Gate by Rk Kanodia

GATE EC BY RK Kanodia

MULTIPLE CHOICE QUESTION

Electronics & Communication EngineeringFifth EditionR. K. KanodiaB.Tech.

NODIA & COMAPNYJAIPUR

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CHAPTER

1.1BASIC CONCEPTS1. A solid copper sphere, 10 cm in diameter is deprived of 1020

(A) 1 A (C) 3 A

(B) 2 A (D) 4 A

electrons by a charging scheme. The charge on (B) -160.2 C (D) -16.02 C

the sphere is (A) 160.2 C (C) 16.02 C

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 bev1

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 (C) 1500 mC (B) 75 C (D) 1.5 C20 W

60 V

3. If 120 C of charge passes through an electric conductor in 60 sec, the current in the conductor is (A) 0.5 A (C) 3.33 mA (B) 2 A (D) 0.3 mA (A) 240 V (C) 60 V 4. The energy required to move 120 coulomb through 3 V is (A) 25 mJ (C) 40 J 5. i = ?1A 5A

100 V

Fig. P.1.1.6

(B) 120 V (D) 30 V

7. In the circuit of the fig P1.1.7, the value of the (B) 360 J (D) 2.78 mJ+ +

voltage source E is0V 2V 1V

+

E+

i 2A 3A

4V

5V 10 V

Fig. P.1.1.74A

+

(A) -16 VFig. P.1.1.5

(b) 4 V (D) 16 VPage 3

(C) -6 Vwww.gatehelp.com

UNIT 1


Networks

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+ 105 V + 65 V 100 V 30 V

12. v1 = ?+ v1 1 kW 7V

2

kW

15 V +

10 V + + 55 V + 35 V + 30 V 8V 5V

+ v1

3

kW4 kW

6V

+

Fig. P1.1.12 Fig. P.1.1.8

(A) -30 V (C) -20 V

(B) 25 V (D) 15 V

(A) -11 V (C) 8 V

(B) 5 V (D) 18 V

9. For the circuit shown in fig P.1.1.9 the value of voltage vo is5W+ vo

13. The voltage vo in fig. P1.1.11 is always equal to1A + vo 5V 4W

15 V

1A

Fig. P1.1.11 Fig. P.1.1.9

(A) 1 V (C) 9 V

(B) 5 V (D) None of the above

(A) 10 V (C) 20 V 10. R1 = ?60 W

(B) 15 V (D) None of the above

14. Req = ?5W 10 W 10 W 10 W

Req

+ R1 100 V R2 + 20 V 70 V

10 W

10 W

10 W

up to

Fig. P1.1.14

(A) 11.86 W (C) 25 W 15. vs = ?

(B) 10 W (D) 11.18 W

Fig. P.1.1.10

(A) 25 W (C) 100 W

(B) 50 W (D) 2000 Wvs

180 W + 60 W 40 W 90 W 20 V

11. Twelve 6 W resistor are used as edge to form a cube. The resistance between two diagonally opposite corner of the cube is 5 (A) W 6 (C) 5 W 6 W 5

180 W

(B)

Fig. P.1.1.15

(D) 6 W

(A) 320 V (C) 240 V

(B) 280 V (D) 200 V

Page 4

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UNIT 1


Networks

24. Let i( t) = 3te-100 t A and v( t) = 0.6(0.01 - t) e -100 t V for the network of fig. P.1.1.24. The power being absorbed by the network element at t = 5 ms isi + v N

28. vab = ?

a

2

R 2 8 Ai1 b 0.3i1

W0.2i1

A 2W

Fig. P.1.1.24

(A) 18.4 mW (C) 16.6 mW

(B) 9.2 mW (D) 8.3 mW (A) 15.4 V (C) -2.6 V

Fig. P.1.1.28

(B) 2.6 V (D) 15.4 V

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 is20 A

29. In the circuit of fig. P.1.1.29 power is delivered by500 W ix 40 V 2ix 200 W 400 W

12 V A B C

Fig. P.1.1.29

Fig. P.1.1.25

(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.1.1.305W

(A) 4 (C) 6

(B) 5 (D) 7

26. In the circuit of fig. P.1.1.26, the power absorbed by the load RL isi1

1V

1W

2i1

RL = 2 W

6

W

20 V

v1

5W

v1 5

Fig. P.1.1.26

(A) 2 W (C) 6 W 27. vo = ?+ v1 0.3v1

(B) 4 W (D) 8 W (A) delivers 80 W (C) absorbs 40 W

Fig. P.1.1.30

(B) delivers 40 W (D) absorbs 80 W

31. In the circuit of fig. P.1.1.31 dependent source0.2 A 5W 8W + v2 5v2 18 W + vo ix + 8V

Fig. P.1.1.27

4A

2ix

(A) 6 V (C) -12 V

(B) -6 V (D) 12 V (A) supplies 16 W (C) supplies 32 Wwww.gatehelp.com

Fig. P.1.1.31

(B) absorbs 16 W (D) absorbs 32 W

Page 6

Basic Concepts


Chap 1.1

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 (C) 0.6 mF (B) 1.33 mF (D) 1.67 mF

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 isi(mA) 5

33. The energy required to charge a 10 mF capacitor to 100 V is (A) 0.10 J (C) 5 10-9

4

t(ms)

Fig. P. 1.1.36

(B) 0.05 J J (D) 10 10-9

v

v 50m

J

5

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 isi(mA)v 250m

4

t(ms)

4

t(ms)

(A)v 50m

(B)

4

t(ms)

4

t(ms)

2

t(ms)

(C)

(D)

Fig. P. 1.1.34v 10 10 v

37. Ceq = ?2.5 mF

2 v 0.2 2

4

t(ms)

2 v 0.2

4

t(ms)1.5 mF Ceq 1 mF 2 mF

(A)t(ms)

(B)t(ms)

4

2

4

(C)

(D)Fig. P.1.1.37

35. The voltage across a 100 mF capacitor is shown in fig. P.1.1.35. The waveform for the current in the capacitor isv 6 t(ms)

(A) 3.5 mF (C) 2.4 mF

(B) 1.2 mF (D) 2.6 mF

38. In the circuit shown in fig. P.1.1.381 2 3

iin ( t) = 300 sin 20 t mA, for t 0.C2 + vin C1 C1 C1 C1 60 mF C2 C2 C2

Fig. P.1.1.35i(mA) 6 t(ms) i(mA) 600

iin1 2 3 1 2 3 t(ms)

(A)i(mA) 6 2 1 3 t(ms) i(mA) 600

(B)

Fig. P. 1.1.38

2 1

3

t(ms)

Let C1 = 40 mF and C2 = 30 mF. All capacitors are initially uncharged. The vin ( t) would be (A) -0.25cos 20t V (C) -36cos 20t mV (B) 0.25cos 20t V (D) 36cos 20t mVPage 7

(C)

(D)

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Basic Concepts


Chap 1.1

SOLUTIONS1. (C) n = 10 20 , Q = ne = e10 20 = 16.02 C Charge on sphere will be positive. 2. (D) DQ = i D t = 15000 100m = 15 C . 3. (B) i = dQ 120 = =2 A dt 60

v3 - 30 = v2

v3 = 65 V v4 = 25 V

105 + v4 - v3 - 65 = 0

v4 + 15 - 55 + v1 = 0 v1 = 15 V 9. (B) Voltage is constant because of 15 V source. 10. (C) Voltage across 60 W resistor = 30 V 30 Current = = 0.5 A 60 Voltage across R1 is = 70 - 20 = 50 V 50 = 100 W R1 = 0.5 11. (C) The current i will be distributed in the cube branches symmetrically

4. (B) W = Qv = 360 J 6. (A)1A 5A i=1A 2A 3A 1A 2A

a

i i 3 i 3

i 3

4A

6A

Fig. S 1.1.5

i 6

6. (A) In order for 600 C charge to be delivered to the 100 V source, the current must be anticlockwise. dQ 600 i= = = 10 A dt 60 Applying KVL we get v1 + 60 - 100 = 10 20 or v1 = 240 V 7. (A) Going from 10 V to 0 V0V+ +

i 3

i

b

Fig. S. 1.1.11

vab =

6i 6i 6i + + = 5 i, 3 6 3 v Req = ab = 5 W i

2V

1V

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

+ 4V+ +

E 5V 10 V

13. (D) It is not possible to determine the voltage across 1 A source. 14. (D) Req = 5 + 10 ( Req + 5) 10 + 5 + Req5W 5W

Fig. S 1.1.7

10 + 5 + E + 1 = 0 or E = -16 V 8. (D) 100 = 65 + v2 + 65 V 100 V + v2

+ 105 V

v2 = 35 V+ v4

15 V +

10 V + + 55 VReq 10 W Req

+ v3 V 30

+ v1

+ 30 V

+

Fig. S 1.1.142 Req + 15 Req = 5 Req + 75 + 10 Req + 50

Fig. S 1.1.8

Req = 125 = 1118 W .Page 9

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Basic Concepts


Chap 1.1

vo - 20 vo 20 + = 5 5 5 Power is P = vo

vo = 20 V

20 v1 = 20 = 80 W 5 5

We can say Cd = 20 mF, Ceq = 20 + 40 = 60 mF 1 1 300 cos 20 t 10 -3 = - 0.25 cos 20 t V vC = idt= 60m 20 C 39. (C) iC1 = iin C1 = 0.8 sin 600 t mA C1 + C2

2 31. (D) Power P = vi = 2 ix ix = 2 ix

ix = 4 A, P = 32 W (absorb) 32. (D) vt 2 - vt1 = 1 Ct2

At t = 2 ms, iC1 = 0.75 mA 12 = 1 2m ( t2 - t1 ) C 40. (B) vC1 = 4 vin vin C2 = C1 + C2 6 + 4 vc1 = 0.4 vin Q = 0.1 F V L = 2 mH

t1

idt

12 C = 2m 10

C = 1.67 mF 41. (D) V = 2 + 3 + 5 = 10, Q = 1 C, C = di dt 200m 4m

1 33. (B) E = Cv 2 = 5 10 -6 100 2 = 0.05 J 2 1 34. (D) vc = c2m

10 10 -3 idt = 100 10 -6 (2 10 ) = 0.2 V 0

-3

42. (A) vL = L

100m = L

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 dv 6 -0 = 100 10 -6 = 600 mA dt 10 -3 - 0

43. (B) vL = L

di = 0.01 2( 377 cos 377 t) V dt

= 7.54 cos 377 t V 1 1 12000 120 cos 3t dt = vdt = sin 377 t 0.01 L 377 12000 120 P = vi = (sin 377 t)(cos 377 t) 377 44. (A) i = = 1910 sin 754 t W 45. (D) vL = L vC = 3vL diL dv , iC = C C dt dt d 2 iL = - 9.6 sin 4 t A dt

For 0 < t < 1 , C

For 1 ms < t < 2 ms, dv 0-6 C = 100 10 -6 = - 600 mA ( 3 - 2)m dt 36. (B) For 0 t 4, vC = 1 1 idt = 200 10 -6 C

4m tdt = 3125 t

5m

iC = 3 LC diL dt

2

At t = 4 ms, vc = 0.05 V It will be parabolic path. at t = 0 t-axis will be tangent. 37. (A) 2 mF is in parallel with 1 mF and this combination is in series with 1.5 mF. 15 (2 + 1) . = 1mF, C1 is in parallel with 2.5 mF C1 = 15 + 2 + 1 . . Ceq = 1 + 2.5 = 35 mF 30 60 30(20 + 40) 38. (A) Ca = = 20 mF, Cb = = 20 mF 30 + 60 30 + 20 + 0Cd + iin vin C1 C1 C1 C1 60 mF C2 Cc C2 Cb C2 Ca C2

46. (B) vL = L For 2 < t 4, For 4 < t 8, For 8 < t 10,

-100 - 0 vL = (0.05) = - 2.5 V 2 100 + 100 vL = (0.05) = 2.5 V 4 0 - 100 vL = (0.05) = - 2.5 V 2

Thus (B) is correct option. 47. (C) Algebraic sum of the current entering or leaving a cutset is equal to 0. 6 16 i2 + i4 + i3 = 0 + + i3 = 0 2 4 i3 = - 7 A, v3 = - 7 3 = - 21 V

Fig. S 1.1.38

*********

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Page 11


CHAPTER

1.2GRAPH THEORY1. Consider the following circuits : Non-planner graphs are (A) 1 and 3 (C) 3 only (B) 4 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(1) (2)

chosen tree, would be (A) 2 (C) 4 (B) 3 (D) 5

4. For the graph shown in fig. P.1.1.4 correct set is

(3)

(4)Fig. P.1.1.4

The planner circuits are (A) 1 and 2 (C) 3 and 4 (B) 2 and 3 (D) 4 and 1 (A) (B) 2. Consider the following graphs (C) (D) Node 4 4 5 5 6 6 6 5

Branch

Twigs 4 3 4 4

Link 2 3 2 1

5. A tree of the graph shown in fig. P.1.2.5 isc a b d 2 e f g

(1)

(2)

1

3

4

h

5

Fig. P.1.2.5

(A) a d e h(3) Page 12 (4)

(B) a c f h (D) a e f g

(C) a f h gwww.gatehelp.com

UNIT 1


Networks

1 -1 0 (A) -1 0 1 1 -1 0 -1 -1 0 (C) 0 1 1 1 0 -1

1 0 -1 (B) -1 -1 0 1 1 0 1 -1 0 (D) 0 1 -1 1 -1 0

2

2

1

4

3

1

4

3

13. The incidence matrix of a graph is as given below -1 1 1 0 0 0 0 0 -1 1 1 0 A = 0 -1 0 -1 0 -1 1 0 0 0 -1 -1 The graph is2 2

5

5

(C)

(D)

15. The incidence matrix of a graph is as given below -1 1 1 0 0 0 0 0 -1 1 1 0 A = 0 -1 0 -1 0 0 1 0 0 0 -1 -1 The graph is

4 1 3 1

4 3

1

2

3

1

2

3

(A)2

(B)2

4

4

(A)4 1 3 1 4 3

(B)

