900 W 14. In a two
element series circuit, the applied voltage and the resulting
current are v t t( ) sin ( )= +60 66 103 V, i t t( ) . sin ( . )= +
2 3 10 68 33 A. The nature of the elements would be (A) R C- (B) L
C- (C) R L- (D) R R- 15. Vo = ? (A) 223 56 - V (B) 223 56 V (C) 124
154 - V (D) 124 154 V 16. v to( ) = ? (A) 315 112. cos ( )t + V (B)
43 2 23. cos ( )t + V (C) 315 112. cos ( )t - V (D) 43 2 23. cos (
)t - V Statement for Q.17-18: The circuit is as shown in fig.
P1.7.17-18 17. i t1( ) = ? (A) 2 36 4 4107. cos ( . )t - A (B) 2 36
4 4107. cos ( . )t + A (C) 1 37 4 4107. cos ( . )t - A (D) 2 36 4
4107. cos ( . )t + A 18. i t2( ) = ? (A) 2 04 4 92 13. sin ( . )t +
A (B) - + 2 04 4 2 13. sin ( . )t A (C) 2 04 4 2 13. cos ( . )t + A
(D) - + 2 04 4 92 13. cos ( . )t A 19. Ix = ? (A) 394 46 28. . A
(B) 4 62 97 38. . A (C) 7 42 92 49. . A (D) 6 78 49 27. . A 20. Vx
= ? (A) 29 11 166. V (B) 29 11 166. - V (C) 43 24 124. V (D) 43 24
124. - V Page 64 UNIT 1 Networks 4 W 0.5Ix Ix j3 W10 30 V o - 2j W
~ Fig. P1.7.19 10sin ( +30 ) Vt o 1 H 3 W 1 F 20cos ( -45 ) Vt o ~
~vo + Fig. P1.7.16 1 H 1 W 1 W 1 H 5cos 4 Vt 10cos (4 -30 ) Vt o i1
i2 1 W 1 F ~ ~ Fig. P1.7.1718 j20 40 W 50 W120 -15 V o - 30j 6 30 A
o ~ ~ Vo Fig. P1.7.15 3 0 A o Vx + 4Vx 20 W j10 W20 W ~ Fig.
P1.7.20 Z 1 Z 2 Z 4 Z 3 ~ Fig. P1.7.13 GATE EC BY RK Kanodia
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46. Statement for Q.2732: Determine the complex power for hte
given values in question. 27. P = 269 W, Q = 150 VAR (capacitive)
(A) 150 269- j VA (B) 150 269+ j VA (C) 269 150- j VA (D) 269 150+
j VA 28. Q = 2000 VAR, pf = 0 9. (leading) (A) 4129 8 2000. + j VA
(B) 2000 4129 8+ j . VA (C) 2000 4129 8- j . . VA (D) 4129 8 2000.
- j VA 29. S = 60 VA, Q = 45 VAR (inductive) (A) 39 69 45. + j VA
(B) 39 69 45. - j VA (C) 45 39 69+ j . VA (D) 45 39 69- j . VA 30.
Vrms = 220 V, P = 1 kW, | |Z = 40 W (inductive) (A) 1000 68125- j .
VA (B) 1000 68125+ j . VA (C) 68125 1000. + j VA (D) 68125 1000. -
j VA 31. Vrms = 21 20 V, Vrms = 21 20 V, Irms = - 8 5 50. A (A) 154
6 89 3. .+ j VA (B) 154 6 89 3. .- j VA (C) 61 167 7+ j . VA (D) 61
167 7- j . VA 32. Vrms = 120 30 V, Z j= +40 80 W (A) 72 144+ j VA
(B) 72 144- j VA (C) 144 72+ j VA (D) 144 72- j VA 33. Vo = ? (A) 7
1 32 29. . kV (B) 42 59 32 29. . kV (C) 38 49 24 39. . kV (D) 38 49
32 29. . kV 34. A relay coil is connected to a 210 V, 50 Hz supply.
If it has resistance of 30 W and an inductance of 0.5 H, the
apparent power is (A) 30 VA (B) 275.6 VA (C) 157 VA (D) 187 VA 35.
In the circuit shown in fig. P1.7.35 power factor is (A) 56.31
(leading) (B) 56.31 (lagging) (C) 0.555 (lagging) (D) 0.555
(leading) 36. The power factor seen by the voltage source is (A)
0.8 (leading) (B) 0.8 (lagging) (C) 36.9 (leading) (D) 39.6
(lagging) 37. The average power supplied by the dependent source is
(A) 96 W (B) -96 W (C) 92 W (D) -192 W 38. In the circuit of fig.
P1.7.38 the maximum power absorbed by ZL is (A) 180 W (B) 90 W (C)
140 W (D) 700 W Page 66 UNIT 1 Networks v1+ 1 W4 W v110cos 2 Vt ~
F3 1 4 3 Fig. P1.7.36 j1.92Ix 4.8 W 8 W1.6Ix2 90 A o ~ Fig. P1.7.37
6 0 A o VO + ~ 20 kW 0.8 pf lagging 16 kW 0.9 pf lagging Fig.
P1.7.33 4 W - 2j - 2j j5 Fig. P1.7.35 10 W j15 - 10j120 0 V o ~ ZL
Fig. P1.7.38 GATE EC BY RK Kanodia www.gatehelp.com
47. 39. The value of the load impedance, that would absorbs the
maximum average power is (A) 12 8 49 6. .- j W (B) 12 8 49 6. .+ j
W (C) 339 86 3. .- j W (D) 339 86 3. .+ j W Statement for Q.4041:
In a balanced Yconnected three phase generator Vab = 400 Vrms 40.
