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GATE EC BY RK Kanodia MULTIPLE CHOICE QUESTION Electronics & Communication Engineering Fifth Edition R. K. Kanodia B.Tech. NODIA & COMAPNY JAIPUR www.gatehelp.com GATE EC BY RK Kanodia MRP 400.00 Price 550.00 www.gatehelp.com GATE EC BY RK Kanodia www.gatehelp.com GATE EC BY RK Kanodia www.g...

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GATE EC BY RK Kanodia

MULTIPLE CHOICE QUESTION

Electronics & Communication Engineering Fifth Edition R. K. Kanodia B.Tech.

NODIA & COMAPNY JAIPUR

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GATE EC BY RK Kanodia

MRP 400.00

Price 550.00

www.gatehelp.com

GATE EC BY RK Kanodia

www.gatehelp.com

GATE EC BY RK Kanodia

www.gatehelp.com

GATE EC BY RK Kanodia

www.gatehelp.com

GATE EC BY RK Kanodia

www.gatehelp.com

GATE EC BY RK Kanodia

www.gatehelp.com

GATE EC BY RK Kanodia

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GATE EC BY RK Kanodia

CHAPTER

1.1 BASIC CONCEPTS

1. A solid copper sphere, 10 cm in diameter is deprived of 10

20

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

(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 v1

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

60 V

20 W

100 V

3. If 120 C of charge passes through an electric

Fig. P.1.1.6

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

(A) 240 V

(B) 120 V

(C) 60 V

(D) 30 V

4. The energy required to move 120 coulomb through 7. In the circuit of the fig P1.1.7, the value of the

3 V is (B) 360 J

(C) 40 J

(D) 2.78 mJ

voltage source E is 0V +

+

(A) 25 mJ

–

–

2V

5. i = ?

+ – +

5V

–

–

4V

i

E

+

1A

1V

10 V

2A

5A

Fig. P.1.1.7 3A

4A

Fig. P.1.1.5

(A) -16 V

(b) 4 V

(C) -6 V

(D) 16 V

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Page 3

GATE EC BY RK Kanodia

UNIT 1

8. Consider the circuit graph shown in fig. P1.1.8. Each

Networks

12. v1 = ?

branch of circuit graph represent a circuit element. The

v1 –

+

value of voltage v1 is

1 kW

2

7V

+ 105 V –

– 15 V +

+

+

55 V + v1 –

–

8V

35 V

–

–

5V

3

6V

kW

+

–

100 V

– 10 V + +

65 V

kW

V

30

V

30

4 kW

+

–

Fig. P1.1.12 Fig. P.1.1.8

(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

(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 1A

voltage vo is 5W + vo

4W

+ vo –

1A

15 V

5V

Fig. P1.1.11

–

Fig. P.1.1.9

(A) 10 V

(B) 15 V

(C) 20 V

(D) None of the above

(A) 1 V

(B) 5 V

(C) 9 V

(D) None of the above

14. Req = ? 5W

10 W

10 W

10 W

10. R1 = ? 60 W

Req

10 W

+

10 W

10 W

up to ¥

R1 100 V R2

+ 20 V –

70 V

Fig. P1.1.14

–

Fig. P.1.1.10

(A) 11.86 W

(B) 10 W

(C) 25 W

(D) 11.18 W

15. vs = ? (A) 25 W

(B) 50 W

(C) 100 W

(D) 2000 W

180 W + 60 W

11. Twelve 6 W resistor are used as edge to form a cube.

vs

The resistance between two diagonally opposite corner of the cube is 5 (A) W 6 (C) 5 W

Page 4

(B)

90 W

6 W 5

(D) 6 W

40 W

20 V –

Fig. P.1.1.15

(A) 320 V

(B) 280 V

(C) 240 V

(D) 200 V

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180 W

GATE EC BY RK Kanodia

UNIT 1

24. Let i( t) = 3te-100 t A and v( t) = 0.6(0.01 - t) e -100 t V for

Networks

28. vab = ?