1

2

3

1

2

3

(C)

(D)

14. The incidence matrix of a graph is as given below - 1 0 0 - 1 1 0 0 0 0 0 1 0 1 1 A = 0 0 1 0 0 0 -1 0 1 0 0 -1 -1 0 1 -1 -1 0 0 0 0 The graph is2 2

4

4

(C)

(D)

16. The graph of a network is shown in fig. P.1.1.16. The number of possible tree are

1

4

3

1

4

3

Fig. P.1.1.16

(A) 85 5

(B) 12 (D) 20

(C) 16www.gatehelp.com

(A)Page 14

(B)

UNIT 1


Networks

22. The fundamental cut-set matrix of a graph is 1 -1 0 0 0 0 -1 1 0 1 QF = 0 0 0 0 -1 0 0 0 1 -1 0 0 1 0 (C)2 3 5 6 4 6 5

The oriented graph of the network is

(D)

1 3 4

2

1

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 1 1 1 0 BF = 1 0 1 11

(A)

(B)

1 3 4 6

2

1 3 5 4 6

2

3

2

4

5

Fig. P.1.2.24

(C)

(D)

The oriented graph will be

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. 1 1 BF = 0 -1 0 0 0 1 03 2 4

0 1 1

1 0 1

0 0 1

(A)

(B)

1

5

6

Fig. P. 1.2.23

(C)

(D)


25. Consider the graph shown in fig. P.1.2.25 in which twigs are solid line and links are dotted line.

1

4 5 3

2

6

(A)

(B)www.gatehelp.com

Fig. P. 1.2.25

Page 16

Graph Theory


Chap 1.2

A fundamental loop matrix for this tree is given as below 1 BF = 0 0 0 1 0 1 0 -1 -1 0 1 0 1 -1 0 1 0

1 0 (D) 0 -1

1 0 0

0 0 1

0 1 1 0

0 0 1 1

0 0 1 0

0 1 0 0

1 0 0 0

1 -1

0 0 0 1

27. Branch current and loop current relation are expressed in matrix form as i1 0 i 0 2 i3 1 i -1 4 = i5 1 i 0 6 i7 0 i8 0 1 -1 0 -1 0 0 1 0 1 0 0 0 0 0 1 0 0 1 -1 I1 0 I 2 0 I 3 0 I 4 0 1


(A)

(B)

where i j represent branch current and I k loop current. The number of independent node equation are (A) 4 (C) 6 (C) (D) (B) 5 (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 (D) n - 1

26. In the graph shown in fig. P.1.2.26 solid lines are twigs and dotted line are link. The fundamental loop matrix isi a c e

(C) b - n + 1 Statement for Q.2930:

h

b

d

f

Branch current and loop current relation are expressed in matrix form as 1 0 i1 0 0 i -1 -1 -1 0 2 1 0 0 I1 i3 0 i 1 0 0 0 I 4 = 2 i5 0 0 -1 -1 I 3 i 1 1 0 -1 I 4 6 i7 1 0 0 0 i8 0 0 0 1 where i j represent branch current and I k loop current. 29. The rank of incidence matrix is (A) 4 (C) 6 (B) 5 (D) 8

g

Fig. P.1.2.26

1 0 (A) 0 -1

1 0 0 0 1 -1 -1 0 0 0 1 -1 0 -1 0 -1

0 0 1 0

0 1 0 0 0 1 0 0

1 0 0 0 1 0 0 0

0 0 0 1 0 0 0 1 0 0 0 1

-1 1 0 0 0 -1 -1 -1 (B) 0 0 0 -1 1 0 1 0 (C) 1 0 0 1 1 1 0 0 0 1 0 1

0 0 0 0 1 -1 1 0 0 0 1 0

0 0 1 0 1 -1 0 -1

0 -1 1 0 0 0 0 0

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Page 17

UNIT 1


Networks

30. The directed graph will be8 5 6 5 8 6

33. The oriented graph for this network is1 2 3 4 1 2 3 4

2 1 4

3

7

2 1 4

3

7

5

5

1 8 5 6 5 8 6

2

3

4

1

2

3

4

5

5

(C)2 1 4 3 7 2 1 4 3 7

(D)

(C)

(D)

************

31. A network has 8 nodes and 5 independent loops. The number of branches in the network is (A) 11 (C) 8 (B) 12 (D) 6

32. A branch has 6 node and 9 branch. The independent loops are (A) 3 (C) 5 Statement for Q.3334: For a network branch voltage and node voltage relation are expressed in matrix form as follows: 1 v1 1 0 0 v 0 1 0 0 2 1 0 V1 v3 0 0 v 0 0 0 1 V2 4 = v5 1 -1 0 0 V3 v 0 1 -1 0 V4 6 1 -1 v7 0 0 v8 1 0 -1 0 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 (C) 6 (B) 5 (D 7www.gatehelp.com

(B) 4 (D) 6

Page 18

Graph Theory


Chap 1.2

SOLUTIONS1. (A) The circuit 1 and 2 are redrawn as below. 3 and 4 can not be redrawn on a plane without crossing other branch.

7. (D) D is not a tree

(A) (1)Fig. S1.2.1

(B)

(2)

2. (B) Other three circuits can be drawn on plane without crossing

(C) (1) (2)Fig. S .1.2.7

(D)

8. (D) it is obvious from the following figure that 1, 3, and 4 are tree2 2 b c d e f 4 4 3 1 d a c e f b 3

(3)Fig. S1.2.1a 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

(1)2 af

(2)2

5. (C) From fig. it can be seen that a f h g is a tree of given grapha b d e c

b c d e f 3 1

a c d e

b 3

1g

f 4

h

4

Fig. S 1.2.5

(3)2 a b c d e f 4 3

(4)

6. (B) From fig. it can be seen that a d f is a tree.c b e

1 f

a

d

(5) Fig. S. 1.2.6 Fig. S. 1.2.8

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Page 19

UNIT 1


Networks

i a c l2 h b l3 d l4 f l1 e

So independent mesh equation =Number of link. 34. (D) We know that [ vb ] = ArT [ Vn ] So reduced 1 0 0 1 Ar = 0 0 0 0 incidence matrix is 0 0 1 0 0 1 0 0 -1 1 0 0 1 0 0 -1 1 -1 0 1 0 0 -1 0

g

Fig. S 1.2.26

At node-1, three branch leaves so the only option is (D).

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 - 1 = t = 4 30. (B) We know the branch current and loop current are related as [ ib ] = [ B T ] [ I L ] So fundamental loop matrix is 1 1 0 0 -1 0 1 0 0 -1 1 0 0 1 0 0 Bf = 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 = b - 7, 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 = 4Page 22www.gatehelp.com


CHAPTER

1.3METHODS OF ANALYSIS1. v1 = ? (A) 120 V (C) 90 Vvs

6R

3R

(B) -120 V (D) -90 V10 W 4W va 1W 4A 2W

4vs

6R

+ v1

4. va = ?12 V

10 V

Fig. P1.3.1

(A) 0.4vs (C) 0.67vs 2. va = ?3A

(B) 1.5vs (D) 2.5vs

Fig. P1.3.4

(A) 4.33 V (C) 8.67 V2W 3W va 1A

(B) 4.09 V (D) 8.18 V

5. v2 = ?+ 30 W

20 W v2

60 W

0.5 A

10 V

30 W

Fig. P1.3.2

(A) -11 V (C) 3 V 3. v1 = ?10 W 30 V 20 W

(B) 11 V (D) -3 V (A) 0.5 V (C) 1.5 V 6. ib = ?3A 30 W

Fig. P1.3.5

(B) 1.0 V (D) 2.0 V

64 W ib 10 V 36 W

37 W

0.5 A 69 W

+

60 W

v1

9A 6A 60 W

Fig. P1.3.6

Fig. P1.3.3

(A) 0.6 A (C) 0.4 Awww.gatehelp.com

(B) 0.5 A (D) 0.3 APage 23

UNIT 1


Networks

7. i1 = ?6A

(A) 20 mA (C) 10 mA 11. i1 = ?10 W

(B) 15 mA (D) 5 mA

8W

i1

2W

50 W

75 V

4W

3W

5W

100 W 6.6 V i1 40 W 60W

0.1 A

Fig. P1.3.7

(A) 3.3 A (C) 1.7 A 8. i1 = ?

(B) 2.1 A (D) 1.1 A

0.06 A

0.1A90 kW 7.5mA 10 kW i1 90 kW i2 75 V 10 kW

Fig. P1.3.11

(A) 0.01 A (C) 0.03 A

(B) -0.01 A (D) 0.02 A

12. The value of the current measured by the ammeterFig. P1.3.8

in Fig. P1.3.12 is2A

(A) 1 mA (C) 2 mA 9. i1 = ?2A

(B) 1.5 mA (D) 2.5 mA7W

4W

Ammeter

3A4W 3V

6W 5W 2W

Fig. P1.3.124A 3W 2W i1

(A)

2 A 3 5 A 6

(B) (D)

5 A 3 2 A 9

Fig. P1.3.9

(C) -

(A) 4 A (C) 6 A 10. i1 = ?2 kW

(B) 3 A (D) 5 A

13. i1 = ?200 W

40 mA

100 W

i1

50 W

10 mA

45 V

i1

500 W

15 mA

Fig. 1.3.13

(A) 10 mAFig. P1.3.10

(B) -10 mA (D) -0.4 mA

(C) 0.4 mAwww.gatehelp.com

Page 24

Methods of Analysis


Chap 1.34W 8W 5W

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 is6W

12 V

i1

2W

i2 8V

2W

i3

20 V

va

4W

vb

3W

vc

Fig. 1.3.17

12 V

1A

2W

0 i1 12 4 -2 (A) -2 8 -2 i2 = -8 5 i3 20 0 -2 -2 0 i1 12 6 (B) 2 -12 2 i2 = 8 2 -7 i3 20 0 0 i1 12 6 -2 (C) -2 12 -2 i2 = 8 7 i3 20 0 -2 0 i1 12 4 -2 2 -8 (D) 2 i2 = 8 2 -5 i3 20 0

Fig. 1.3.14

(A) 19.8 W (C) 46.9 W 15. i1 , i2 , i3 = ?2W

(B) 27.3 W (D) 54.6 W

3W

i1

9W

15 V

i2

6W

i3

21 V

18. For the circuit shown in Fig. P1.3.18 the mesh equation are6 kW

Fig. P1.3.156 kW i3 6 kW

(A) 3 A, 2 A, and 4 A (C) 1 A, 3 A, and 4 A 16. vo = ?

(B) 3 A, 3 A, and 8 A (D) 1 A, 2 A, and 8 A6V i1 i2 5 mA 6 kW

Fig. 1.3.184 mA 1 kW 2 kW 1 kW + 1 kW 1 mA 2 kW vo 2 mA

6 k -12 k -12 k i1 -6 -6 k i = 0 (A) 6 k -18 k 2 -1k 0 k i3 -1k 5 6 k 12 k -12 k i1 -6 (B) -6 k -6 k 18 k i2 = 0 1k 0 k i3 -1k 5 -6 k -12 k 12 k i1 -6 (C) 6 k -6 k 18 k i2 = 0 1k 0 k i3 1k 5 -6 k 12 k -12 k i1 -6 (D) -6 k 6 k -18 k i2 = 0 1k 0 k i3 -1k 5

Fig. P1.3.16

(A) (C)

6 V 5 6 V 7

(B) (D)

8 V 5 5 V 7

17. The mesh current equation for the circuit in Fig. P1.3.17 are

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Page 25

Methods of Analysis


Chap 1.3

R1

(A) 66.67 mA (C) 23.12 mAR3

(B) 46.24 mA (D) 33.33 mA

R2

i3

i1 v1

29. va = ?R4 25i2 i2 v2

10 W 4A

50 W

va

40 W

Fig. P1.3.2510 A

The value of R4 is (A) 40 (C) 5 26. va = ?2.5 kW 10 kW 10 kW va

200 W

5A

100 W

20 W

20 A

(B) 15 (D) 20 (A) 342 V (C) 198 V 30. ia = ?50 W 150 W

Fig. P1.3.29

(B) 171 V (D) 396 V

20 V

10 kW

5 kW

4 mA

225 W ia

100 W

200 W

Fig. P1.3.26

2V 75 W

4V 50 W

8V

(A) 26 V (C) 13 V 27. v = ?2A

(B) 19 V (D) 18 V (A) 14 mA (C) 7 mA20 W

Fig. P1.3.30

(B) -6.5 mA (D) -21 mA

10 W

v

31. v2 = ?50 W v2 +

4A10 V

100 W

0.04v2

15 W

5W

Fig. P.3.1.27

Fig. P1.3.31

(A) 60 V (C) 30 V 28. i1 = ?300 W

(B) -60 V (D) -30 V

(A) 5 V (C) 3 V 32. i1 = ?2W

(B) 75 V (D) 10 V

0.5i1

4W

40 V

i1

500 W

0.4i1

8V i1

4A

6V

Fig. P1.3.28

Fig. P1.3.32

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Page 27

UNIT 1


Networks

(A) -1.636 A (C) -2.314 A 33. vx = ?

(B) -3.273 A (D) -4.628 A

37. va = ?0.8va

16 A

va

2W

+ 1.6 A 100 W 0.02vx 50 W vx

2.5 W

5W

10 A

Fig. P1.3.37 Fig. P1.3.33

(A) 32 V (C) 12 V 34. ib = ?1 kW va

(B) -32 V (D) -12 V

(A) 25.91 V (C) 51.82 V

(B) -25.91 V (D) -51.82 V

38. For the circuit of Fig. P1.3.38 the value of vs , that will result in v1 = 0, is3 kW ib

3A

2A

0.1v16V 2 kW 4va

10 W + vs 40 W v1

20 W 48 V

Fig. P1.3.34

(A) 4 mA (C) 12 mA 35. vb = ?ia 4 kW vb

(B) -4 mA (D) -12 mA (A) 28 V (C) 14 V 39. i1 , i2 = ?4W2V 2 kW 5ia

Fig. P1.3.38

(B) -28 V (D) -14 V

2ix ix

2W

15 V

i1

6W

i2

18 V

Fig. P1.3.35

(A) 1 V (C) 4 V 36. vx = ?