If phase sequence is abc then phase voltage V V Va b c, , and are
respectively (A) 231 0 231 120 231 240 , , (B) 231 30 231 150 231
90 - - , , (C) 231 30 231 150 231 90 - , , (D) 231 60 231 180 231
60 - , , 41. If phase sequence is acb then phase voltage are (A)
231 0 231 120 231 240 , , (B) 231 30 231 150 231 90 - - , , (C) 231
30 231 150 231 90 - , , (D) 231 60 231 180 231 60 - , , 42. A
balanced three-phase Y-connected load has one phase voltage Vc =
277 45 V. The phase sequence is abc. The line to line voltage VAB
is (A) 480 45 V (B) 480 45 - V (C) 339 45 V (D) 339 45 - V 43. A
three-phase circuit has two parallel balanced D loads, one of the 6
W resistor and one of 12 W resistors. The magnitude of the total
line current, when the line-to-line voltage is 480 Vrms , is (A)
120 Arms (B) 360 Arms (C) 208 Arms (D) 470 Arms 44. In a balanced
three-phase system, the source has an abc phase sequence and is
connected in delta. There are two parallel Y-connected load. The
phase impedance of load 1 and load 2 is 4 4+ j W and 10 4+ j W
respectively. The line impedance connecting the source to load is 0
3 0 2. .+ j W. If the current in a phase of load 1 is I = 10 20
Arms, the current in source in ab branch is (A) 15 122 - Arms (B) 8
67 122. - Arms (C) 15 27 9 . Arms (D) 8 67 57 9. . - Arms 45. An
abc phase sequence 3-phase balanced Y-connected source supplies
power to a balanced D connected load. The impedance per phase in
the load is 10 8+ j W. If the line current in a phase is IaA = - 28
10 28 66. . Arms and the line impedance is zero, the load voltage
VAB is (A) 207 8 140. - Vrms (B) 148 3 40. Vrms (C) 148 3 40. -
Vrms (D) 207 8 40. Vrms 46. The magnitude of the complex power
supplied by a 3-phase balanced Y-Y system is 3600 VA. The line
voltage is 208 Vrms . If the line impedance is negligible and the
power factor angle of the load is 25, the load impedance is (A) 5
07 10 88. .+ j W (B) 10 88 5 07. .+ j W (C) 432 14 6. .+ j W (D) 14
6 432. .+ j W *********** Chap 1.7Sinusoidal Steady State Analysis
Page 67 j100 3 20 A o 80 W - 40j ~ ZL Fig. P1.7.39 GATE EC BY RK
Kanodia www.gatehelp.com
48. SOLUTIONS 1. (D) Z j j= + = + = 3 25 300 3 7 5 8 08 68 2(
)( ) . . .m W I = = - 20 0 8 08 68 2 2 48 68 2 . . . . A i t t( ) .
cos ( . )= - 2 48 300 68 2 A 2. (A) Y j j= + = + = 1 2 1 10 0 5 112
63 433 ( )( ) . . .m VC = = - ( ) . . . . 1 0 112 63 43 0 89 63 43
V v t tC( ) . cos ( . )= - 0 89 10 63 433 V 3. (A) Z j j= + - = - =
- 5 0 1 2 5 5 5 5 45 ( . )( ) VC = - - = - ( ) ( )1 0 5 90 5 2 45 1
2 45 V v tC( ) = 1 2 2 45cos ( )t - V 4. (D) Z j j j= + + - = +9 3
5 50 5 9 11( )( ) ( ) ( )m = Z 14 21 50 71. . W VC = = ( )( ) . . .
. 8 0 4 90 14 21 50 71 2 25 140 71 - V v t tC( ) . cos ( . )= - 2
25 5 140 71 V 5. (B) V j j j a = + - + + = + 10 0 1 1 1 1 2 1 4 8
10 0 105 0 4. . V I V j j a = + = + = - 4 8 10 0 1 10 1 84 23. A i
t t( ) cos ( . )= - 2 84 23 A 6. (D) w p p= = 2 10 10 2 103 4 Y j j
= + - +( )( ) ( )( ) 1 2 10 160 2 10 1 36 4 4 m p m p = -0 0278 0
0366. .j S Z Y j= = + 1 1316 17 33. . W 7. (C) Z j j= - + w m w ( )
||( ( ) ) 22 6 27m = - + + - -j j j 10 22 6 27 10 6 27 10 22 6 3 6
w w w w ( ) ( )m = - + - 27 10 22 6 10 22 6 27 10 22 3 6 6 2 j j w
w w m - - - = j j36 10 22 27 10 22 27 10 22 0 6 3 6 2 w w w m =w
1278 f = w p2 Hz = 1278 2 203 p = Hz 8. (C) Vs = 477 68. V, V2 7
51= 35. V V Vs1 2 7 68 7 51 159= - = 47 - 35 = 125. . . 9. (B) vin
= + - =3 14 10 52 2 ( ) 10. (C) I1 744 118= - mA, I2 540 100= mA I
I I= + = - + 1 2 744 118 540 5 100. = - 460 164 i t t( ) cos ( )= -
460 3 164 mA 11. (A) 2 4 20 5 10 45 = - + - 0 + V j V j C C ( )( )(
) ( )1 4 10 5 10 5 4 8+ - + = + - +j j j V j j jC - = +60 100 10j V
jC( ) VC = - 64.711 6. 12. (D) X X XL C= + = 0 So reactive power
drawn from the source is zero. 13. (B) Z Z Z Z1 4 3 2= 300 300 600
200 1004Z j j= - +( )( ) = -Z j4 400 300 14. (A) R C- causes a
positive phase shift in voltage Z Z= | | q , - < 0 is (A) 9 t V
(B) 9e t- V (C) 9 V (D) 0 V 15. A unit step current of 1 A is
applied to a network whose driving point impedance is Z s V s I s s
s ( ) ( ) ( ) ( ) ( ) = = + + 3 2 2 The steady state and initial