a

the network of fig. P.1.1.24. The power being absorbed

2 8

–

b 0.3i1

A 2

W

W

N

0.2i1

i1

A

6

+

W

R

i v

2

by the network element at t = 5 ms is

Fig. P.1.1.24

(A) 18.4 mW

(B) 9.2 mW

(C) 16.6 mW

(D) 8.3 mW

Fig. P.1.1.28

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

(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

when lit. The additional A bulbs in parallel to this

500 W

circuit, that would be required to blow the fuse is

400 W

ix

20 A

200 W

2ix

40 V 12 V C

B

A

Fig. P.1.1.29

(A) dependent source of 192 W

Fig. P.1.1.25

(A) 4

(B) 5

(B) dependent source of 368 W

(C) 6

(D) 7

(C) independent source of 16 W

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

(D) independent source of 40 W

the load RL is

30. The dependent source in fig. P.1.1.30

i1

5W RL = 2 W

2i1

1W

1V

v1

20 V

v1 5

5W

Fig. P.1.1.26

(A) 2 W

(B) 4 W

(C) 6 W

(D) 8 W

Fig. P.1.1.30

27. vo = ?

(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 0.2 A

5W

+ v1 0.3v1 –

8W

+ v2 –

5v2

18 W

+ vo –

+ 8V – ix

Page 6

(A) 6 V

(B) -6 V

(C) -12 V

(D) 12 V

2ix

4A

Fig. P.1.1.27

Fig. P.1.1.31

(A) supplies 16 W

(B) absorbs 16 W

(C) supplies 32 W

(D) absorbs 32 W

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Basic Concepts

Chap 1.1

32. A capacitor is charged by a constant current of 2 mA

36. The waveform for the current in a 200 mF capacitor

and results in a voltage increase of 12 V in a 10 sec

is shown in fig. P.1.1.36 The waveform for the capacitor

interval. The value of capacitance is

voltage is

(A) 0.75 mF

(B) 1.33 mF

(C) 0.6 mF

(D) 1.67 mF

i(mA) 5

33. The energy required to charge a 10 mF capacitor to 100 V is

Fig. P. 1.1.36

(A) 0.10 J (C) 5 ´ 10

t(ms)

4

(B) 0.05 J

-9

v

(D) 10 ´ 10

J

-9

J

v 50m

5

34. The current in a 100 mF capacitor is shown in fig.

t(ms)

4

P.1.1.34. If capacitor is initially uncharged, then the

4

(A)

(B)

v

waveform for the voltage across it is

t(ms)

v 50m

250m

i(mA) t(ms)

4

2

4

(C)

t(ms)

t(ms)

(D)

Fig. P. 1.1.34 v

37. Ceq = ?

v 10

10

2

4

t(ms)

2

(A)

v

2.5 mF

4

t(ms) 1.5 mF

(B)

v

2

4

1 mF

Ceq

0.2

0.2 t(ms)

2

(C)

4

2 mF

t(ms)

(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

(A) 3.5 mF

(B) 1.2 mF

capacitor is

(C) 2.4 mF

(D) 2.6 mF

v 6

38. In the circuit shown in fig. P.1.1.38 1

2

3

t(ms)

iin ( t) = 300 sin 20 t mA, for t ³ 0.

Fig. P.1.1.35 i(mA) 6

C2

C2

C2

C2

+

i(mA) 600

vin C1

iin 1

2

3

t(ms)

1

2

3

C1

C1

C1

60 mF

t(ms)

–

(A)

(B)

i(mA) 6

Fig. P. 1.1.38

i(mA) 600 2 1

(C)

3

t(ms)

2 1

3

t(ms)

Let C1 = 40 mF and C2 = 30 mF. All capacitors are initially uncharged. The vin ( t) would be

(D)

(A) -0.25cos 20t V

(B) 0.25cos 20t V

(C) -36cos 20t mV

(D) 36cos 20t mV

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GATE EC BY RK Kanodia

Basic Concepts

Chap 1.1

v3 - 30 = v2

SOLUTIONS

Þ

v3 = 65 V

105 + v4 - v3 - 65 = 0 Þ

v4 = 25 V

v4 + 15 - 55 + v1 = 0 Þ v1 = 15 V

1. (C) n = 10 20 , Q = ne = e10 20 = 16.02 C Charge on sphere will be positive.