(B) 1.5 V (D) 6 V (A) 2.6 A, 1.4 A (C) 1.6 A, 1.35 A50 W iy +

Fig. P1.3.39

(B) 2.6 A, -1.4 A (D) 1.2 A, -1.35 A

40. v1 = ?3W + vy 14 V + v1

2A

100 W

25iy

50 W

vx

0.2vx

3W 6W

2A

Fig. P1.3.36

7A 2W

2vy

(A) -3 V (C) 10 VPage 28

(B) 3 V (D) -10 Vwww.gatehelp.com

Fig. P1.3.40

Methods of Analysis


Chap 1.3

(A) 10 V (C) 7 V 41. vx = ? vx +

(B) -10 V (D) -7 V

SOLUTIONS1. (B) Applying the nodal analysis v 4 vs + s 6 R 3R v1 = = 15 vs . 1 1 1 + + 6 R 3R 6 R 2. (C) va = 2( 3 + 1) + 3 (1) = 11 V 3. (D) v1 -v + 1 + 6 =9 60 60 v1 = - 90 V

0.5vx

500 W

500 W 0.6 A 600 W 900 W 0.3 A

Fig. P1.3.41

(A) 9 V (C) 10 V

(B) -9 V (D) -10 V

4. (C)

va - 10 va + =4 4 2 v2 v2 + 10 + = 0.5 20 30

va = 8.67 V

42. The power being dissipated in the 2 W resistor in the circuit of Fig. P1.3.42 is5W ia 2W 2A 2.5 A 4W 6ia 30 V

5. (D)

v2 = 2 V

6. (B) Using Thevenin equivalent and source transform8W 3 i1 2W va 10 W 60 V

3W

25 V

3W

5W

Fig. P1.3.42

Fig. S.1.3.6

(A) 76.4 W (C) 52.5 W 43. i1 = ?500 W + vx 100 W

(B) 305.6 W (D) 210.0 W

25 60 + + 2 15 va = = 15.23 V 3 1 1 + + 14 3 15 25 - 15.23 = 2.09 A i1 = 14 38 3

0.6 A

180 V 400 W

+ vy

7. (A) ib =0.001vy

10 + 0.5 = 0.6 A 64 + 36

100 W

8. (B) 75 = 90 ki1 + 10 k( i1 - 7.5m) 150 = 100 ki1 i1 = 15 mA . i1 = 3 A

0.005vy

9. (B) 3 = 2 i1 + 3( i1 - 4)

Fig. P1.3.43

10. (B) 45 = 2 ki1 + 500 ( i1 + 15m) (A) 0.12 A (C) 0.36 A (B) 0.24 A (D) 0.48 A i1 = 15 mA

11. (D) 6.6 = 50 i1 + 100( i1 + 0.1) + 40( i1 - 0.06) + 60( i1 - 0.1) i1 = 0.02 Awww.gatehelp.com

*****************

Page 29

UNIT 1


Networks

38. (D) If v1 = 0, the dependent source is a short circuit v1 v - vs v1 - 48 + 1 + =2 - 3 40 10 20 48 v - s = - 1 vs = - 14 V 10 203A 2A

ia =

v1 = 0

30 + 7.5 + 2 = 329 A . 12

6 ia = 19.75 V

voltage across 2 W resistor 30 - 19.75 = 10.25 V, P= (10.25) 2 = 52.53 W 2

43. (A) vx = 500 i110 W vs 40 W + v1 20 W 48 V

v y = 400( i1 - 0.001vx ) = 400 ( i1 - 0.5 i1 ) = 200 i1 180 = 500 i1 + 100( i1 - 0.6) + 200 i1 + 100( i1 + 0.005 v y) 180 = 900 i1 - 60 + 100 0.005 200 i1 i1 = 0.12 A

Fig. S1.3.38 ************

39. (D) ix = i1 - i2 15 = 4 i1 - 2( i1 - i2 ) + 6( i1 - i2 ) 8 i1 - 4 i2 = 15 3i1 - 4 i2 = 9 K(i) K(ii) -18 = 2 i2 + 6( i2 - i1 ) . i1 = 12 A, i2 = -1.35 A 40. (B) 14 = 3i1 + v y + 6( i1 - 2 - 7) + 2 v y + 2( i1 - 7) v y = 3( i1 - 2) 14 = 3i1 + 9( i1 - 2) + 6( i1 - 9) + 2( i1 - 7) 14 = 20 i1 - 18 - 54 - 143W + vy 14 V + v1 7A 2vy

i1 = 5 A

v1 = 6(5 - 2 - 7) + 2 3(5 - 2) + 2(5 - 7) = -10 V

3W 6W

2A

i1

2W

Fig. S1.3.40

41. (D) Let i1 and i2 be two loop current 0.5 vx = 500 i1 + 500( i1 - i2 ), vx = -500 i1 5 i1 - 2 i2 = 0 K(i) K(ii) 500( i2 - i1 ) + 900( i2 + 0.3) + 600( i2 - 0.6) = 0 -5 i1 + 20 i2 = 0.9 i1 = 20 mA, vx = -500 20m = -10 V 42. (C) 30 = 5 ia + 3( ia - 2.5) + 4( ia - 2.5 + 2)

Page 32

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CHAPTER

1.4NETWORKS THEOREM1. vTH , RTH = ? 4. A simple equivalent circuit of the 2 terminal network shown in fig. P1.4.4 isR

3W

2W

6V

6W

vTH, RTH

Fig. P.1.4.1

i

v

(A) 2 V, 4 W (C) 4 V, 5 W 2. i N , R N = ?2W

(B) 4 V, 4 W (D) 2 V, 5 W2W

Fig. P.1.4.4

R R v

15 V

4W

iN, RN

(A)

(B)

Fig. P.1.4.2

R R i i

(A) 3 A,

10 W 3

(B) 10 A, 4 W (D) 1.5 A, 4 W2W

(C) 1,5 A, 6 W 3. vTH , RTH = ?

(C)

(D)

5. i N , R N = ?1W vTH, RTH 6A 4W

2W

2A

3W

3W

iN RN

Fig. P.1.4.3

(A) -2 V,

6 W 5

(B) 2 V,

5 W 6

Fig. P.1.4.5

5 (C) 1 V, W 6

6 (D) -1 V, W 5

(A) 4 A, 3 W (C) 2 A, 9 Wwww.gatehelp.com

(B) 2 A, 6 W (D) 4 A, 2 WPage 33

UNIT 1


Networks

6. vTH , RTH = ?30 W 20 W 5V 5A 25 W

The value of the parameter are vTH (A)vTH, RTH

RTH 2 W 2 W 1.2 W 5 W

iN 2 A 2 A30 3 8 5

RN 2 W 3 W 1.2 W 5 W

4 V 4 V 8 V 8 V

(B) (C) (D)

A A

Fig. P.1.4.6

10. v1 = ? (A) -100 V, 75 W (C) 155 V, 37 W 7. RTH = ?6W

(B) 155 V, 55 W (D) 145 V, 75 W8V

2W

1W

1W

3W

2W

6W

+ v1

6W

18 V

2A

6W RTH 5V

Fig. P.1.4.10

(A) 6 V (C) 8 VFig. P.1.4.7

(B) 7 V (D) 10 V

11. i1 = ?4 kW i1 4 kW 20 V 6 kW

(A) 3 W (C) 6 W

(B) 12 W (D) 12 V

8. The Thevenin impedance across the terminals ab of the network shown in fig. P.1.4.8 isa 3W

4 kW

3 kW

24 V

Fig. P.1.4.11

(A) 3 A2A 6W 8W b 2V 8W

(B) 0.75 mA (D) 1.75 mA

(C) 2 mA Statement for Q.1213:

Fig. P.1.4.8

A circuit is given in fig. P.1.4.1213. Find the Thevenin equivalent as given in question..10 W x 16 W y

(A) 2 W (C) 6.16 W

(B) 6 W 4 (D) W 3

5V

40 W

8W

1A

9. For In the the circuit shown in fig. P.1.4.9 a network and its Thevenin and Norton equivalent are given2W 3W RTH 4V 2A vTH iN RN

x

y

Fig. P.1.4.1213

12. As viewed from terminal x and x is (A) 8 V, 6 W (C) 5 V, 32 W (B) 5 V, 6 W (D) 8 V, 32 W

Fig. P.1.4.9

Page 34

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Network Theorems


Chap 1.4

13. As viewed from terminal y and y is (A) 8 V, 32 W (C) 5 V, 6 W (B) 4 V, 32 W (D) 7 V, 6 W

19. vTH , RTH = ?6W i1

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 (C) 30.3 kW Statement for Q.1516: In the circuit of fig. P.1.4.1516 when R = 0 W , the current iR equals 10 A.4W 2W 2W

3i1

4W

iN,

RN

Fig. P1.4.19

(B) 45 kW (D) 40 kW (A) 0 W (C) 2.4 W 20. vTH , RTH = ?4V + 0.1v1 5W v1 vTH RTH

(B) 1.2 W (D) 3.6 W

E

4W

R iR

2W

4A

Fig. P.1.4.1516.

Fig. P.1.4.20

15. The value of R, for which it absorbs maximum power, is (A) 4 W (C) 2 W (B) 3 W (D) None of the above

(A) 8 V, 5 W (C) 4 V, 5 W 21. RTH = ?2W

(B) 8 V, 10 W (D) 4 V, 10 W

3W + vx 4 vx RTH

16. The maximum power will be (A) 50 W (C) 200 W (B) 100 W (D) value of E is required4V

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 (C) 8 W 18. i N , R N = ?10 W i1 20i1 30 W iN, RN

Fig. P.1.4.21

(A) 3 W (C) 5 W

(B) 1.2 W (D) 10 W

(B) 4 W (D) 12 W5W

22. In the circuit shown in fig. P.1.4.22 the effective resistance faced by the voltage source is4W

vs

i 4

i

Fig. P.1.4.22 Fig. P.1.4.18

(A) 2 A, 20 W (C) 0 A, 20 W

(B) 2 A, -20 W (D) 0 A, -20 W

(A) 4 W (C) 2 Wwww.gatehelp.com

(B) 3 W (D) 1 WPage 35

UNIT 1


Networks

23. In the circuit of fig. P1.4.23 the value of RTH at terminal ab is0.75va

26. The value of RL will beix

16 V

0.9 A 2W

3W

RL

8W a 9V 4W va + b

Fig. P.1.4.2627

(A) 2 W (C) 1 W 27. The maximum power is (A) 0.75 W (C) 2.25 W 28. RTH = ?

(B) 3 W (D) None of the above

Fig. P.1.4.23

(A) -3 W (C) 8 W 3

(B)

9 W 8

(B) 1.5 W (D) 1.125 W

(D) None of the above

24. RTH = ?200 W

-2ix

va 100

va +

(A) (C) 3 W 125

25. In the circuit of fig. P.1.4.25, the RL will absorb maximum power if RL is equal to40 W i 6V 200 W 3i 100 W

(A) (C)

400 W 3 800 W 3

Statement for Q.2627: In the circuit shown in fig. P1.4.2627 the maximum power transfer condition is met for the load RL .Page 36www.gatehelp.com

100 W 50 W

RTH

0.01vx 100 W

100 W 300 W ix 800 W

+ vx RTH

Fig. P.1.4.24

(B) 0 (D) 125 W 3 (A) 100 W (C) 200 W

Fig. P.1.4.28

(B) 136.4 W (D) 272.8 W

29. Consider the circuits shown in fig. P.1.4.29ia 2W 6W 2W 6W 2W 12 V 12 V 8V 6W

RL

Fig. P.1.4.25

(B) (D)

2 kW 9 4 kW 92W

ib

2W 6W 6W 2W

18 V

6W

3A

12 V

Fig. P.1.4.29a & b

Network Theorems


Chap 1.4

The relation between ia and ib is (A) ib = ia + 6 (C) ib = 15 ia . 30. Req = ?12 W 4W

33. If vs1 = 6 V and vs 2 = - 6 V then the value of va is (A) 4 V (C) 6 V (B) -4 V (D) -6 V

(B) ib = ia + 2 (D) ib = ia

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, theReq 6W 18 W 9WN

2W 6W

power consumed by R will be

R

N

R

N

Fig. P.1.4.30 Fig. P.1.4.34

(A) 18 W 36 (C) W 13

72 (B) W 13 (D) 9 W

(A) equal to P (C) between P and 4P

(B) less than P (D) more than 4P

31. In the lattice network the value of RL for the maximum power transfer to it is7W6 W

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

RLW 5

simultaneously, then the power consumed by R is (A) P1 P2 (B) P1 P2

9W

(C) ( P1 P2 ) 2 (B) 9 W (D) 8 W

(D) ( P1 P2 ) 2

Fig. P.1.4.31

(A) 6.67 W (C) 6.52 W Statement for Q.3233:

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 (C) 112.5 W (B) 1800 W (D) 228 W results were obtained from

A circuit is shown in fig. P.1.4.3233.12 W 1W 3W 3W + vs1 1W va vs2 1W

37.