values of the voltage developed across the source would be
respectively (A) 3 4 V, I V (B) 1 4 V, 3 4 V (C) 3 4 V, 0 V (D) 1
V, 3 4 V 16. In the circuit of Fig. P1.8.16 i( )0 1= A, vC( )0 8= V
and v e u tt 1 2 10 2 4 = - ( ). The i t( ) is (A) 1 15 10 2 10 4
10 10 3 22 4 4 4 [ ] ( )e e e u tt t t- - - - - A (B) 1 15 10 2 10
4 10 10 3 22 4 4 4 [ ] ( )- + +- - - e e e u tt t t A (C) 1 3 10 2
10 4 10 10 3 22 4 4 4 [ ] ( )e e e u tt t t- - - + + A (D) 1 3 10 2
10 4 10 10 3 22 4 4 4 [ ] ( )- + -- - - e e e u tt t t A 17. In the
circuit shown in Fig. P1.8.18 v( )0 8- = V and i t tin ( ) ( )= 4d
. The v tC( ) for t 0 is (A) 164e t- V (B) 208e t- V (C) 208 1 3 (
)- - e t V (D) 164 3 e t- V Page 74 UNIT 1 Networks A 2 W Vo( )s +
- 2Vo( )s a b s 1 ( +1)s 4 Fig. P1.8.12 1 F vC2 + - 1 FvC1 + - t=0
Fig. P1.8.13 10 V S1 5 V Va 1 V S2 5 V2 F + - 6 V +- 3 F 4 F + -
Fig. P1.8.14 50 W 1m H v1 2.5 Fm i vC + Fig. P1.8.16 iin 20 mF50 W
vC + Fig. P1.8.17 GATE EC BY RK Kanodia www.gatehelp.com
55. 18. The driving point impedance Z s( ) of a network has the
pole zero location as shown in Fig. P1.8.18. If Z( )0 3= , the Z s(
) is (A) 4 3 12 ( )s s s + + + (B) 2 3 2 22 ( )s s s + + + (C) 2 3
2 22 ( )s s s + + + (D) 4 3 22 ( )s s s + + + Statement for
Q.19-21: The circuit is as shown in the fig. P1.8.1921. All initial
conditions are zero. 19. I s I s o in ( ) ( ) = ? (A) ( )s s + 1 2
(B) 2 1 1 s s( )+ - (C) ( )s s+ - 1 1 (D) s s( )+ - 1 1 20. If i t
tin ( ) ( )= 4d then i to( ) will be (A) 4d( ) ( )t e u tt - - A
(B) 4 4d( ) ( )t e u tt - - A (C) 4 4e u t tt- -( ) ( )d A (D) e u
t tt- -( ) ( )d A 21. If i t tu tin ( ) ( )= then i to( ) will be
(A) e u tt- ( ) A (B) ( ) ( )1 - - e u tt A (C) u t( ) A (D) ( ) (
)2 - - e u tt A 22. The voltage across 200 mF capacitor is given by
V s s s s C( ) ( ) = + + 2 6 3 The steady state voltage across
capacitor is (A) 6 V (B) 0 V (C) (D) 2 V 23. The transformed
voltage across the 60 mF capacitor is given by V s s s s C( ) ( )(
) = + + + 20 6 10 3 4 The initial current through capacitor is (A)
0.12 mA (B) - 0 12. mA (C) 0.48 mA (D) - 0 48. mA 24. The current
through an 4 H inductor is given by I s s s L ( ) ( ) = + 10 2 The
initial voltage across inductor is (A) 40 V (B) 20 V (C) 10 V (D) 5
V 25. The amplifier network shown in fig. P1.8.25 is stable if (A)
K 3 (B) K 3 (C) K 1 3 (D) K 1 3 26. The network shown in fig.
P1.8.26 is stable if (A) K 5 2 (B) K 5 2 (C) K 2 5 (D) K 2 5 Chap
1.8Circuit Analysis in the s-Domain Page 75 iin io 1 H 1 W 1 F 1 W
Fig. P1.8.1921 4 W 1 H1 F 2 W v1 v2 + - Amplifier gain=K + -
Fig.P1.8.25 Kv2 1 W v2 + - 2 F1 W 1 F Fig.P1.8.26 jw 1 -1 -1-3 s
Fig. P1.8.18 GATE EC BY RK Kanodia www.gatehelp.com
56. 27. A circuit has a transfer function with a pole s = - 4
and a zero which may be adjusted in position as s a= - The response
of this system to a step input has a term of form Ke t-4 . The K
will be (H= scale factor) (A) H a 1 4 - (B) H a 1 4 + (C) H a 4 4 -
(D) H a 4 4 + 28. A circuit has input v t t u tin ( ) cos ( )= 2 V
and output i t t u to( ) sin ( )= 2 2 A. The circuit had no
internal stored energy at t = 0. The admittance transfer function
is (A) 2 s (B) s 2 (C) s (D) 1 s 29. A two terminal network
consists of a coil having an inductance L and resistance R shunted
by a capacitor C. The poles of the driving point impedance function
Z of this network are at - 1 2 3 2 j and zero at -1. If Z( )0 1=
the value of R, L, C are (A) 3 W, 3 H, 1 3 F (B) 2 W, 2 H, 1 2 F
(C) 1 W, 2 H, 1 2 F (D) 1 W, 1 H, 1 F 30. The current response of a
network to a unit step input is I s s s s o = + + + 10 2 11 302 ( )
( ) The response is (A) Under damped (B) Over damped (C) Critically
damped (D) None of the above Statement for Q.31-33: The circuit is
shown in fig. P1.8.31-33. 31. The current ratio transfer function I
I o s is (A) s s s s ( )+ + + 4 3 42 (B) s s s s ( ) ( )( ) + + + 4
1 3 (C) s s s s 2 3 4 4 + + +( ) (D) ( )( ) ( ) s s s s + + + 1 3 4
32. The response is (A) Over damped (B) Under damped (C) Critically
damped (D) cant be determined 33. If input is is 2u t( ) A, the