9. (B) Voltage is constant because of 15 V source.

2. (D) DQ = i ´ D t = 15000 ´ 100m = 15 . C

10. (C) Voltage across 60 W resistor = 30 V 30 Current = = 0.5 A 60

3. (B) i =

dQ 120 = =2 A dt 60

4. (B) W = Qv = 360 J

Voltage across R1 is = 70 - 20 = 50 V 50 = 100 W R1 = 0.5

6. (A)

11. (C) The current i will be distributed in the cube branches symmetrically 1A

i=1A

a

5A

2A 3A

4A

i 3

i i 3

6A

1A

i 3

2A

i 6

Fig. S 1.1.5

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

0V

Fig. S. 1.1.11

+

+

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

E

13. (D) It is not possible to determine the voltage across

–

–

+

+

5V

1 A source.

10 V

Fig. S 1.1.7

14. (D) Req = 5 +

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

v2 = 35 V

+ 105 V –

+

30 +

Req

Fig. S 1.1.14

–

–

10 W

–

V

v2

Req

55 V + v1 –

5W

– 10 V +

30

V

–

+ v3 –

10 + 5 + Req 5W

+

v4

10 ( Req + 5)

+

– +

– 15 V +

+

65 V 100 V

b

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

7. (A) Going from 10 V to 0 V

4V

i

vab =

v1 + 60 - 100 = 10 ´ 20 or v1 = 240 V

2V

i 3

Þ Req2 + 15 Req = 5 Req + 75 + 10 Req + 50 Fig. S 1.1.8

Þ

Req = 125 = 1118 . W

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GATE EC BY RK Kanodia

Basic Concepts

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

Þ

We can say Cd = 20 mF, Ceq = 20 + 40 = 60 mF 1 1 æ 300 ö cos 20 t ÷ ´ 10 -3 = - 0.25 cos 20 t V vC = ò idt= ç60m è 20 C ø

vo = 20 V

20 v1 = 20 ´ = 80 W 5 5

39. (C) iC1 =

31. (D) Power P = vi = 2 ix ´ ix = 2 ix2 ix = 4 A, P = 32 W (absorb) 32. (D) vt 2 - vt1 = Þ

1 C

t2

ò idt Þ

iin C1 = 0.8 sin 600 t mA C1 + C2

At t = 2 ms, iC1 = 0.75 mA

Þ

12 =

t1

12 C = 2m ´ 10

Chap 1.1

1 2m ( t2 - t1 ) C

40. (B) vC1 =

4 vin vin C2 = C1 + C2 6 + 4

Þ

vc1 = 0.4 vin

C = 1.67 mF 41. (D) V = 2 + 3 + 5 = 10, Q = 1 C, C =

1 33. (B) E = Cv 2 = 5 ´ 10 -6 ´ 100 2 = 0.05 J 2

42. (A) vL = L

di Þ dt

This 0.2 V increases linearly from 0 to 0.2 V. Then

43. (B) vL = L

di = 0.01 ´ 2( 377 cos 377 t) V dt

current is zero. So capacitor hold this voltage.