The

following

measurements taken between the two terminal of a resistive network Terminal voltage Terminal current 32. If vs1 = vs 2 = 6 V then the value of va is (A) 3 V (C) 6 V (B) 4 V (D) 5 V (C) 0 12 V 0A 0V 1.5 A

Fig. P.1.4.3233

The Thevenin resistance of the network is (A) 16 Wwww.gatehelp.com

(B) 8 W (D) Page 37

UNIT 1


Networks

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 16 200 (A) V, kW 3 3 (C) 18 V, 2 MW 15 32 V, 3 1 MW 15 200 kW 3

SOLUTIONS1. (B) vTH = ( 6)( 6) = 4 V, 3+ 6 | RTH = ( 3| 6) + 2 = 4 W 2. (A)2W 2W isc 15 V 4W

v1

(B)

(D) 36 V,

39. Consider the network shown in fig. P.1.4.39. RN = 2 | 4 + 2 = |+ Linear Network RL vab

Fig. S.1.4.2

10 W, 3

Fig. P.1.4.39

15 2 v1 = =6 V 1 1 1 + + 2 2 4 v isc = i N = 1 = 3 A 2 3. (C) vTH = (2)( 3)(1) = 1 V, 3+ 3 5 = 1| 5 = W | 6

The power absorbed by load resistance RL is shown in table : RL P 10 kW 3.6 MW 30 kW 4.8 MW 4. (B) After killing all source equivalent resistance is R The value of RL , that would absorb maximum power, is (A) 60 kW (C) 300 W (B) 100 W (D) 30 kW6A 4W

RTH

Open circuit voltage = v1 5. (D) The short circuit current across the terminal is2W isc

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 isi(mA)

3W

Fig. S1.4.5+ Resistive Network v a

30 20 10-4 -3 -2 -1 0 1 2

vab b

isc =

6 4 = 4 A = iN , 4+2

| 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

Fig. P.1.4.40

(A) 300 W (C) 100 W

(B) -300 W (D) -100 W

***********

Page 38

www.gatehelp.com

Network Theorems


Chap 1.4

7. (C) After killing the source, RTH = 6 W6W

12. (B) We Thevenized the left side of xx and source transformed right side of yy8W x 16 W y 8W

6W RTH4V 8V

Fig. S.1.4.7

x

y

Fig. S1.4.12

8. (B) After killing all source, | | RTH = 3| 6 + 8 | 8 = 6 W 9. (D) voc = 2 2 + 4 = 8 V = vTH RTH = 2 + 3 = 5 W = R N , iN = vTH 8 = A RTH 5 vxx = vTH

4 8 + = 8 24 = 5 V, 1 1 + 8 24

RTH = 8 | 16 + 8) = 6 W |( 13. (D) Thevenin equivalent seen from terminal yy is v yy = vTH 4 8 + 24 8 = 7 V, = 1 1 + 24 8

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 equivalent1W 1W 1W 2W

| RTH = ( 8 + 16)| 8 = 6 W 14. (A)12 Vi r RL

+ 4V 6W v1

Fig. S1.4.10

4 12 + v1 = 1 + 1 1 + 2 = 6 V 1 1 1 + + 1+1 6 1+2 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 equivalent2 kW i1 4 kW 20 V 2 kW

Fig. S1.4.14

ir 50 = 20 k, r + 50 r = 100 W Pmax =

2

ir 200 = 20 k r + 200

2

( r + 200) 2 = 4( r + 50) 2 i = 30 A, ( 30) 2 100 = 22.5 kW 4

15. (C) Thevenized the circuit across R, RTH = 2 W4W 2W 2W

6V

8V

4W

2W

Fig. S1.4.15 Fig. S1.4.11

i1 =

20 - 6 - 8 = 0.75 mA 2k + 4k + 2k

16. (A) isc = 10 A, RTH = 2 W, 10 Pmax = 2 = 50 W 2 www.gatehelp.com2

Page 39

UNIT 1


Networks

Now in this circuit all straight-through connection have been cut as shown in fig. S1.4.32b6W 1W 3W 2W + va 6V

i 2 R = ( P1 P2 ) 2 36. (C) r = P= voc = 1. 2 W isc

24 2 2 = 112.5 W (1. 2 + 2) 2 voc 12 = =8W isc 15 .

37. (B) RTH =

Fig. S.1.4.32b

va =

6 (2 + 3) =5 V 2 + 3+1

38. (A) Let

1 1 = = 50 mA sensitivity 20 k

33. (B) Since both source have opposite polarity, hence short circuit the all straight-through connection as shown in fig. S.1.4.336W 1W 3W 2W + va 6V

For 0 -10 V scale Rm = 10 20 k = 200 kW For 0 -50 V scale Rm = 50 20 k = 1 MW 4 For 4 V reading i = 50 = 20 mA 10 vTH = 20mRTH + 20m 200 k = 4 + 20mRTH 5 For 5 V reading i = 50m = 5 mA 50 vTH = 5m RTH + 5m 1M = 5 + 5mRTH Solving (i) and (ii) 16 200 V, RTH = vTH = kW 3 3 39. (D) v10 k = 10 k 3.6m = 6 v30 k = 30 k 4.8m = 12 V 6 = 12 = 10 vTH 10 + RTH 30 vTH 30 + RTH 10 vTH = 6 RTH + 60 ...(i)

...(ii)

Fig. S1.4.33

6 ( 6 | 3) | va = = -4 V 2 +1 34. (C) Let Thevenin equivalent of both networkRTH RTH RTH

5 vTH = 2 RTH + 60

vTH

R

vTH

R

vTH

RTH = 30 kW 40. (D) At v = 0 , isc = 30 mA

Fig. S1.4.34

VTH P = R + R R TH V TH P = R + RTH 2 VTH R = 4 2R + R TH 2

2

At i = 0, voc = - 3 V v -3 RTH = oc = = - 100 W isc 30m2

R

************

Thus P < P < 4 P 35. (C) i1 = P1 P2 and i2 = R R P1 R P2 Rwww.gatehelp.com

using superposition i = i1 + i2 =

Page 42


CHAPTER

1.6THE RLC CIRCUITS1. The natural response of an RLC circuit is described by the differential equation d 2v dv dv(0) +2 + v = 0, v(0) = 10, = 0. dt 2 dt dt The v( t) is (A) 10(1 + t) e - t V (C) 10e - t V (B) 10(1 - t) e - t V (D) 10te - t V (A) iL ( t) + 1100 i ( t) + 11 108 iL ( t) = 108 is ( t) L (B) i ( t) + 1100 i ( t) + 11 108 iL ( t) = 108 is ( t) L L (C) (D) . L iL ( t) 11i ( t) . + + 11iL ( t) = is ( t) 108 10 4 i ( t) 11i ( t) L L + + 11iL ( t) = is ( t) 108 10 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(0) dt = -3000 V s. The v( t) for t > 0 is1H

2. The differential equation for the circuit shown in fig. P1.6.2. is2W 1 mH v

vs

100 W

10 mF

vs

80 W

25 mF

+ vC

Fig. P1.6.2

(A) v( t) + 3000 v( t) + 102 108 v( t) = 108 vs ( t) . (B) v( t) + 1000 v ( t) + 102 10 v( t) = 10 vs ( t) .8 8

Fig. P1.6.4

(C) (D)

2 v ( t) v( t) . + + 102 v( t) = vs ( t) 108 10 5 2 v ( t) v( t) . + + 198 v( t) = vs ( t) 8 10 10 5

(A) -2 e -100 t + 8 e -400 t V (C) 6 e -100 t - 8 e -400 t V

(B) 6 e -100 t + 8 e -400 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 > ist=0

3. The differential equation for the circuit shown in fig. P1.6.3 is10 W is 100 W 10 mF

1H

3 4W

1 3F

+ vC

Fig. P.1.6.5iL

(A) 5 e - 7 ewww.gatehelp.com

-t

-3t

V

(B) 7 e - t - 5 e -3t V (D) 3e - t - e -3t V

Fig. P.1.6.3

(C) - e - t + 3e -3t V

Page 54

The RLC Circuits


Chap 1.6

Statement for Q.67: Circuit is shown in fig. P.1.6. Initial conditions are i1 (0) = i2 (0) = 11 Ai1 2H i2

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 is3W t=0

6V

1W

1 2H

1 4F

+ vC

1W

3H

2W

Fig. P1.6.10 Fig. P1.6.67

(A) 4 e -2 t sin 2 t V (C) 4 e -2 t cos 2 t V

(B) -4 e -2 t sin 2 t V (D) -4 e -2 t cos 2 t V

6. i1 (1 s) = ? (A) 0.78 A (C) 2.56 A 7. i2 (1 s) = ? (A) 0.78 A (C) 2.56 A 8. vC ( t) = ? for t > 025 mH

(B) 1.46 A (D) 3.62 A

11. In the circuit of fig. P1.6.23 the switch is opened at t = 0 after long time. The current iL ( t) for t > 0 isiL 4H

(B) 1.46 A (D) 3.62 A2W 1 4F

t=0 8W 4W 7A

Fig. P1.6.11

(A) e -2 t (2 cos t + 4 sin t) A30u(-t) mA 100 W 10 mF + vC

(B) e -2 t ( 3 sin t - 4 cos t) A (D)e -2 t (2 sin t - 4 cos t) A

(C) e -2 t ( -4 sin t + 2 cos t) A Statement for Q.1214:

Fig. P1.6.8

In the circuit shown in fig. P1.6.1214 all initial condition are zero.iL isu(t) A 100 W 65 1 mF 10 mH + vL

(A) 4 e -1000 t - e -2000 t V (C) 2 e -1000 t + e -2000 t V

(B) ( 3 + 6000 t) e -2000 t V (D) ( 3 - 6000 t) e -2000 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 vC ( t) for t > 0 is

Fig. P1.5.12-14

12. If is ( t) = 1 A, then the inductor current iL ( t) is (A) 1 A (B) t A (D) 0 A then iL ( t) is A A (B) 2 t - 3250 A (D) 2 t + 3250 A (C) t + 1 A 13. If is ( t) = 0.5 t A, (A) 0.5 t + 3. 25 10 (C) 0.5 t - 0. 25 10-3 -3

0.8 H + vC

250 W t=0

500 W

5 mF

9V

Fig. P1.6.9

(A) -9 e (B) e (C) e

-400 t

+ 12 e

-300 t

-400 t -300 t

[ 3 cos 300 t + 4 sin 300 t ] [ 3 cos 400 t + 4 sin 300 t ]

14. If is ( t) = 2 e -250 t A then iL ( t) is 4000 -250 t 4000 -250 t (A) A (B) A te e 3 3 (C) 200 -250 t A e 7 (D) 200 -250 t A te 7

(D) e -300 t [ 3 cos 400 t + 2. 25 sin 300 t ]

www.gatehelp.com

Page 55

UNIT 1


Networks

15. The forced response for the capacitor voltage v f ( t) is100 W vx +

19. In the circuit shown in fig. P 1.5.19 v( t) for t > 0 is2u(-t) A

iL 20 mH

avx

50 W

1H

0.04 F + vC

Fig. P1.6.15

4W

2W

(A) 0. 2 t + 117 10 -3 V . (C) 117 10 -3 t - 0. 2 V .

(B) 0. 2 t - 117 10 -3 V . (D) 117 10 -3 t + 0. 2 V .50u(t) V

Fig. P1.6.19

16. For a RLC series circuit R = 20 W , L = 0.6 H, the value of C will be [CD =critically damped, OD =over damped, UD =under damped]. CD (A) C = 6 mF (B) C = 6 mF (C) C >6 mF (D) C < 6 mF OD C >6 mF C < 6 mF C = 6 mF C =6 mF UD C 6 mF C < 6 mF C > 6 mF

(A) 50 - ( 46.5 sin 3t + 62 cos 3t) e -4 t V (B) 50 + ( 46.5 sin 3t + 62 cos 3t) e -4 t V (C) 50 + ( 62 cos 4 t + 46.5 sin 4 t) e -3t V (D) 50 - ( 62 cos 4 t + 46.5 sin 4 t) e -3t 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 is1 16 F + vC 20 V 5W iL 1 4H

17. The circuit shown in fig. P1.6.17 is critically damped. The value of R is

t=0

R

120 W

Fig. P1.6.20

10 mF 4H

(A) -10 sin 8 t A (C) -10 cos 8 t A

(B) 10 sin 8 t A (D) 10 cos 8 t A

21. In the circuit of fig. P1.6.21 switch is moved from 8Fig. P1.6.17

V to 12 V at t = 0. The voltage v( t) for t > 0 is2W t=0 8V 12 V 1H 1 F 6 + vC

(A) 40 W (C) 120 W

(B) 60 W (D) 180 W

18. The step response of an RLC series circuit is given by d 2 i( t) 2 di( t) di(0 + ) = 4. + + 5 i( t) = 10, i(0 + ) = 2, dt dt dt The i( t) is (A) 1 + e - t cos 4 t A (C) 2 + e - t sin 4 t A (B) 4 - 2 e - t cos 4 t A (D) 10 + e - t sin 4 t A

Fig. P1.6.21

(A) 12 - ( 4 cos 2 t + 2 sin 2 t) e - t V (B) 12 - ( 4 cos 2 t + 8 sin 2 t) e - t V (C) 12 + ( 4 cos 2 t + 8 sin 2 t) e - t V (D) 12 + ( 4 cos 2 t + 2 sin 2 t) e - t Vwww.gatehelp.com

Page 56

The RLC Circuits


Chap 1.6

22. In the circuit of fig. P1.5.22 the voltage v( t) is5H 1W

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 is5W 20 W t=0 100 V 5W 1H i 1 F 25

3u(t) A

5W 20 V

0.2 F 2W

+ vC

Fig. P1.6.22

(A) 40 - (20 cos 0.6 t + 15 sin 0.6 t) e -0 .8 t V (B) 35 + (15 cos 0.6 t + 20 sin 0.6 t) e (D) 35 - 15 cos 0.6 t e-0 .8 t -0 .8 t -0 .8 t

Fig. P1.6.25

V V

(A) 0.73e -2 t sin 4.58 t A (B) 0.89 e -2 t sin 6.38 t A (C) 0.73e -4 t sin 4.58 t A (D) 0.89 e -4 t sin 6.38 t A 26. The switch is closed after long time in the circuit of fig. P1.6.26. The v( t) for t > 0 is2A

(C) 35 - (15 cos 0.6 t + 20 sin 0.6 t) e 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 is2A

3 4H 1 3F 5W 10 W

t=0

1W 6W10 W

1H

4V

t=0

1 25 F

+ vC

Fig. P1.6.23

(A) e-2 .306 t + e-0 . 869t A (B) -e -2 .306 t + 2 e -0 . 869t A (C) e -4 .431 t + e -0 .903t A (D) 2 e -4 .431 t - e -0 .903t A 24. In the circuit of fig. P1.6.24 switch is moved from position a to b at t = 0. The iL ( t) for t > 0 is0.02 F b 2 H t=0 2W a iL 6W 12 V 14 W