output current io is (A) ( ) ( )2 3 3 e te u tt t- - - A (B) ( ) (
)3 3 te e u tt t- - - A (C) ( ) ( )3 3 e e u tt t- - - A (D) ( ) (
)e e u tt t- - -3 3 A 34. In the network of Fig. P1.8.34, all
initial condition are zero. The damping exhibited by the network is
(A) Over damped (B) Under damped (C) Critically damped (D) value of
voltage is requires 35. The voltage response of a network to a unit
step input is V s s s s o( ) ( ) = + + 10 8 162 The response is (A)
under damped (B) over damped (C) critically damped (D) cant be
determined 36. The response of an initially relaxed circuit to a
signal vs is e u tt-2 ( ). If the signal is changed to ( )vs dv dt
s + 2 , the response would be (A) 5 2 e u tt- ( ) (B) - - 3 2 e u
tt ( ) (C) 4 2 e u tt- ( ) (D) - - 4 2 e u tt ( ) Page 76 UNIT 1
Networks is 4 W 1 H io 3 1 F Fig. P1.8.31-33 vs 2 H 2 W vo + - 1 4
F Fig. P1.8.34 GATE EC BY RK Kanodia www.gatehelp.com
57. 37. Consider the following statements in the circuit shown
in fig. P1.8.37 1. It is a first order circuit with steady state
value of vC = 10 3 , i = 5 3 A 2. It is a second order circuit with
steady state of vC = 2 V , i = 2 A 3. The network function V s I s
( ) ( ) has one pole. 4. The network function V s I s ( ) ( ) has
two poles. The true statements are (A) 1 and 3 (B) 1 and 4 (C) 2
and 3 (D) 2 and 4 38. The network function s s s s 2 2 10 24 8 15 +
+ + + represent a (A) RC admittance (B) RL impedance (C) LC
impedance (D) None of the above 39. The network function s s s s s
( ) ( )( )( ) + + + + 4 1 2 3 represents an (A) RC impedance (B) RL
impedance (C) LC impedance (D) None of these 40. The network
function s s s s ( ) ( )( ) 3 8 1 3 + + + represents an (A) RL
admittance (B) RC impedance (C) RC admittance (D) None of the above
41. The network function ( )( ) ( )( ) s s s s s + + + + 1 4 2 5 is
a (A) RL impedance function (B) RC impedance function (C) LC
impedance function (D) Above all 42. The network function s s s 2 7
6 2 + + + is a (A) RL impedance function (B) RL admittance (C) LC
impedance function (D) LC admittance 43. A valid immittance
function is (A) ( )( ) ( )( ) s s s s + + + - 4 8 2 5 (B) s s s s (
) ( )( ) + + + 1 2 5 (C) s s s s s ( )( ) ( )( ) + + + + 2 3 1 4
(D) s s s s s ( )( ) ( )( ) + + + + 2 6 1 4 44. The network
function s s s s 2 2 8 15 6 8 + + + + is a (A) RLadmittance (B) RC
admittance (C) LC admittance (D) Above all 45. A impedance function
is given as Z s s s s s ( ) ( )( ) ( ) = + + + 3 2 4 3 The network
for this function is ************ Chap 1.8Circuit Analysis in the
s-Domain Page 77 i 4 W2 H 10 V 1 W vC + 2 1 F Fig. P1.8.37 8 F 1 F
3 W8 1 F 3 1 W 1 F 3 W 3 1 W 3 W 1 F 3 1 W 8 1 F 3 W 1 H 3 1 W 8 1
H (C) (D) (A) (B) GATE EC BY RK Kanodia www.gatehelp.com
58. 18. (B) Z s K s s j s j K s s s ( ) ( ) ( ( ))( ( )) ( ) =
+ - - + - - = + + + 3 1 1 3 2 22 Z K K( )0 3 2 3 2= = = 19. (D) I s
I s s s s s s s s o in ( ) ( ) = + + + + = + 1 1 1 1 1 20. (B) I
sin ( ) = 4 I s s s s o( ) = + = - + 4 1 4 4 1 i t t e u to t ( ) (
) ( )= - - 4 4d 21. (B) I s s in ( ) = 1 2 , I s s s s s o( ) ( ) =
+ = - + 1 1 1 1 1 i t u t e u to t ( ) ( ) ( )= - - = - - ( ) ( )1
e u tt 22. (D) v sV s s s C s C s ( ) lim ( ) lim = = + + = 0 0 2 6
3 2 V 23. (D) v sV sC s C( ) lim ( )0+ = = + + + = s s s s ( ) ( )(
) 20 6 10 3 4 2 V i Cdv dt C C = I s C sV s vC C C( ) [ ( ) ( )]= -
+ 0 = + + + - - 60 10 20 6 10 3 4 26 s s s s ( ) ( )( ) = - + + + -
480 10 10 3 10 43 12 6 2 ( )s s s i sI sC s C( ) lim ( ) .0 480 10
0 486+ - = = - = - mA 24. (A) v L d i dt V s L sI s iL L L L L= = -
+ ( ) [ ( ) ( )]0 i sI s s L s L( ) lim ( )0 10 2 0+ = = + = V s s
s s s L ( ) ( ) = + = + 40 2 40 2 v sV s s s L s L( ) lim ( )0 40 2
40+ = = + = 25. (A) V s KV s2 1( ) ( )= V s V s KV s s s 1 1 1 2 4
1 0 ( ) ( ) ( ) + - + + = 4 1 2 2 0+ + + - =s s K s K s2 6 2 1 0+ -
+ =( ) ( )6 2 0- >K K < 3 26. (B) Let v1 be the node voltage
of middle node V s KV s sV s s s 1 2 22 1 2 ( ) ( ) ( ) = + + + + =
+( ) ( ) ( ) ( )3 1 21 2s V s s K V s V s sV s s 2 12 2 1 ( ) ( ) =
+ + =( ) ( ) ( )2 1 22 1s V s sV s + + = +( )( ) ( )3 1 2 1 2 2s s
s s K 2 5 2 1 02 s K s+ - + =( ) , 5 2 0- >K , K < 5 2 27.
