Solutions Manual for GATE Electronics and Communication Engineering 2022 by Trishna

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Networks Chapter 1 N etwork Elements and Basic Laws

1.3

Chapter 2 Network Theorems

1.21

Chapter 3

Transient Analysis (Ac and Dc) 1.40

Chapter 4 Two-Port Networks

1.73

Chapter 5 Resonance 1.93

U n i t 1

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Hints/Solutions | 1.3

Chapter 1 Network elemeNts aNd BasiC laws exerCises Practice Problems 1 Directions for questions 1 to 21: Select the correct alternative from the given choices.

1. The current in the 2 W resistor ‘I’ is 1Ω

1Ω

+

V I

10V –

2

1Ω

2Ω

1 (A) A 8

3 (B) A 8

+ I – 8V

1A

21 (C) A 8

23 (D) A 8

2. The voltage eo in the figure is

10 Ω

IA

16 V

12 Ω

6Ω

R1

R1 R2

3 Ω 2

(C) 1 A

(D)

8 A 3

+

Vx

25 A 3

2Ω

2 Vx

+ 25 V –

(A)

–

1Ω

1A

(B)

25 A 6

(C) 2 A

(D) 3 A

8. In the circuit shown in the figure, R2

A

 2 (C) 3 +  Ω (D)  3

2 Ω 3

(B)

4

(B) 4 A

B

(A)

4

4 4

I

A R2

2

4

3Ω

o+ eo o–

(A) 48 V (B) 24 V (C) 36 V (D) 28 V 3. The driving point impedance of the infinite ladder network shown in the figure is _____.

Given R1 = 3 W and R2 = 2 W R1

(A) 2 A

2

7. In the circuit shown in the figure, the current I is

2Ω

8A

(A) 132 J (B) 98 J (C) 144 J (D) 168 J 6. In the circuit of the given figure, the source current ‘I’ is

2Ω

2Ω

2Ω 6Ω

21

4. The current wave formed in a pure resistor of 5 W is shown in the figure. The power dissipated in the resistor is IA

B

2Ω 6Ω

RAB =?

(A)

6

6Ω

21 W 4

(B)

5 W 6

(C) 10 W

(D) 8 W

9. What should be the value of current ‘I’ to have zero current flowing through AB? 1Ω 0

2

+ – 5V

6 t sec

4

(A) 20 W (B) 45 W (C) 60 W (D) 90 W 5. Figure shows the waveform of the current passing through an inductor of resistance 1 W and inductance 2 H. The energy absorbed by the inductor in the first four seconds is _____.

I

1V

A 1Ω

B

– + 1 A 1Ω

IA

(A) 2 A (B) 4 A (C) −4 A (D) −2 A 10. The two electrical sub-networks N1 and N2 are connected through three resistors, as shown in the figure.

The voltage across the 2 W resistor is 8 V and the 3 W resistor is 6 V. The voltage across the 4 W resistor is +

6A N1

8Ω – 2Ω

4Ω

N2 3Ω

2s

4s

+

t in sec

(A) 24 V

6Ω –

(B) −24 V

(C) 8 V

(D) −8 V

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1.4 || Part Networks 3.6 III • Unit 1 • Networks 11. In the circuit shown in the figure, the voltages V1, V2, and V3 are 3Ω

6Ω 1Ω

V1

V2

6Ω

16. The current i4 in the circuit of the figure equal to

6Ω

i0 = 7 A

1Ω V3

8 16 , ,4 3 3

16 8 (C) , ,4 3 3

17.

5∠ 0 ° A

12. In the network shown in the figure, the current in resistor R is

B 2Ω 10 V

3Ω

5A

(A) 2 A

4A

D

(B) 3 A

3Ω

10 ∠ 60° A

10 3 10 3 ∠ − 90°Amp ∠90° A (B) 2 2 (C) 5∠60° A (D) 5∠ − 60° A 18. A delta-connected network with its Wye equivalent is shown in the given figure. The resistance R1, R2, and R3 (in Ohm) are, respectively, (A)

R C

i1

For the circuit shown in the figure, the instantaneous current i1(t) is

3A

2A

−j 2 Ω

j 2Ω

16 8 (D) 4, , 3 3

A

(B) −12 A (D) None of these

(A) 12 A (C) 4 A

16 8 8 , , 3 3 3

(B)

i3 = 4 A

i 4 =?