Fig. P1.6.26

(A) -8 + 6 e -3t sin 4 t V (B) -12 + 4 e -3t cos 4 t V (C) -12 + ( 4 cos 4 t + 3 sin 4 t) e -3t V (D) -12 + ( 4 cos 4 t + 6 sin 4 t) e -3t V 27. i( t) = ?2 kW i 12u(t) V 5 mF 8 mH

Fig. P1.6.27

(A) 6 - ( 6 cos 500 t + 6 sin 5000 t) e -50 t mA4A

(B) 8 - ( 8 cos 500 t + 0.06 sin 5000 t) e -50 t mA (C) 6 - ( 6 cos 5000 t + 0.06 sin 5000 t) e -50 t mA (D) 6 e -50 t sin 5000 t mAwww.gatehelp.com

Fig. P1.6.24

(A) ( 4 - 6 t) e 4 t A (C) ( 3 - 9 t) e -5t A

(B) ( 3 - 6 t) e -4 t A (D) ( 3 - 8 t) e -5t A

Page 57

UNIT 1


Networks

28. In the circuit of fig. P1.6.28 i(0) = 1 A and v(0) = 0. The current i( t) for t > 0 isi 4u(t) A 1H 2W 0.5 F

SOLUTIONS1. (A) s 2 + 2 s + 1 = 0-t

s = -1, - 1,

v( t) = ( A1 + A2 t) e dv(0) v(0) = 10 V, = 0 = - 1 A1 + A2 dt A1 = A2 = 10 2. (A) iL = v dv + 10 10 -6 100 dtiL 2W 1 mH v

Fig. P1.6.28

(A) 4 + 6.38 e -0 .5t A

(B) 4 - 6.38 e -0 .5t A

(C) 4 + ( 3 cos 1.32 t + 113 sin 1.32 t) e -0 .5t A . (D) 4 - ( 3 cos 1.32 t + 113 sin 1.32 t) e -0 .5t A . 29. In the circuit of fig. P1.6.29 a steady state has been established before switch closed. The vo( t) for t > 0 is10 W

vs

100 W

10 mF

t=0 3A 5W 10 mF 1H

+ vo

Fig. S1.6.2

vs = 2 iL + 10 -3

diL +v dt

Fig. P1.6.29

(A) 100 te -10 t V (C) 400 te -50 t V

(B) 200 te -10 t V (D) 800 te -50 t V

2 1 dv dv v -6 d v + 10 -3 = 2 + 10 -6 10 - t 100 dt + 10 10 dt 2 + v dt 100

108 vs ( t) = v( t) + 3000 v( t) + 102 v( t) . t > 0 is vC dvC + iL + 10m 100 dt diL vC = 10 iL + 10 -3 dt diL d di is = 0.1iL + 10 -5 + iL + 10 -5 (10 iL + 10 -3 L ) dt dt dt 3. (C) is = = 0.1iL + 10 -5 diL di d 2 iL + iL + 10 -4 L + 10 -8 dt dt dt 2 . i ( t) 11 L . + iL ( t) + 11iL ( t) = is ( t) 108 10 4 v dv + 25m + ( v - vs ) dt = 0 80 dt

30. In the circuit of fig. P1.6.30 a steady state has been established before switch closed. The i( t) fori 1H 1 4F

1W t=0

2W

6V

Fig. P1.6.30

(A) 2 e

-2 t

sin 2 t A-2 t

(B) -e

-2 t

sin 2 t A

(C) -2(1 - t) e

A

(D) 2(1 - t) e -2 t A

4. (A)

31. In the circuit of fig. P1.6.31 a steady state has been established. The i( t) for t > 0 isi 3A 10 W 6u(t) A 10 mF 40 W 4H

d 2v dv + 500 + 40000 = 0 2 dt dt s = -100, - 400, B = 8, A = -2 1 s + =0 RC LC

s 2 + 500 s + 40000 = 0 v( t) = Ae -100 t + Be -400 t

Fig. P1.6.31

A + B = 6, -100 A - 400 B = -300010 t

(A) 9 + 2 e

-10 t

- 8e

-2 .5t

A-2 .5t

(B) 9 - 8 e A

+ 2e

-2 .5t

A 5. (C) The characteristic equation is s 2 + After putting the values, v( t) = Ae - t + Be -3t ,www.gatehelp.com

(C) 9 + (2 cos 10 t + sin 10 t) e

(D) 9 + (cos 10 t + 2 sin 10 t) e -2 .5t A***************

s2 + 4 s + 3 = 0

Page 58

The RLC Circuits


Chap 1.6

v(0 + ) = 2 V iL (0 + ) = 0 -C

A + B =2 2 8 = , 34 3 dv(0 + ) = - 8, dt

9. (B) vC (0 + ) = 3 V , iL (0 + ) = -12 mA vC dvC + iL + 5 10 -6 =0 250 dt dvC (0 + ) 3 dvC (0 + ) - 12m + 5 10 -6 =0 =0 dt 250 dt 1 s + =0 s2 + -6 250 5 10 0.8 5 10 -6 s 2 + 800 s + 25 10 4 = 0 s = -400 j 300

iR (0) =

dv(0 + ) 8 = dt 3

- A - 3B = -8, B = 3, A = -1 di1 di - 3 2 = 0, dt dt 3di2 di 2 i2 + - 3 1 =0 dt dt 6. (D) i1 + 5 (1 + 5 s) i1 - 3si2 = 0, -3si1 + (2 + 3s) i2 = 0 ( 3s)( 3s) i1 (1 + 5 s) i1 =0 2 + 3s 6 s 2 + 13s + 2 = 0 1 s = - , -2 61 - t 6

vC ( t) = e-400 t ( A1 cos 300 t + A2 sin 300 t) dvC (0) A1 = 3, = -400 A1 + 300 A2 , A2 = 4 dt 10. (B) v(0 + ) = 0, s2 + 4 s + 8 = 0-2 t

iL (0 + ) = 2 A, s = - 2 j2

1 dvC (0 + ) = -2 4 xdt

vC ( t) = e ( A1 cos 2 t + A2 sin 2 t) A1 = 0, dvC (0 + ) = -8 = -2 (0 + 0) + (0 + 2 A2 ), A2 = -4 dt vC (0 + ) = 8 ViL 4H

i1 = A e

+ Be

-2 t

, i(0) = A + B = 11

In differential equation putting t = 0 and solving di1 (0 + ) 33 di2 (0 + ) 143 ==, dt 2 dt 6 A 33 - 2 B = - , A = 3, B = 8, 6 2 i1 = 3et 6

11. (D) iL (0 + ) = -4,

2W

1 4F

+ 8e ,1 6

-2 t

+ vC

8W

i1 (1 s) = 3e

+ 8 e -2 = 3.62 At 6

Fig. S1.6.11

7. (A) i2 = Ce

+ De

-2 t

4

i2 (0) = 11 = C + D, C = -1 i2 = - et 6

di2 (0) -143 C = = - - 2D dt 6 61 6

diL (0 + ) diL (0 + ) = 8 - ( -4) 8 = 10 dt dt 1 s vC + vC + iL = 0, vC = 4 siL + 8 iL 4 2

and D = 12 + 12 e-2 t

s 2 iL + 4 siL + 5 = 0, s = -2 j-

A, i2 (1 s) = e

+ 12 e

-2

= 0.78 A

iL ( t) = e-2 t ( A1 cos t + A2 sin t) A1 = -4, diL (0 + ) = 10 = -2 ( A1 + 0) + A2 , A2 = 2 dt v dv di + 10 -3 + iL , v = 10 10 -3 L 100 65 dt dt

8. (B) vC (0 + ) = 30m 100 = 3 V dvC (0 - ) dvC (0 + ) = iL (0 - ) = 0 = iL (0 + ) = C dt dt 100 1 s2 + s+ 25 10 -3 25 10 -3 10 10 -6 C s = -2000, -2000 vC ( t) = ( A1 + A2 t) e -2000 t dvC ( t) = A2 e -2000 t + ( A1 + A2 t) e -2000 t ( -2000) dt dvC (0) = A2 - 2000 3 = 0 vC (0 + ) = A1 = 3, dt A2 = 6000

12. (A) is = is =

di d 2 iL 65 (10 10 -3) L + 10 -3(10 10 -3) + iL = 0 dt dt 100

d 2 iL di + 650 L + 10 5 iL = 10 5 is dt dt Trying iL ( t) = B 0 + 0 + 10 5 B = 10 5, B = 1, iL = 1 A

13. (A) Trying iL ( t) = At + B, 0 + 650 A + ( At + B)10 5 = 10 5(0.5 t), A = 0.5www.gatehelp.com

Page 59

The RLC Circuits


Chap 1.6

di(0 + ) -16 = = -4.431 A - 0.903B dt 3 A = 1, B = 1 4 6 =3 6 +2 dvC (0) dv (0) = 150 0.02 C = iL (0) = 3 dt dt 6 + 14 1 a= = 5, wo = =5 22 2 0.02 24. (C) vC (0) = 0, iL (0) = a = wo critically damped v( t) = 12 + ( A + Bt) e 0 = 12 + A,-5t

28.(D) i(0 + ) = 1 A, v(0 + ) = a= 1 = 0.5, 2 2 0.5

Ldi(0 + ) dt Wo = 1 1 - 0.5 = 2

s = -0.5 0.5 2 - 2 = 0.5 j1.323 i( t) = 4 + ( A cos 1.32 t + B sin 1.32 t) e -0 .5t 1 = 4 + A, A = -3 di(0) = 0 = 0.5 A + 1.32 B, dt B = -113 .

29. (B) Vo(0 + ) = 0 ,iL (0 + ) = 1 A A = -12,-5t

150 = -5 A + B-5t -5t

B = 90-5t

v( t) = 12 + (90 t - 12) e

iL ( t) = 0.02( -5) e (90 t - 12) + 0.02(90) e 100 5 50 , 25. (A) v(0 ) = = 5 + 5 + 20 3+

= ( 3 - 9 t) e

diL (0 + ) = v1 (0) = 0 dt 1 a= = 10, 2 5 0.01 s = -10, - 10 i( t) = 3( A + Bt) e -10 t , di(0 ) = -10 A + B dt iL ( t) = 3 - (2 + 20 t) e -10 t ,+

Wo =

1 1 0.01

= 10

a = Wo, so critically damped response iL (0 ) = 0+

if = 0 A diL (0 + ) 50 10 = 20 = dt 3 3 4 1 a= = 2, wo = =5 21 1 1 25 s = -2 4 - 25 = -2 j 4.58 i( t) = ( A cos 4.58 t + B sin 4.58 t) e -2 t 26.(A) iL (0 + ) = 0, vL (0 + ) = 4 - 12 = -8 1 dvL (0 ) = iL (0 + ) = 0 25 dt 6 1 a = = 3, Wo = =5 2 1 1 / 25 b = -3 9 - 25 = -3 j 4 v1 ( t) = -12 + ( A cos 4 t + B sin 4 t) e-3t +

i(0) = 1 = 3 + A

vo =

LdiL ( t) = 200 te -10 t dt

di(0 + ) -6 vc (0 + ) = 2 1 = 2 = = -2 A, dt 1+2 1 1 1 a= = = 2, Wo = =2 2 RC 2 1 0.25 LC 30. (C) i(0 + ) = a = Wo, critically damped response s = -2 , -2 A = -2 i( t) = ( A + Bt) e -2 t , di( t) = ( -2 + Bt) e 2 t ( -2) + (0 + B) e -2 t dt At t = 0, B = -2 31. (A) i(0 + ) = 3 A, vC (0 + ) = 0 V = is = 9 A, R = 10| 40 = 8 W | 1 1 a= = = 6.25 2 RC 2 8 0.01 1 1 Wo = = =5 LC 4 0.01 a > Wo, so overdamped response s = -6.25 6.25 2 - 25 = -10, -2.5 i( t) = 9 + Ae -10 t + Be -2 .5t 3 = 9 + A + B, 0 = -10 A - 2.5 B On solving, A = 2, B = -8 ************www.gatehelp.com

vL (0) = -8 = 12 + A, A =4 dvL (0) = 0 = -3 A + 4 B, B=3 dt 27. (C) a = Wo = 1 1 = = 50 2 RC 2 2 k 54 1 1 = = 5000 LC 8 m 5m

4 di(0 + ) dt

a < Wo, underdamped response. s = -50 50 2 - 5000 2 = -50 j5000 i( t) = 6 + ( A cos 5000 t + B sin 5000 t) e -50 t mA i(0) = 6 = 6 + A, A = -6 di(0) = -50 A + 5000 B = 0, B = -0.06 dt

Page 61


CHAPTER

1.7SINUSOIDAL STEADY STATE ANALYSIS1. i( t) = ? (A) (C)25 mH

1 2 1 2

i

3W

cos (2 t - 45 ) V sin (2 t - 45 ) V

(B) (D)

1 2 1 2

cos (2 t + 45 ) V sin (2 t + 45 ) V

20cos 300t V

~Fig. P1.7.1

4. vC ( t) = ?3H

(A) 20 cos ( 300 t + 68. 2 ) A (B) 20 cos( 300 t - 68. 2 ) A (C) 2.48 cos( 300 t + 68. 2 ) A (D) 2.48 cos( 300 t - 68. 2 ) A 2. vC ( t) = ? (A) 2. 25 cos (5 t + 150 ) Vcos 10 t A3

8cos 5t V

~

50 mF 9W

+ vC

Fig. P1.7.4

~

2W

1 mF

+ vC

(B) 2. 25 cos (5 t - 150 ) V (C) 2. 25 cos (5 t + 140.71 ) V (D) 2. 25 cos (5 t - 140.71 ) V 5. i( t) = ?1W 4W i 10cos 2t V

Fig. P1.7.2

(A) 0.89 cos (10 3 t - 63.43 ) V (B) 0.89 cos (10 3 t + 63.43 ) V (C) 0.45 cos (10 t + 26.57 ) V3

(D) 0.45 cos (10 t - 26.57 ) V3

~Fig. P1.7.5

0.25 F

4H

3. vC ( t) = ?5W

cos 2t V

~

0.1 F

+ vC

(A) 2 sin (2 t + 5.77 ) A (C) 2 sin (2 t - 5.77 ) A

(B) cos (2 t - 84. 23 ) A (D) cos (2 t + 84. 23 ) A

Fig. P1.7.3

Page 62

www.gatehelp.com

UNIT 1


Networks

13. In the bridge shown in fig. P1.7.13, Z1 = 300 W, Z 2 = ( 300 - j 600) W, Z 3 = (200 + j 100)W. The Z 4 balance is1 3

Statement for Q.17-18: The circuit is as shown in fig. P1.7.17-181H 1W 1W 1H

at

Z

Z

5cos 4t V2

~

1W i1 1F i2

~

10cos (4t-30 ) V

o

Z

Z

4

Fig. P1.7.1718

17. i1 ( t) = ? (A) 2.36 cos ( 4 t - 4107 ) A . (B) 2.36 cos ( 4 t + 4107 ) A . (C) 1.37 cos ( 4 t - 4107 ) A . (D) 2.36 cos ( 4 t + 4107 ) A . 18. i2 ( t) = ? (A) 2.04 sin ( 4 t + 92.13 ) A (B) -2.04 sin ( 4 t + 2.13 ) A (C) 2.04 cos ( 4 t + 2.13 ) A (D) -2.04 cos ( 4 t + 92.13 ) A 19. I x = ?