= 7.54 cos 377 t V

1 34. (D) vc = c

35. (D) i = C

-3

10 ´ 10 -3 ò0 idt = 100 ´ 10 -6 (2 ´ 10 ) = 0.2 V

2m

dv dt

For 0 < t < 1 , C

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

1 1 12000 120 cos 3t dt = vdt = sin 377 t 0.01 ò Lò 377 12000 ´ 120 P = vi = (sin 377 t)(cos 377 t) 377

45. (D) vL = L

1 1 idt = ò 200 ´ 10 -6 C

5m

ò 4m tdt = 3125 t

vC = 3vL Þ

iC = 3 LC

46. (B) vL = L

It will be parabolic path. at t = 0 t-axis will be tangent.

combination is in series with 1.5 mF. 15 . (2 + 1) = 1mF, C1 is in parallel with 2.5 mF C1 = 15 . +2+1 . mF Ceq = 1 + 2.5 = 35

diL dt æ -100 - 0 ö vL = (0.05)ç ÷ = - 2.5 V 2 è ø

For 4 < t £ 8,

æ 100 + 100 ö vL = (0.05)ç ÷ = 2.5 V 4 è ø

For 8 < t £ 10,

æ 0 - 100 ö vL = (0.05)ç ÷ = - 2.5 V 2 è ø

Thus (B) is correct option.

30 ´ 60 30(20 + 40) 38. (A) Ca = = 20 mF, Cb = = 20 mF 30 + 60 30 + 20 + 0 C2

Cc

C2

Cb

C2

Ca

47. (C) Algebraic sum of the current entering or leaving a cutset is equal to 0. 6 16 i2 + i4 + i3 = 0 Þ + + i3 = 0 2 4

C2

+

i3 = - 7 A, C1

d 2 iL = - 9.6 sin 4 t A dt

For 2 < t £ 4,

37. (A) 2 mF is in parallel with 1 mF and this

vin

diL dv , iC = C C dt dt

2

At t = 4 ms, vc = 0.05 V

iin

L = 2 mH

= 1910 sin 754 t W

36. (B) For 0 £ t £ 4,

Cd

Þ

44. (A) i =

For 1 ms < t < 2 ms, dv 0-6 C = 100 ´ 10 -6 ´ = - 600 mA ( 3 - 2)m dt

vC =

100m = L

200m 4m

Q = 0.1 F V

C1

C1

C1

v3 = - 7 ´ 3 = - 21 V

60 mF

–

Fig. S 1.1.38

*********

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GATE EC BY RK Kanodia

CHAPTER

1.2 GRAPH THEORY

1. Consider the following circuits :

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 (1)

(2)

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

(3)

(4) Fig. P.1.1.4

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

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 c

(1)

(2)

b

3 g

e

d

4

f

2

a

1

5

h

Fig. P.1.2.5

(3) Page 12

(4)

(A) a d e h

(B) a c f h

(C) a f h g

(D) a e f g

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GATE EC BY RK Kanodia

UNIT 1

é 1 -1 0 ù (A) ê-1 0 1ú ê ú 1 -1úû êë 0 é-1 -1 0 ù (C) ê 0 1 1ú ê ú êë 1 0 -1ûú

é 1 0 -1ù (B) ê-1 -1 0 ú ê ú 1 1úû êë 0 1ù é-1 0 ê (D) 0 1 -1ú ê ú êë 1 -1 0 úû

2

2

4

1

3

4

1

5

13. The incidence matrix of a graph is as given below

3

5

(C)

é-1 1 1 0 0 0 ù ê 0 0 -1 1 1 0 ú ú A =ê ê 0 -1 0 -1 0 -1ú ê 1 0 0 0 -1 -1ú ë û

(D)

15. The incidence matrix of a graph is as given below é-1 1 1 0 0 0 ù ê 0 0 -1 1 1 0 ú ú A =ê ê 0 -1 0 -1 0 0 ú ê 1 0 0 0 -1 -1ú ë û

The graph is 2

Networks

2

The graph is 4

4 3

1

3

1

(A)

1

2

3

1

2

3

(B)

2

2

4

4

(A) 4

(B)

4 3

1

3

1

(C)

1

2

3

1

2

3

(D)

14. The incidence matrix of a graph is as given below é- ...