(A) H s H s a s ( ) ( ) = + + 4 R s H s a s s Ha s H a s ( ) ( ) (
) = + + = + - +4 4 1 4 4 r t Ha u t H a e t ( ) ( )= + - - 4 1 4 4
28. (A) V s s s in ( ) = +2 1 , I s s o( ) = + 2 12 , I s V s s o
in ( ) ( ) = 2 29. (D) Z s sL R sC sL R sC ( ) ( ) = + + + 1 1 = +
+ + 1 12 C s R L s R L LC Z s K s s j s j ( ) ( ) = + + + + - 1 1 2
3 2 1 2 3 2 = + + + K s s s ( ) ( ) 1 12 Since Z K( ) ,0 1 1= =thus
1 1 1 1 1 C R L LC = = =, , C L R= = =1 1 1, , 30. (B) The
characteristic equation is s s s2 2 11 30 0( )+ + = s s s2 6 5( ) (
)+ + =0 s = - -6 5, , Being real and unequal, it is overdamped.
Page 80 UNIT 1 Networks sL R Cs 1Z s( ) Fig. S1.8.29 GATE EC BY RK
Kanodia www.gatehelp.com
59. 31. (B) I I s s s s s s s o s = + + + = + + + 4 4 3 4 1 3 (
) ( )( ) 32. (A) The characteristic equation is ( ) ( )s s+ + =1 3
0. Being real and unequal root, it is overdamped response. 33. (C)
i u ts = 2 ( ) I s s s( ) = 2 I s s s s s s o ( ) ( ) ( )( ) = + +
+ = + - + 2 4 1 3 3 1 1 3 i e e u to t t = -- - ( ) ( )3 3 A 34.
(B) V s V s s s s s o s ( ) ( ) = + + = + + 2 4 2 2 1 22 The roots
are imaginary so network is underdamped. 35. (C) The characteristic
equation is s s s( )2 8 16 0+ + = , ( )s + =4 02 , s = - -4 4,
Being real and repeated root, it is critically damped. 36. (B) v e
u to t = -2 ( ) V s H s V s s o s( ) ( ) ( )= = + 1 2 = +v v dv dt
s s s2 = +V s s V ss s( ) ( ) ( )1 2 = = +V s H s V s s V s H so s
s( ) ( ) ( ) ( ) ( ) ( )1 2 = + + = - + V s s s s o( ) 1 2 2 2 3 2
= - - v s e u to t 2 3 2 d( ) ( ) 37. (C) It is a second order
circuit. In steady state i = + = 10 4 1 2 A , v = =2 1 2 V I s s s
s s ( ) ( ) ( ) = + + + = + + + 10 2 4 1 1 1 2 5 2 2 12 V s s s s s
( ) ( ) ( ) = + + + + = + + 10 1 1 2 2 4 1 1 1 2 10 2 12 V s I s s
( ) ( ) = + 2 2 , It has one pole at s = -2 38. (D) s s s s s s s s
2 2 10 24 8 15 4 6 3 5 + + + + = + + + + ( )( ) ( )( ) The
singularity near to origin is pole. So it may be RC impedance or RL
admittance function. 39. (D) Poles and zero does not interlace on
negative real axis so it is not a immittance function. 40. (C) The
singularity nearest to origin is a zero. So it may be RL impedance
or RC admittance function. Because of (D) option it is required to
check that it is a valid RC admittance function. The poles and
zeros interlace along the negative real axis. The residues of Y s s
RC ( ) are real and positive. 41. (B) The singularity nearest to
origin is a pole. So it may be RC impedance or RL admittance
function. 42. (A) s s s s s s 2 7 6 2 1 6 2 + + + = + + + ( )( ) (
) The singularity nearest to origin is at zero. So it may be RC
admittance or RL impedance function. 43. (D) (A) pole lie on
positive real axis (B) poles and zero does not interlace on axis.
(C) poles and zero does not interlace on axis. (D) is a valid
immittance function. 44. (A) s s s s s s s s 2 2 8 15 6 8 3 5 2 4 +
+ + + = + + + + ( ) ( ) ( ) ( ) The singularity nearest to origin
is a pole. So it may be a RL admittance or RC impedance function.
45. (A) The singularity nearest to origin is a pole. So this is RC
impedance function. Z s s s s s ( ) = + + + = + + + 3 8 1 3 3 8 1 3
1 3 ************** Chap 1.8Circuit Analysis in the s-Domain Page 81
GATE EC BY RK Kanodia www.gatehelp.com
60. Statement for Q.1-2: In the circuit of fig. P1.9.1-2 i t1 4
2= sin A, and i2 0= . 1. v1 = ? (A) -16 2cos t V (B) 16 2cos t V
(C) 4 2cos t V (D) -4 2cos t V 2. v2 = ? (A) 2 2cos t V (B) -2 2cos
t V (C) 8 2cos t V (D) -8 2cos t V Statement for Q.3-4: Consider
the circuit shown in Fig. P1.9.3-4 3. If i1 0= and i t2 2 4= sin A,
the voltage v1 is (A) 24 4cos t V (B) -24 4cos t V (C) 15 4. cos t
V (D) -15 4. cos t V 4. If i e t 1 2 = - V and i2 0= , the voltage
v2 is (A) - - 6 2 e t V (B) 6 2 e t- V (C) 15 2 . e t- V (D) - - 15
2 . e t V Statement for Q.5-6: Consider the circuit shown in fig.