Fig. P1.7.13

(A) 400 + j 300 W (C) j100 W

~(B) 400 - j 300 W (D) - j900 W 14. In a two element series circuit, the applied voltage and the resulting current are v( t) = 60 + 66 sin (10 3 t) V, i( t) = 2.3 sin (10 3 t + 68.3 ) A. The nature of the elements would be (A) R - C (C) R - L 15. Vo = ?j20 40 W Vo

(B) L - C (D) R - R

4W

0.5Ix Ix

120-15o V

~

-j30

50 W

Fig. P1.7.15

(A) 223 - 56 V (C) 124 - 154 V 16. vo( t) = ?1H

(B) 22356 V (D) 124 154 V

3W

10sin (t+30o) V

~

1F

+ vo

~

20cos (t-45 ) V

Fig. P1.7.16 Fig. P1.7.20

(A) 315 cos ( t + 112 ) V . (B) 43. 2 cos ( t + 23 ) V (C) 315 cos ( t - 112 ) V . (D) 43. 2 cos ( t - 23 ) VPage 64www.gatehelp.com

~

630 Ao

1030 Vo

~

-j2 W

j3 W

Fig. P1.7.19

(A) 394 46. 28 A . (C) 7.42 92.49 A 20. Vx = ?20 Wo

(B) 4.62 97.38 A (D) 6.78 49. 27 A

j10 W

4Vx

30 Ao

~

20 W

+ Vx

(A) 29.11166 V (C) 43. 24 124 V

(B) 29.11 - 166 V (D) 43. 24 - 124 V

UNIT 1


Networks

Statement for Q.2732: Determine the complex power for hte given values in question. 27. P = 269 W, Q = 150 VAR (capacitive) (A) 150 - j269 VA (C) 269 - j150 VA (B) 150 + j269 VA (D) 269 + j150 VA

35. In the circuit shown in fig. P1.7.35 power factor is4W -j2

-j2

j5

Fig. P1.7.35

28. Q = 2000 VAR, pf = 0.9 (leading) (A) 4129.8 + j2000 VA (C) 2000 - j 4129.8 VA . (B) 2000 + j 4129.8 VA (D) 4129.8 - j2000 VA

(A) 56.31 (leading) (C) 0.555 (lagging)

(B) 56.31 (lagging) (D) 0.555 (leading)

36. The power factor seen by the voltage source is 29. S = 60 VA, Q = 45 VAR (inductive) (A) 39.69 + j 45 VA (C) 45 + j 39.69 VA (B) 39.69 - j 45 VA (D) 45 - j 39.69 VA10cos 2t V 4W + v1 1W

30. Vrms = 220 V, P = 1 kW, |Z |= 40 W (inductive) (A) 1000 - j 68125 VA . (C) 68125 + j1000 VA . (B) 1000 + j 68125 VA . (D) 68125 - j1000 VA . (A) 0.8 (leading) 31. Vrms = 2120 V, Vrms = 2120 V, I rms = 8.5 - 50 A (A) 154.6 + j 89.3 VA (C) 61 + j167.7 VA (B) 154.6 - j 89.3 VA (D) 61 - j167.7 VA

~Fig. P1.7.36

3v 4 1

1 3F

(B) 0.8 (lagging) (D) 39.6 (lagging)

(C) 36.9 (leading)

37. The average power supplied by the dependent source isIx j1.92

32. Vrms = 120 30 V, Z = 40 + j 80 W (A) 72 + j144 VA (C) 144 + j72 VA 33. Vo = ?+

(B) 72 - j144 VA (D) 144 - j72 VA290 Ao

~

4.8 W

1.6Ix

8W

Fig. P1.7.3720 kW 0.8 pf lagging

60 Ao

~

16 kW 0.9 pf lagging

VO

(A) 96 W (C) 92 W

(B) -96 W (D) -192 W

Fig. P1.7.33

38. In the circuit of fig. P1.7.38 the maximum power absorbed by Z L is10 W j15

(A) 7.132. 29 kV (C) 38.49 24.39 kV

(B) 42.59 32.29 kV (D) 38.49 32. 29 kV1200o V

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 (C) 157 VA (B) 275.6 VA (D) 187 VA (A) 180 W (C) 140 WPage 66www.gatehelp.com

~

-j10

ZL

Fig. P1.7.38

(B) 90 W (D) 700 W

Sinusoidal Steady State Analysis


Chap 1.7

39. The value of the load impedance, that would absorbs the maximum average power isj100

The line impedance connecting the source to load is 0.3 + j0.2 W. If the current in a phase of load 1 is I = 10 20 A rms , the current in source in ab branch is (A) 15 - 122 A rms (B) 8.67 - 122 A rms (D) 8.67 - 57.9 A rms sequence 3-phase balanced (C) 15 27.9 A rms 45. An abc phase

320o A

~

80 W

-j40

ZL

Fig. P1.7.39

Y-connected source supplies power to a balanced D connected load. The impedance per phase in the load is 10 + j 8 W. If the line current in a phase is I aA = 28.10 - 28.66 A rms and the line impedance is zero, the load voltage V AB is (A) 207.8 - 140 Vrms (C) 148.3 - 40 Vrms (B) 148.340 Vrms (D) 207.8 40 Vrms

(A) 12.8 - j 49.6 W (C) 339 - j 86.3 W . Statement for Q.4041:

(B) 12.8 + j 49.6 W (D) 339 + j 86.3 W .

In a balanced Yconnected three phase generator Vab = 400 Vrms 40. If phase sequence is abc then phase voltage Va , Vb , and Vc are respectively (A) 2310 , 231120 , 231240 (B) 231 - 30 , 231 - 150 , 23190 (C) 23130 , 231150 , 231 - 90 (D) 23160 , 231180 , 231 - 60 41. If phase sequence is acb then phase voltage are (A) 2310 , 231120 , 231240 (B) 231 - 30 , 231 - 150 , 23190 (C) 23130 , 231150 , 231 - 90 (D) 23160 , 231180 , 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 V AB is (A) 480 45 V (C) 339 45 V (B) 480 - 45 V (D) 339 - 45 V

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 + j10.88 W (C) 432 + j14.6 W . (B) 10.88 + j5.07 W (D) 14.6 + j 432 W .

***********

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 A rms (C) 208 A rms (B) 360 A rms (D) 470 A rms

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 + j 4 W and 10 + j 4 W respectively.www.gatehelp.com

Page 67

UNIT 1


Networks

SOLUTIONS1. (D) Z = 3 + j(25m)( 300) = 3 + j7.5 W = 8.08 68. 2 I= 20 0 = 2.48 - 68. 2 A 8.08 68. 2 i( t) = 2.48 cos ( 300 t - 68. 2 ) A 1 + j(1m)(10 3) = 0.5 + j = 112 63.43 . 2 (10) VC = = 0.89 - 63.43 V 112 63.43 . 2. (A) Y = vC ( t) = 0.89 cos (10 3 t - 63.43 ) V -j 3. (A) Z = 5 + = 5 - j5 = 5 5 - 45 (0.1)(2) VC = (10) (5 - 90 ) 5 2 - 45 1 2 = 1 2 - 45 V

-j 7. (C) Z = w(22m ) | 6 + j(27m) w) |( 27 10 3 j 6 106 - j106 ( 6 + j27 10 -3 w) 22 22 w = 22 w = 6 10 106 6 + j(27mw ) 6 + jw 27m 22 w 22 w2 j27 10 3 - j 36 106 22 w22 106 w 27m 22 w2 =0

w = 1278 1278 w Hz = = 203 Hz f = 2p 2p 8. (C) Vs = 7.68 47 V, V2 = 7.5135 V1 = Vs - V2 = 7.68 47 - 7.5135 = 159 125 . 9. (B) vin = 32 + (14 - 10) 2 = 5 10. (C) I1 = 744 - 118 mA, I 2 = 540 100 mA I = I1 + I 2 = 744 - 118 + 540.5 100 = 460 - 164 i( t) = 460 cos ( 3t - 164 ) mA 11. (A) 2 45 = VC V - 20 0 + C - j4 j5 + 10j50 10 W

vC ( t) =

cos (2 t - 45 ) V -j = 9 + j11 (50m) (5)

4. (D) Z = 9 + j( 3)(5) +

Z = 14.2150.71 W ( 8 0)( 4 - 90 ) VC = = 2. 25 140.71 V 14. 2150.71

vC ( t) = 2. 25 cos (5 t - 140.71 ) V 10 0 10 0 1 5. (B) Va = V = 1 1 1 105 + j0.4 . + + 1 - j2 4 + j 8 I= Va 10 0 = = 1 - 84. 23 A 4 + j 8 1 + j101W Va 4W I 100 Vo

245 Ao

~

+ -j4 VC

~

200 Vo

Fig. S1.7.11

(1 + j)( - j 4)(10 + j5) = VC (10 + j5 - j 4) + j 8 60 - j100 = VC (10 + j) VC = 11.6 - 64.7

~

-j2

j8

12. (D) X = X L + X C = 0Fig. S1.7.5

So reactive power drawn from the source is zero. 13. (B) Z1 Z 4 = Z 3Z 2 300 Z 4 = ( 300 - j 600)(200 + j100) Z 4 = 400 - j 300 14. (A) R - C causes a positive phase shift in voltage Z =|Z |q , -90 < q < 0 , I= V V = -q Z |Z |

i( t) = cos (2 t - 84.23 ) A 6. (D) w = 2 p 10 10 3 = 2 p 10 4 -j 1 Y = j(1m )(2 p 10 4 ) + + 4 (160m )(2 p 10 ) 36 = 0.0278 - j0.0366 S 1 Z= = 1316 + j17.33 W . Y

Page 68

www.gatehelp.com



Chap 1.7

120 15 - 6 30 40 + j20 15. (C) Vo = = 124 - 154 1 1 1 + + 40 + j20 - j 30 50 16. (C) 10 sin ( t + 30 ) = 10 cos ( t - 60 ) 10 - 60 20 - 45 + j 3 Vo = 1 1 1 + + j -j 3 = 30 - 150 +20 - 45 . Vo = 315 - 112 Vj1 3W

20. (B) Let Vo be the voltage across current source Vo - 4 Vx Vo - Vx + =3 20 j10 Vo(20 + j10) - (20 + j 40) Vx = j 600 Vo(20) V Vx = Vo = x (2 + j) 20 + j10 2 (2 + j)(20 + j10) Vx = - 20(1 + j2) = j 600 2 j 600 Vx = = 29. 22 - 166 -5 - j20 j V - V2 21. (A) I1 = V3 + 3 = j0.1V2 + j0.4 V3 j10 2

10-60 Vo

~

+ -j1 W Vo

~

= (0.190 )(0.757 66.7 ) + (0.4 90 )(0.606 - 69.8 )20-45 Vo

I1 = 0.196 35.6 Vo Vo - 3Vo + = 4 - 30 2 j4 Vo = 5.65 - 75

22. (A)Fig. S.1.7.16

j j 17. (C) 5 0 = I1 j 4 + 1 + 1 - - I 2 1 - 4 4 j4 1W 1W j4

Vo(0.5 + j0.5) = 3.46 - j2

23. (D) I 2 = 4 90 , I 3 = 2 0 -j3 j4

50 Vo

~

1W I1 I2 -j0.25 W

~

10-30 Vo

120 Vo

~

+ I1 2W Vo I2

~

490 Vo

Fig. S.1.7.17

2W

2W

( 8 + j15) I1 - ( 4 - j) I 2 = 20 0 j j -10 - 30 = I 2 (1 + j 4 + 1 - ) - I1 (1 - ) 4 4 ( 4 - j) I1 - ( 8 + j15) I 2 = 40 - 30 2 2

...(i)

20o A

Fig. P1.7.23

...(ii) 12 0 = I1 ( - j 3 + 2 + 2) + 8 90 -4 0 . I1 = 352 + j0.64 . Vo = 2( 352 + j0.64 + j 4) = 11.65 52.82 V 24. (D) I 2 = 30 A , I 4 - I 3 = 6 0 AIo

I1 [( 8 + j15) - ( 4 - j) ] = (20 0)( 8 + j15) - ( 40 - 30 )( 4 - j) I1 ( -176 + j248) = 41.43 + j 414.64 . . I1 = 103 - j0.9 = 1.37 - 4107

( 8 + j15)(103 - j0.9) - 20 0 . 18. (B) I 2 = 4-j = -0.076 + j2.04 I 2 = 2.04 92.131590o V

~

~

2W -j4

I2 j2

~

30o A

19. (B) 10 30 = 4 I1 - 0.5 I x + ( - j2) I x ( - j2) I x = ( I1 - I x ) j 3, I1 = 4 10 30 = - 0.5 - j2 I x 3 Ix 3 Ix = 10 30 2.17 - 67.38 www.gatehelp.com1W I3