P19.5-6 5. If current i t1 3 4= cos A and i2 0= , then voltage v1
and v2 are (A) v t1 24 4= - sin V, v t2 24 4= - sin V (B) v t1 24
4= sin V, v t2 36 4= - sin V (C) v t1 15 4= . sin V, v t2 4= sin V
(D) v t1 15 4= - . sin V, v t2 4= -sin V 6. If current i1 0= and i
t2 4 3= sin A, then voltage v1 and v2 are (A) v t1 24 3= cos V, v
t2 36 3= cos V (B) v t1 24 3= cos V, v t2 36 3= - cos V (C) v t1 24
3= - cos V, v t2 36 3= cos V (D) v t1 24 3= - cos V, v t2 36 3= -
cos V CHAPTER 1.9 MAGNETICALLY COUPLED CIRCUITS Page 82 i1 i2 3 H -
+ v2 - + v1 3 H 4 H Fig. P1.9.5-6 i1 i2 2 H - + v2 - + v1 2 H 3 H
Fig. P1.9.5-6 i1 i2 1 H - + v2 - + v1 2 H 1 H Fig. P1.9.1-2 GATE EC
BY RK Kanodia www.gatehelp.com
61. Statement for Q.7-8: In the circuit shown in fig. P1.9.7-8,
i t1 3 3= cos A and i t2 4 3= sin A. 7. v1 = ? (A) 6 2 3( cos sin
)- +t t V (B) 6 2 3( cos sin )t t+ V (C) - +6 2 3( cos sin )t t V
(D) 6 2 3( cos sin )t t- V 8. v2 = ? (A) 3 8 3 3( cos sin )t t- V
(B) 6 2 3( cos sin )t t+ V (C) 3 8 3 3 3( cos sin )t t+ V (D) 6 2
3( cos sin )t t- V Statement for Q.9-10: In the circuit shown in
fig. P1.9.9-10, i t1 5 3= sin A and i t2 3 3= cos A 9. v1 =? (A) 9
5 3 3 3( cos sin )t t+ V (B) 9 5 3 3 3( cos sin )t t- V (C) 9 4 3 5
3( cos sin )t t+ V (D) 9 5 3 3 3( cos sin )t t- V 10. v2 = ? (A) 9
4 3 5 3( sin cos )- +t t V (B) 9 4 3 5 3( sin cos )t t- V (C) 9 4 3
5 3( sin cos )- -t t V (D) 9 4 3 5 3( sin cos )t t+ V 11. In the
circuit shown in fig. P1.9.11 if current i t1 5 500 20= - cos ( )
mA and i t2 20 500 20= - cos ( ) mA, the total energy stored in
system at t = 0 is (A) 151.14 mJ (B) 45.24 mJ (C) 249.44 mJ (D)
143.46 mJ 12. Leq = ? (A) 4 H (B) 6 H (C) 7 H (D) 0 H 13. Leq = ?
(A) 2 H (B) 4 H (C) 6 H (D) 8 H 14. Leq = ? (A) 8 H (B) 6 H (C) 4 H
(D) 2 H 15. Leq = ? (A) 0.4 H (B) 2 H (C) 1.2 H (D) 6 H 16. The
equivalent inductance of a pair of a coupled inductor in various
configuration are (a) 7 H after series adding connection (b) 1.8 H
after series opposing connection (c) 0.5 H after parallel
connection with dotted terminal connected together. Chap
1.9Magnetically Coupled Circuits Page 83 i1 i2 3 H - + v2 - + v1 3
H 4 H Fig. P1.9.9-10 i1 i2 k=0.6 - + v2 - + v1 0.4 H2.5 H Fig.
P1.9.11 i1 i2 1 H - + v2 - + v1 2 H 2 H Fig. P1.9.7-8 3.6 H 1 H 1.4
H Leq Fig. P1.9.12 4 H 2 H 2 H Leq Fig. P1.9.13 Leq 4 H 4 H 6 H
Fig. P1.9.14 Leq 2 H 4 H 2 H Fig. P1.9.15 GATE EC BY RK Kanodia
www.gatehelp.com
62. The value of L L1 2, and M are (A) 3 H, 1.6 H, 1.2 H (B)
1.6 H, 3 H, 1.4 H (C) 3.7 H, 0.7 H, 1.3 H (D) 2 H, 3 H, 3 H 17. Leq
= ? (A) 0.2 H (B) 1 H (C) 0.4 H (D) 2 H 18. Leq = ? (A) 1 H (B) 2 H
(C) 3 H (D) 4 H 19. In the network of fig. P1.9.19 following
terminal are connected together (i) none (ii) A to B (iii) B to C
(iv) A to C The correct match for equivalent induction seen at
terminal a b- is (i) (ii) (iii) (iv) (A) 1 H 0.875 H 0.6 H 0.75 H
(B) 13 H 0.875 H 0.6 H 0.75 H (C) 13 H 7.375 H 6.6 H 2.4375 H (D) 1
H 7.375 H 6.6 H 2.4375 H 20. Leq = ? (A) 1 H (B) 2 H (C) 3 H (D) 4
H 21. Leq = ? (A) 41 5 H (B) 49 5 H (C) 51 5 H (D) 39 5 H Statement
for Q.22-24: Consider the circuit shown in fig. P1.9.2224. 22. The
voltage VAG of terminal AD is (A) 60 V (B) -60 V (C) 180 V (D) 240
V 23. The voltage vBG of terminal BD is (A) 45 V (B) 33 V (C) 69 V
(D) 105 V 24. The voltage vCG of terminal CD is (A) 30 V (B) 0 V
(C) -36 V (D) 36 V Page 84 UNIT 1 Networks 3 H 5 H 3 H Leq Fig.
P1.9.18 a b A C B 2 H 4 H 1 H 3 H 2 H 5 H Fig. P1.9.19 1 H 1 H 2 H
2 H 3 H Leq Fig. P1.9.20 3 H 2 H 4 H Leq 2 H 3 H Fig. P1.9.21 6 At
20 H 15 At A C B 3 H 5 H D 6 H 4 H Fig. P1.9.2224 2 H 4 H 2 H Leq
Fig. P1.9.17 GATE EC BY RK Kanodia www.gatehelp.com
63. 33. In the circuit of fig. P1.9.33 the w = 2 rad s. The
resonance occurs when C is (A) 1 F (B) 1 2 F (C) 1 3 F (D) 1 6 F
34. In the circuit of fig. P1.9.34, the voltage gain is zero at w =
333.33 rad s. The value of C is (A) 100 mF (B) 75 mF (C) 50 mF (D)
25 mF 35. In the circuit of fig. P1.9.35 at w = 333.33 rad s, the
voltage gain v vout in is zero. The value of C is (A) 3.33 mF (B)
33.33 mF (C) 3.33 mF (D) 33.33 mF 36. The Thevenin equivalent at
terminal ab for the network shown in fig. P1.9.36 is (A) 6 V, 10 W
(B) 6 V, 4 W (C) 0 V, 4 W (D) 0 V, 10 W 37. In the circuit of fig.
P1.9.37 the maximum power delivered to RL is (A) 250 W (B) 200 W
(C) 150 W (D) 100 W 38. The average power delivered to the 8 W load
in the circuit of fig. P1.9.38 is (A) 8 W (B) 1.25 kW (C) 625 kW
(D) 2.50 kW 39. In the circuit of fig. P1.9.39 the ideal source
supplies 1000 W, half of which is delivered to the 100 W load. The
value of a and b are (A) 6, 0.47 (B) 5, 0.89 (C) 0.89, 5 (D) 0.47,
6 40. I2 = ? (A) 1.65 Arms (B) 0.18 Arms (C) 0.66 Arms (D) 5.90
Arms Page 86 UNIT 1 Networks C 2 HZin 4 W2 H 2 H Fig. P1.9.33 0.09
H vin C 2 F - + vout 0.12 H 0.27 H 20 W 40 W ~ Fig. P1.9.34 C k=0.5
0.12 H 0.27 Hvin - + vout 20 W 40 W ~ 20 W Fig. P1.9.35 a b 20 W
20Ix Ix 60 W 1 : 4 Fig. P1.9.36 100Vrms 10 W 1 : 4 RL~ Fig. P1.9.37
-0.04V2 8 W - + V1 I1 I2 50Vrms 300 W 5 : 1 ~ - + V2 Fig. P1.9.38
100 W100Vrms 4 W 1 : a ~ 1 : b 25 W Fig. P1.9.39 3 W I2 50Vrms 3 :
1 ~ 4 : 3 25 W 2 W Fig. P1.9.40 GATE EC BY RK Kanodia
www.gatehelp.com
64. 41. V2 = ? (A) -12.31 V (B) 12.31 V (C) -9.231 V (D) 9.231
V 42. The power being dissipated in 400 W resistor is (A) 3 W (B) 6
W (C) 9 W (D) 12 W 43. Ix = ? (A) 1921 57 4. . A (B) 2 931 59 4. .
A (C) 1 68 43 6. . A (D) 179 43 6. . A 44. Zin = ? (A) 46 3 6 8. .+
j W (B) 432 1 0 96. .+ j W (C) 10 8 9 6. .+ j W (D) 615 4 0 38. .+
j W ******************** SOLUTIONS 1. (B) v di dt di dt di dt 1 1 2
1 2 1 2= + = = 16 2cos t V 2. (C) v di dt di dt di dt 2 2 1 1 1 1=
+ =( ) ( ) = 8 2cos t V 3. (B) v di dt di dt di dt 1 1 2 2 3 3 3= -
= - = - 24 4cos t V 4. (C) v di dt di dt 2 2 1 4 3= - = - = - 3 61
2di dt e t V 5. (A) v di dt di dt di dt 1 1 2 1 2 2 2= - = = -24
4sin t V v di dt di dt di dt t2 2 1 1 3 2 2 24 4= - + = = - sin V
6. (D) v di dt di dt di dt 1 1 2 2 2 2 2= - = - = - 24 3cos t V v
di dt di dt di dt t2 2 1 2 3 2 3 36 3= + = - = - cos V 7. (A) v di
dt di dt 1 1 2 2 1= + = - + = -18 12 6 2 3sin cos ( cos sin )t t t
t V 8. (A) v di dt di dt 2 2 1 2 1= + = - = -24 3 9 3 3 8 3 3 3cos
sin ( cos sin )t t t t V 9. (A) v di dt di dt 1 1 2 3 3= - = +45 3
27 3cos sint t = +9 5 3 3 3( cos sin )t t V 10. (D) v di dt di dt 2
2 1 4 3= - + = +36 3 45 3sin cost t = +9 4 3 5 3( sin cos )t t V
11. (A) W L i L i Mi i= + + 1 2 1 2 1 1 2 2 2 2 1 2 At t = 0, i1 4
20 4 7= - =cos ( ) . mA i2 20 20 18 8= - =cos ( ) . mA , M = =0 6 2
5 0 4 0 6. . . . W = + + 1 2 2 5 4 7 1 2 0 4 18 8 0 6 4 7 18 82 2 (
. )( . ) ( . )( . ) . ( . )( . ) = 151 3. mJ 12. (C) L L L Meq = +
+ =1 2 2 7 H Chap 1.9Magnetically Coupled Circuits Page 87 48 W 400
W10Vrms 1 : 2 ~ 1 : 5 1 W 4 W Fig. P1.9.42 - 4j Ix j6100 0 V o 8 W
10 W ~ 2 : 1 Fig. P1.9.43 j16 6 W Zin 1 : 5 4 : 1 6 W 24 W - 10j
Fig. P1.9.44 50 W 10 W - + V2 80Vrms 40 W 5 : 2 ~ Fig. P1.9.41 GATE
EC BY RK Kanodia www.gatehelp.com
65. 13. (A) L L L Meq = + -1 2 2 = + - =4 2 2 2 2 H 14. (C) L L
L M L L M eq = - + - 1 2 2 1 2 2 = - + - = 24 16 6 4 8 4 H 15. (A)
L L L M L L M eq = - + + 1 2 2 1 2 2 = - + = 8 4 6 4 0 4. H 16. (C)
L L M1 2 2 7+ + = , L L M1 2 2 1 8+ - = . + =L L1 2 4 4. , M = 1 3.
L L M L L M 1 2 2 1 2 2 0 5 - + - = . , L L1 2 2 1 3 0 5 1 8- = . .