~

60 A I4o

1W

Fig. S.1.7.24

I 3(1) + ( I 3 - I o)( - j 4) + ( I 4 + I 2 )( j2) + I 4 = 0Page 69

UNIT 1


Networks

I 3 + ( I 3 - I o)( - j 4) + ( I 3 + 6 0 +30 )( j2) + I 3 + 60 = 0 I 3(2 - j2) + I o( j 4) = -18 j - 6 -I ( j2) - 3 - 9 j I3 = o (1 - j) I3 = Io + 3 - j6 j15 = 2 I o + 6 + ( j 4)( 3 - j 6) ( j10)( 8 - j5) = = 9 + j 4.4 8 + j10 - j5 15 90 = ( I o + 30 )(2) + ( I o - I 3)( - j 4)

32. (A) S =

2 |V | (120) 2 = = 72 + j144 VA Z* 40 - j 80

33. (A) S1 = 16 + j S2 = 20 + j

16 sin (cos -1 (0.9)) = 16 + j7.75 0.9

20 sin (cos -1 (0.8)) = 20 + j15 0.8 Vo = 7.132.29

S = S1 + S2 = 36 + j2.75 = 42.59 32.29 S = VoI * = 6 Vo

25. (A) Z TH VTH =

34. (B) Z = 30 + j(0.5)(2 p)(50) = 30 + j157, S=2 |V | (210) 2 = * Z 30 - j157

( 32 0 )( j10) = 339 58 V . 8 + j10 - j5 ...(i)

26 (D) ( 600 - j 300) I1 + j 300 I 2 = 9 300 I 2 = 3V1 , V1 = ( - j 300)( I1 - I 2 ) I 2 = - j 3( I1 - I 2 ) 3I1 = ( 3 + j) I 2

Apparent power =|S |=

(210) 2 30 2 + 152 2

= 275.6 VA

...(ii)

35. (D) Z = 4 +

Solving (i) and (ii) I 2 = 12.36 - 16 mA . Voc = 300 I 2 = 371 - 16 -2 V1 - V1 = 0 V1 = 0 9 0 I sc = = 15 0 mA 600 311 - 16 . V Z TH = oc = = 247 - 16 W I sc 15 0 10 -3 27. (C) S = P - jQ = 269 - j150 VA

( - j2)( j5 - j2) - j2 + j5 - j2

= 4 - j 6 = 7.21 - 56.31, pf = cos 56.31 = 0.555 leading 36. (A) 10 - V1 V1 3 + V1 = 4 4 1 - j15 .4W I1 100 Vo

V1 = 4 36.9 ,1W

+ V1

28. (D) pf = cos q = 0.9 Q = S sin q S=

q = 25.84

~

3V 4 1

-j1.5

Q 2000 = = 4588.6 VA sin q sin 25.84

Fig. S.1.7.36

P = S cos q = 4129.8, S = 4129.8 - j2000 29. (A) Q = S sin q q = 48.59 , sin q = Q 45 or = S 60

I1 = 136.9 (136.9 )(10 0 ) S= = 5 - 36.9 2 pf = cos 36.9 = 0.8 leading 37. (A) (2 - 90 ) 4.8 = - I x ( 4.8 + j192) + 0.6 I x ( 8) .Ix j1.92 Va

P = S cos q = 39.69, S = 39.69 + j 45 VA2 |V | (220 2 ) 30. (B) S = rms = = 1210 |Z | 40

4.8 W 1.6Ix (290o)4.8 V 8W

P 1000 cos q = = = 0.8264 or q = 34.26 , S 1210 Q = S sin q = 68125, . S = 1000 + j 68125 VA .* 31. (C) S = Vrms I rms = (2120 )( 8.5 50 )

~Fig. S.1.7.37

I x = 5 0 , Va = 0.6 5 8 = 24 0 , 1 Pave = 24 1.6 5 = 96 2www.gatehelp.com

= 61 + j167.7 VAPage 70



Chap 1.7

38. (A) Z TH = VTH =

( - j10)(10 + j15) = 8 - j14 W 10 + j15 - j10

45. (D) I AB =

I aA 3

( q + 30 ) = 16.22 1.34 A rms

120( - j10) = 107.3 - 116.6 V 10 + j5

V AB = I AB Z D = (16. 22 1.340 )(10 + j 8) = 207.8 40 Vrms 46. (B) |S |= 3VL I L ZY = 208 10 3 IL = 3600 208 3 = 10 A rms

107.3 - 116.6 IL = = 6.7 - 116.6 16 1 PLmax = ( 6.7) 2 8 = 180 W 2 39. (B) Z TH = ( - j 40)( 80 + j100) = 12.8 - j 49.6 W 80 + j 60 - 30 = 231 - 30 V

25 = 12 25 = 10.88 + j5.07 W

40. (B) Va =

400 3

********

Vb = 231 - 150 V, Vc = 231 - 270 V 41. (C) For the acb sequence Vab = Va - Vb = Vp 0 - Vp 120 1 3 400 = Vp 1 + - j 2 2 400 Vp = 30 3 = Vp 3 - 30

Va = Vp 0 = 23130 V, Vb = Vp 120 = 231150 V Vc = Vp 240 = 231 - 90 V 42. (B) V A = 277 ( 45 -120 ) = 277 - 75 V VB = 277 ( 45 + 120 ) = 277 165 V V AB = V A - VB = 480 - 45 V 43. (C) Z A = 6 | 12 = 4, | IP = 480 = 120 A rms 4

I L = 3I P = 208 A rms 44 (B) I = I aA (10 + j 4) = 10 20 (10 + j 4) + ( 4 + j 4)a IaA

Iac Ibc

Iab IbB b icC

c

Fig. S.1.7.44

I aA = 15 - 27.9 A rms |I | I ab = - aA ( q + 30 ) = 8.67 - 122.1 A rms 3www.gatehelp.com

Page 71


CHAPTER

1.8CIRCUIT ANALYSIS IN THE S-DOMAIN1. Z ( s) = ?1F 2H

(A)

s2 + 1 s2 + 2 s + 1 2 s2 + 1 (C) 2 s + 2s + 2

(B)

2( s 2 + 1) ( s + 1) 2 s2 + 1 (D) 3s + 2

Z(s)

1W

1W

4. Z ( s) = ?

1W

Fig. P1.8.1

s 2 + 15 s + 1 . (A) s( s + 1) 2 s 2 + 3s + 2 (C) s( s + 1) 2. Z ( s) = ?1F Z(s) 1W

s 2 + 3s + 1 s( s + 1) 2 s 2 + 3s + 1 (D) 2 s( s + 1) (B) (A)

2

Z(s)

W

1H

1W

0.5 F

Fig. P1.8.4

1W

1H

3s 2 + 8 s + 7 s(5 s + 6) 3s 2 + 7 s + 6 (C) s(5 s + 6)

s(5 s + 6) 3s 2 + 8 s + 7 s(5 s + 6) (D) 3s 2 + 7 s + 6 (B)

5. The s-domain equivalent of the circuit of Fig.P1.8.5. isFig. P1.8.2t=02

3W

s + s+1 s( s + 1) s( s + 1) (C) 2 s2 + s + 12

(A)

(B)

2s + s + 1 s( s + 1) s( s + 1) (D) 2 s + s+13W

6V

3F

+ vC

3. Z ( s) = ?

Fig. P1.8.53W +1H Z(s) 1F 2W

1 3sVC(s)

1 3s

6 V s

+ VC(s) -

2A

-

Fig. P1.8.3

(A) (C) Both A and Bwww.gatehelp.com

(B) (D) None of these

Page 72

Circuit Analysis in the s-Domain


Chap 1.8

6. The sdomain equivalent of the circuit shown in Fig. P1.8.6 ist=0

9. For the network shown in fig. P1.8.9 voltage ratio transfer function G12 is1H 1H

+ 2A 12 W 2H vL v1 1F

1F

1F + v2 -

1F

1F

Fig. P1.8.6 Fig. P1.8.9

(A)+ 2s 12 W VL 4V -

(B)+ 2A s

(A) (C)

( s 2 + 2) 5 s + 5 s2 + 14

(B) (D)

s2 + 1 5 s + 5 s2 + 14

12 W

2s

VL -

( s 2 + 2) 2 5 s4 + 5 s2 + 1

( s 2 + 1) 2 5 s4 + 5 s2 + 1

10. For the network shown in fig. P1.8.10, the admittance transfer function is Y12 = K ( s + 1) ( s + 2)( s + 4)

(C) Both A and B Statement for Q.7-8:

(D) None of these

The circuit is as shown in fig. P1.8.78. Solve the problem and choose correct option.+

3 W 2

i1

1W

i2 +

is

1W

1H

io +

v1 -

2F3

2F

6

1W

v2 -

vs

1F

1F

1W

vo

Fig. P1.8.10

Fig. P1.8.78

The value of K is (A) -3 (C) 1 3 (B) 3 (D) 1 3

V ( s) 7. H1 ( s) = o =? Vs ( s) (A) s( s3 + 2 s2 + 3s + 1) -1 (B) ( s 3 + 3s 2 + 2 s + 1) -1 (C) ( s 3 + 2 s 2 + 3s + 2) -1 (D) s( s 3 + 3s 2 + 2 s2 + 2) -1 8. H 2 ( s) = (A) I o( s) =? Vs ( s)2

11. In the circuit of fig. P1.8.11 the switch is in position 1 for a long time and thrown to position 2 at t = 0. The equation for the loop currents I1 ( s) and I 2 ( s) are1 2 12 V t=0 i1 3H 2W i2 1F 1F

-s ( s + 3s + 2 s + 1)3

(B) -( s 3 + 3s 2 + 2 s + 1) -1 (D) ( s 3 + 2 s 2 + 3s + 2) -1

(C)

-s 3 2 ( s + 2 s + 3s + 1)

Fig. P1.8.11

1 2 + 3s + s (A) -3s

- 3s 1 2+ s

12 I1 ( s) s I ( s) = 2 0

www.gatehelp.com

Page 73

UNIT 1


Networks

5V

1V

10 V

2F

+ 6V -

1 -3s 2 + 3s + s (C) 2 + 3s + -3s 1 -3s 2 + 3s + s (D) 2 + 3s + -3s

12 I1 ( s) - s = 1 I 2 ( s) 0 s 12 I1 ( s) s = 1 I 2 ( s) 0 s

Fig. P1.8.14

(A)

9 t

V

(B) 9e - t V (D) 0 V

(C) 9 V

15. A unit step current of 1 A is applied to a network whose driving point impedance is V ( s) ( s + 3) = I ( s) ( s + 2) 2

12. In the circuit of fig. P1.8.12 at terminal ab Thevenin equivalent is1s a + Vo(s) -

Z ( s) =

The steady state and initial values of the voltage developed across the source would be respectively (A)2Vo(s)3 4 3 4

(s+1)

4 A

V, I V V, 0 V

(B)

2W

1 4

V,

(C)b

(D) 1 V,

16. In the circuit of Fig. P1.8.16 i(0) = 1 A, vC (0) = 8 V and v1 = 2 e -2 10 t u( t). The i( t) isi 50 W 1m H4

Fig. P1.8.12

(A) VTH ( s) = (B) VTH ( s) = (C) VTH ( s) =

-8( s + 2) -(2 s + 1) , Z TH ( s) = 3s( s + 1) 3s 8( s + 2) (2 s + 1) , Z TH ( s) = 3s( s + 1) 3s 4( s + 3) (2 s + 1) , Z TH ( s) = 3s( s + 1) 6s (A) (B)1 15 1 15

v1

2.5 mF

Fig. P1.8.16

-4( s + 3) -(2 s + 1) (D) VTH ( s) = , Z TH ( s) = 3s( s + 1) 6s 13. In the circuit of fig. P1.8.13 just before the closing of switch at t = 0, the initial conditions are known to be vC1 (0 ) = 1 V, vC2 (0 ) = 0. The voltage vC1 ( t) ist=0 1F 1F + vC2 -

[10 e-10

4

t4

- 3e-2 10t

4

t

- 22 e-4 10 t ]u( t) A4

4

[ -10 e -104

+ 3e -2 104

t

+ 22 e -4 10 t ]u( t) A4

4

(C) 1 [10 e-10 3

t4

+ 3e-2 10t

t4

+ 22 e-4 10 t ]u( t) At

(D) 1 [ -10 e -10 3

+ 3e -2 10

- 22 e -4 10 t ]u( t) A

4

17. In the circuit shown in Fig. P1.8.18 v(0 - ) = 8 V and+ vC1 -

iin ( t) = 4d( t). The vC ( t) for t 0 is

iin

50 W

20 mF

Fig. P1.8.13

(A) u( t) V (C) 0.5 e-t

(B) 0.5 u( t) V V (D) e - t V-

Fig. P1.8.17

(A) 164e- t V (C) 208(1 - e -3t ) V

(B) 208e- t V (D) 164 e -3t V

14. The initial condition at t = 0 of a switched capacitor circuit are shown in Fig. P1.8.14. Switch S1 and S2 are closed at t = 0. The voltage va ( t) for t > 0 isPage 74

www.gatehelp.com

5V3 4 3 4

+

+

-

1 2 + 3s + s (B) -3s

- 3s 1 2+ s

12 I1 ( s) - s I ( s) = 2 0

S1

3F

Va

4F

S2

V V

+ vC

+ vC



Chap 1.8

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) isjw 1

The steady state voltage across capacitor is (A) 6 V (C) 23. The transformed (B) 0 V (D) 2 V voltage across the 60 mF

-3

-1 -1

s

capacitor is given by VC ( s) = 20 s + 6 (10 s + 3)( s + 4) (B) - 0.12 mA (D) - 0.48 mA

Fig. P1.8.18

The initial current through capacitor is 2( s + 3) 2 s + 2s + 2 (A) 0.12 mA (C) 0.48 mA

(A)

4( s + 3) s2 + s + 1

(B)

2( s + 3) (C) 2 s + 2s + 2 Statement for Q.19-21:

4( s + 3) (D) 2 s + s+2

24. The current through an 4 H inductor is given by I L ( s) = 10 s( s + 2)

The circuit is as shown in the fig. P1.8.1921. All initial conditions are zero.io

The initial voltage across inductor is (A) 40 V (C) 10 V (B) 20 V (D) 5 V

iin

1H

1W

1F

1W

25. The amplifier network shown in fig. P1.8.25 is stable if4W 1F 1H

Fig. P1.8.1921

19. (A)

I o( s) =? I in ( s) ( s + 1) 2s (B) 2 s( s + 1) -1 (D) s( s + 1) -1 (A) K 3 (C) K 1 3

2W

+ + v1 Amplifier v 2 gain=K -

(C) ( s + 1) s -1

Fig.P1.8.25

(B) K 3 (D) K 1 3

20. If iin ( t) = 4d( t) then io( t) will be (A) 4d( t) - e- t u( t) A (B) 4 d( t) - 4 e- t u( t) A (C) 4 e - t u( t) - 4 d( t) A (D) e - t u( t) - d( t) A 21. If iin ( t) = tu( t) then io( t) will be (A) e - t u( t) A (C) u( t) A (B) (1 - e - t ) u( t) A (D) (2 - e - t ) u( t) A

26. The network shown in fig. P1.8.26 is stable if1W 2F + Kv2 1F 1W v2 -

Fig.P1.8.26

22. The voltage across 200 mF capacitor is given by VC ( s) = 2s + 6 s( s + 3)

5 2 2 (C) K 5 (A) K

5 2 2 (D) K 5 (B) K

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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 -4 t . The K will be a (A) H 1 - 4 a (C) H 4 - 4 (H= scale factor) a (B) H 1 + 4 a (D) H 4 + 4

31. The current ratio transfer function (A) s( s + 4) s + 3s + 42

Io is Is

(B)

s( s + 4) ( s + 1)( s + 3) ( s + 1)( s + 3) s( s + 4)

(C)

s 2 + 3s + 4 s( s + 4)

(D)

32. The response is (A) Over damped (C) Critically damped (B) Under damped (D) cant be determined

28. A circuit has input vin ( t) = cos 2 t u( t) V and output io( t) = 2 sin 2 t u( t) A. The circuit had no internal stored energy at t = 0. The admittance transfer function is 2 s (A) (B) s 2 (C) s (D) 1 s

33. If input is is 2u( t) A, the output current io is (A) (2 e - t - 3te -3t ) u( t) A (C) ( 3e - t - e -3t ) u( t) A (B) ( 3te - t - e -3t ) u( t) A (D) ( e -3t - 3e - t ) u( t) A

34. In the network of Fig. P1.8.34, all initial condition are zero. The damping exhibited by the network is

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 j 2 Z(0) = 1 the value of R, L, C are 1 (A) 3 W, 3 H, F 3 1 (C) 1 W, 2 H, F 2 1 (B) 2 W, 2 H, F 2Fig. P1.8.343 2

1F4

2H

and zero at -1. Ifvs 2W

+ vo -

(D) 1 W, 1 H, 1 F

(A) Over damped (B) Under damped

30. The current response of a network to a unit step input is 10( s + 2) Io = 2 s ( s + 11s + 30) The response is (A) Under damped (C) Critically damped (B) Over damped (D) None of the above

(C) Critically damped (D) value of voltage is requires 35. The voltage response of a network to a unit step input is Vo( s) = The response is 10 s( s 2 + 8 s + 16)

Statement for Q.31-33: The circuit is shown in fig. P1.8.31-33.

(A) under damped (C) critically damped

(B) over damped (D) cant be determined

36. The response of an initially relaxed circuit to a1H is 4W io

signal vs is e -2 t u( t). If the signal is changed to vs + , the response would be

(

2 dvs dt

)

1F3

(A) 5 e -2 t u( t) (C) 4 e -2 t u( t)

(B) -3e -2 t u( t) (D) -4 e -2 t u( t)

Fig. P1.8.31-33 Page 76

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Chap 1.8

37. Consider the following statements in the circuit shown in fig. P1.8.37i 2H 4W

42. The network function

s2 + 7 s + 6 is a s+2 (B) RL admittance (D) LC admittance

(A) RL impedance function (C) LC impedance function

10 V

1W

2

1F

+ vC

43. A valid immittance function is ( s + 4)( s + 8) s( s + 1) (A) (B) ( s + 2)( s - 5) ( s + 2)( s + 5) (C) s( s + 2)( s + 3) ( s + 1)( s + 4) (D) s( s + 2)( s + 6) ( s + 1)( s + 4)

Fig. P1.8.37

1. It is a first order circuit with steady state value of 10 5 , i= A vC = 3 3 2. It is a second order circuit with steady state of vC = 2 V , i = 2 A 3. The network function 4. The network function V ( s) has one pole. I ( s) V ( s) has two poles. I ( s)

44. The network function (A) RLadmittance (C) LC admittance

s 2 + 8 s + 15 is a s2 + 6 s + 8 (B) RC admittance (D) Above all

45. A impedance function is given as Z ( s) = 3( s + 2)( s + 4) s( s + 3)

The true statements are (A) 1 and 3 (C) 2 and 3 38. The network function (A) RC admittance (C) LC impedance 39. The network function an (A) RC impedance (C) LC impedance (B) 1 and 4 (D) 2 and 4 s 2 + 10 s + 24 represent a s 2 + 8 s + 15 (B) RL impedance (D) None of the above s( s + 4) represents ( s + 1)( s + 2)( s + 3) (B) RL impedance (D) None of these3 1F3W

The network for this function is1F3

1W3W 8F

3

1W

8

1F

1F

(A)

(B)

1H 3

1W

8

1F3W

1W

8

1H3W

s( 3s + 8) 40. The network function represents an ( s + 1)( s + 3) (A) RL admittance (C) RC admittance 41. The network function (B) RC impedance (D) None of the above ( s + 1)( s + 4) is a s( s + 2)( s + 5) (C)

(D)

************

(A) RL impedance function (B) RC impedance function (C) LC impedance function (D) Above all

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Networks

18. (B) Z ( s) = 3K Z (0) = =3 2

K ( s + 3) K ( s + 3) = 2 ( s - ( -1 + j))( s - (1 - j)) s + 2 s + 2 K =2 s s+1

26. (B) Let v1 be the node voltage of middle node V1 ( s) = KV2 ( s) + 2 sV2 ( s) 1 + 2s + s

I ( s) 19. (D) o = I in ( s)

s 1 + s+1 s+1

=

s s+1

( 3s + 1) V1 ( s) = (2 s + K ) V2 ( s) 2 sV1 ( s) V2 ( s) = 2s + 1 (2 s + 1) V2 ( s) = 2 sV1 ( s) ( 3s + 1)(2 s + 1) = 2 s(2 s + K ) 2 s 2 + (5 - 2 K ) s + 1 = 0, 5 5 - 2 K > 0, K < 2 27. (A) H ( s) = H ( s + a) s+4

20. (B) I in ( s) = 4 I o( s) = 4s 4 =4s+1 s+1 io( t) = 4 d( t) - 4 e - t u( t)

1 , s2 1 1 1 I o( s) = = s( s + 1) s s + 1 21. (B) I in ( s) = io( t) = u( t) - e - t u( t) = (1 - e - t ) u( t) 22. (D) vC ( ) = lim sVC ( s) = lims 0 s 0

a H 1 - H ( s + a) Ha 4 R( s) = = + s( s + 4) 4s s+4 r ( t) = Ha a u( t) + H 1 - e - 4 t 4 4 I o( s) 2 s 2 , I o( s) = 2 , = s2 + 1 s + 1 Vin ( s) s

2s + 6 =2 V s+3

28. (A) Vin ( s) = 23. (D) vC (0 + ) = lim sVC ( s)=s

s(20 s + 6) =2 V (10 s + 3)( s + 4)+

CdvC iC = dt

I C ( s) = C[ sVC ( s) - vC (0 )]

s(20 s + 6) -480 10 -6 (10 s + 3) = 60 10 -6 (10 s + 3)( s + 4) - 2 = 10 s 2 + 43s + 12 iC (0 + ) = lim sI C ( s) = - 480 10 -6 = - 0.48 mAs

R 1 1 s + C L sC = 29. (D) Z ( s) = R 1 1 s2 + sL + R + + L LC sC K ( s + 1) K ( s + 1) Z ( s) = = 2 ( s + s + 1) 1 3 1 3 s + + j s + - j 2 2 2 2 ( sL + R)

24. (A) vL = L

d iL VL ( s) = L [ sI L ( s) - iL (0 + )] dt 10 iL (0 + ) = lim sI L ( s) = =0 s s+2 VL ( s) = 40 s 40 = s( s + 2) s + 2s

sL Z(s) R

1Cs

vL (0 + ) = lim sVL ( s) =

s 40 = 40 s+2

Fig. S1.8.29

25. (A) V2 ( s) = KV1 ( s) V1 ( s) V1 ( s) - KV1 ( s) + =0 1 2 4+s+ s 1 4 + s + + 2 - 2K = 0 s s2 + ( 6 - 2 K ) s + 1 = 0 K 0

Since Z (0) = 1, thus K = 1 1 R 1 = 1, = 1, =1 C L LC C = 1, L = 1, R = 1

30. (B) The characteristic equation is s 2 ( s 2 + 11s + 30) = 0 s 2 ( s + 6) ( s + 5) =0 s = -6, - 5, Being real and unequal, it is overdamped.

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Chap 1.8

31. (B)

Io s+4 s( s + 4) = = 3 ( s + 1)( s + 3) Is s + 4 + s

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 valid RC admittance function. Because of (D) option it is required to check that it is a admittance function. The poles and zeros interlace along the negative real axis. The residues of YRC ( s) are real and positive. s 41. (B) The singularity nearest to origin is a pole. So it may be RC impedance or RL admittance function. 42. (A) s 2 + 7 s + 6 ( s + 1)( s + 6) = s+2 ( s + 2)

32. (A) The characteristic equation is ( s + 1) ( s + 3) = 0. Being real and unequal root, it is overdamped response. 33. (C) is = 2 u( t) I o ( s) = 2 I s ( s) = s

2( s + 4) 3 1 = ( s + 1)( s + 3) s+1 s+ 3

io = ( 3e- t - e-3t ) u( t) A V ( s) 2 1 34. (B) o = = 2 4 Vs ( s) + 2s + 2 s + s + 2 s The roots are imaginary so network is underdamped. 35. (C) The characteristic equation is s( s 2 + 8 s + 16) = 0, ( s + 4) 2 = 0, s = -4, - 4 Being real and repeated root, it is critically damped. 36. (B) vo = e -2 t u( t) v = vs + s 2 dvs dt Vo( s) = H ( s) Vs ( s) = 1 s+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 2 + 8 s + 15 ( s + 3) ( s + 5) = s 2 + 6 s + 8 ( s + 2) ( s + 4)

Vs( s) = (1 + 2 s) Vs ( s)

Vo( s) = H ( s) Vs( s) = (1 + 2 s) Vs ( s) H ( s) Vo( s) = 1 + 2s 3 =2 s+2 s+2 v = 2 d( s) - 3e -2 t u( t) o

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. 8 1 8 13 Z ( s) = 3 + + =3+ + s s+3 s 1+ s 3**************

37. (C) It is a second order circuit. In steady state i= 10 =2 A , v =2 1 =2 V 4+1 10 2s + 4 + 1 1 1+ s 2 = 5( s + 2) ( s + 2) 2 + 1

I ( s) =

V ( s) =

10 1 1+ s 2 1 (2 s + 4) + 1 1+ s 2

=

10 ( s + 2) 2 + 1

V ( s) 2 , It has one pole at s = -2 = I ( s) s + 2 38. (D) s 2 + 10 s + 24 ( s + 4)( s + 6) = s 2 + 8 s + 15 ( s + 3)( s + 5)

The singularity near to origin is pole. So it may be RC impedance or RL admittance function.www.gatehelp.comPage 81


CHAPTER

1.9MAGNETICALLY COUPLED CIRCUITSStatement for Q.1-2: In the circuit of fig. P1.9.1-2 i1 = 4 sin 2 t A, and i2 = 0.i1 + v1 2H 1H i2 + 1H v2 +

4. If i1 = e -2 t V and i2 = 0, the voltage v2 is (A) -6 e -2 t V (C) 15 e -2 t V . Statement for Q.5-6: Consider the circuit shown in fig. P19.5-6i1 2H i2 + 3H v2 -

(B) 6 e -2 t V (D) -15 e -2 t V .

Fig. P1.9.1-2v1 2H

1. v1 = ? (A) -16 cos 2 t V (C) 4 cos 2 t V 2. v2 = ? (A) 2 cos 2 t V (C) 8 cos 2 t V Statement for Q.3-4: Consider the circuit shown in Fig. P1.9.3-4i1 + v1 3H 3H i2 + 4H v2 -

(B) 16 cos 2 t V (D) -4 cos 2 t V

-

Fig. P1.9.5-6

5. If current i1 = 3 cos 4 t A and i2 = 0, then voltage v1 and (B) -2 cos 2 t V (D) -8 cos 2 t V v2 are (A) v1 = -24 sin 4 t V, (B) v1 = 24 sin 4 t V, (C) v1 = 15 sin 4 t V, . (D) v1 = -15 sin 4 t V, . v2 = -24 sin 4 t V v2 = -36 sin 4 t V v2 = sin 4 t V v2 = -sin 4 t V

6. If current i1 = 0 and i2 = 4 sin 3t A, then voltage v1 and v2 are (A) v1 = 24 cos 3t V, (B) v1 = 24 cos 3t V, (C) v1 = -24 cos 3t V, (D) v1 = -24 cos 3t V,www.gatehelp.com

v2 = 36 cos 3t V v2 = -36 cos 3t V v2 = 36 cos 3t V v2 = -36 cos 3t V

Fig. P1.9.5-6

3. If i1 = 0 and i2 = 2 sin 4 t A, the voltage v1 is (A) 24 cos 4 t V (C) 15 cos 4 t V .Page 82

(B) -24 cos 4 t V (D) -15 cos 4 t V .

Magnetically Couple

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