. L L1 2 2 59= . , ( ) . .L L1 2 2 2 4 4 4 2 59 9- = - = L L1 2 3-
= , L1 37= . , L2 0 7= . 17. (D) L L M L eq = -1 2 2 = - =4 4 2 2 H
18. (B) L L M L eq = -1 2 2 = - =5 9 3 2 H 19. (A) 20. (D) V sI sI
sIL1 1 1 2= + = V sI sI sI sIL2 2 1 2= + - = , V sI sI sIL3 3 2= -
= V V V V sIL L L L= + + =1 2 3 4 Leq = 4 H 21. (B) Let I1 be the
current through 4 H inductor and I2 and I3 be the current through 3
H, and 2 H inductor respectively I I I1 2 3= + , V V2 3= 3 3 2 22 1
3 1sI sI sI sI+ = + + =3 22 1 3I I I 4 2 3I I= = =I I I I2 1 3 1 5
4 5 , V sI sI sI sI s I= + + + +4 3 2 3 31 2 3 2 1 = + + 7 6 5 2 4
5 1 1 3sI s I s I V sI= 49 5 1 , Leq = 49 5 H 22. (C) v d t dt d t
dt AG = + =20 6 4 15 180 ( ) ( ) V 23. (B) v d t dt d t dt d t dt
BG = + - =3 15 4 6 6 6 33 ( ) ( ) ( ) V 24. (C) v d t dt CG = - =
-6 6 36 ( ) V 25. (B) Z Z M Z = +11 2 2 22 w = + + + 4 50 1 10 50 1
5 5 50 1 2 2 2 j j ( ) ( ) ( ) = +4 77 115. .j W 26. (B) V j js = -
( . ) ( . ) ( . )( )( )0 8 10 12 0 0 2 10 2 0 + + + [ ( . )( )]( .
)3 0 5 10 12 0 2 0j = + = 9 6 21 6 26 64 66 04. . . .j V 27. (A) [
( ) ( ) ] ( )( . ) ( )j I j100 2 10 100 0 4 2 0 02p p+ + = = - -I
j2 0 4 0 0064. . , V I jo = = - -10 4 0 0642 . = - 4 179 1. = - v
to 4 100 179 1cos ( . )p V 28. (B) 30 30 6 8 4 12 4 10 = - + - + -
+I j j j j j( ) = + = -I j j 30 30 10 6 2 57 0 043 ( ) . . V I j jo
= - +( )12 4 10 = - +( . . )( )2 57 0 043 10 8j j = +26 067 20 14.
.j = 32 9 37 7. . V 29. (A) ( )- + + - = -j j I jI j2 4 31 2 ( )j I
jI4 2 12 302 1+ - = - V I j1 1 45 0 56= - -. . , V I jx = - = +2 2
9 1121 . . = 311 2112. . V Page 88 UNIT 1 Networks 3 H 2 H 5 H -1 H
2 H Fig. S.1.9.19 I1 I2 3 -90 A o +- Vx -j j4 j 12 30 V o 2 W 2 W
j4 ~~ Fig. S1.9.29 GATE EC BY RK Kanodia www.gatehelp.com
66. 30. (D) Z j j j j j j j eq = + + + - + + - 10 8 14 10 2 6
14 10 2 6 ( ) ( ) = +112 112. .j W 31. (C) Z j Zin o= -( )||( )6 Z
j j j j o = + + - + 20 12 30 5 2 4 2 ( ) ( ) = +0 52 15 7. .j Z j j
j j in = - + - + + ( )( . . ) ( . . ) 6 0 52 15 7 6 0 52 15 7 = -0
20 9 7. .j W 32. (D) M k L L= 1 2 , M 2 12 160 10= - Z j L M Z j L
in L = + + w w w 1 2 2 2 = + + + - - j j j 250 10 2 10 250 10 160
10 2 10 250 1 3 6 3 2 12 ( ) 0 80 103 6 - Z jin = +0 02 0 17. . W
33. (D) V jI C j I j I1 1 1 2 2 4 2 2= - + + 0 4 4 2 22 1= + +( )j
I j I I j I j 2 12 2 1 = - +( ) V I j C j j 1 1 2 4 2 1 = - + + + =
- + + -j j C C j C C 8 2 2 2 Z j j C C j C C in = - + + -8 2 2 2 Im
( )Zin = 0 - + - =j j C j C8 2 0 C = 1 6 34. (A) j j C 30 3 1000 0-
= , C = 100 mF 35. (D) The p equivalent circuit of coupled coil is
shown in fig. S1.9.35 L L M M L L k k 1 2 2 1 2 2 1- = -( ) = = 0
12 0 27 10 5 0 5 0 27 2 . . ( . . ) . . Output is zero if - + = j
jC 0 27 0 . w w C = = 1 0 27 33 332 . . w mF 36. (C) Applying 1 V
test source at ab terminal, Vab = 1 V, Ix = = 1 20 0 05. A, V2 4= V
, 4 60 20 0 052= + I . I2 0 05= . A I I I I Iin x x= + = + =1 24 0
25. A R I TH in = = 1 4 W , VTH = 0 37. (A) Impedance seen by RL =
=10 4 1602 W For maximum power RL = 160 W, Zo = 10 W PLmax = + =
100 10 10 10 250 2 W 38. (B) I V 2 2 8 = , I I V 1 2 2 5 40 = = , V
V1 25= Chap 1.9Magnetically Coupled Circuits Page 89 I1 I2 j4V1 4
W~ j4 2j2 2C -j Fig. S.1.9.33 0.03 H 0.18 H 2 F - + Vout - 3j Vin
1000C 20 W j30 ~ 40 W 0.09 H Fig. P1.9.34 M L L M1 2 - 2 L L M1 2 -
2 L L M1 2 - 2 M L1 L2 L M2 - L M1 - Fig. S1.9.35 20Ix Ix 60 W 1 V
20 W 1 : 4 Fig. S1.9.36 j10 j18 j20 j10 j10j8 Fig. S.1.9.30 GATE EC
BY RK Kanodia www.gatehelp.com