12V

(A)

i2 = 3 A

i1 = 5 A

(C) 4 A

(D) 9 A

a

13. In the following circuit, the voltage AB is A

50 V

5Ω

40 V

4Ω

30V

3Ω

a B

5Ω b

R1 30Ω

⇒ R2

c

15 Ω

R3 c

b 2Ω

(A) 35 V

(B) 28.2 V

(C) 38.3 V

(D) 42.6 V

(A) 1.5, 3, and 9 (B) 3, 9, and 1.5 (C) 9, 3, and 1.5 (D) 3, 1.5, and 9 19. The voltage e0 in the figure 2Ω

14. A network contains linear resistors that are connected in series across an ideal voltage source. If all the resistances are halved and the voltage is doubled, then the voltage across each resistor becomes (A) doubled (B) halved (C) not changed (D) None of these 15. Obtain the equivalent capacitance of the network given

+ 10 Ω

8A

(A) 48 V 20.

5Ω

1µF

1µF

12 Ω

1µF

1µF

(B) 0.8 mF

(D) 28 V

x 2Ω

I +

6Ω

3Ω

50 V y

1µF

1µF

b

(A) 1mF

(C) 36 V

–

1µF 1µF

8Ω

12 Ω e 0 −

6Ω

(B) 24 V

a 1µF

16 V

(C) 1.9 mF

(D) 2.6 mF

The current I supplied by the source 50 V is (A) 25 A (B) 13.7 A (C) 9.8 A (D) 3.66 A 21. The voltages VC1, VC2, and VC3 across the capacitors in the circuit in the given figure, under steady state are, respectively,

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Hints/Solutions Chapter 1 • Network Elements and Basic Laws || 1.5 3.7 2F

~

1F

(A) 80 V, 32 V, 48 V (C) 20 V, 8 V, 12 V

25 k Ω

+ − VC2 40 k Ω

1H

10 kΩ VC1 + −

2H

+ VC3 −

3F

Practice Problems 2

6. The current ‘I’ supplied by the source in the figure is

Directions for questions 1 to 19: Select the correct alternative from the given choices.

1. A network contains linear resistors that are connected in series across an ideal voltage source. If all the resistances are halved and the voltage is doubled, then the voltage across each resistor becomes (A) doubled (B) halved (C) not changed (D) None of these 2. Twelve similar conductors of 1 W resistance form a cubical framework. Then, the resistance between two adjacent corners, two opposite corners of one face, and two opposite corners of the cube are (A)

3 5 7 , , 4 6 12

(B)

7 5 3 , , 4 6 4

(C)

7 3 5 , , 12 4 6

(D)

12 4 6 , , 7 3 5

3. In the network shown in the figure, the voltage at node 2 is

10 A

1

2Ω

1Ω

4Ω

2 5A 4Ω

(A) 2 V (B) 10 V (C) 6 V (D) 4 V 4. In the network shown in the figure, the voltage AF is 3Ω

A

10 v

2Ω

20 v

2Ω

3A

5Ω F

1Ω

4Ω

(A) 2 A

1Ω

(B) 3 A

(C) 1 A

4Ω 4Ω

6V

4Ω

3Ω

1Ω

3 2 A (D) A 2 3 7. A resistance of 10 W is connected in series with two resistances of 20 W arranged in parallel. What resistance should be shunted across this parallel combination so that the total current taken shall be 2 A with 30 V applied?

(A) 5 W (B) 10 W (C) 20 W (D) 25 W 8. The resistance of a strip of conductor is RW. If the strip is elongated such that its length is doubled, the resistance of the strip is given by R (A) 4 R (B) 2 R (C) (D) R 2 9. The current i(t) through a 10 W resistor in series with an inductance is given by i(t) = 3 + 4 sin(100t + 45°) + 4 sin(300t + 60°) A. The RMS value of the current and the power dissipated in the circuit are (A) 2 A

(A)

(B) 3 A

(C)

41A, 410 W, respectively

(B) 35A, 350 W, respectively (C) 5 A, 250 W, respectively (D) 11 A, 1210 W, respectively 10. The nodal method of circuit analysis is based on (A) KVL and Ohm’s law (B) KCL and Ohm’s law (C) KCL and KVL (D) KCL, KVL, and Ohm’s law 11. In the given circuit, the voltage v(t) is 1Ω

2Ω I 3Ω

6Ω

I

(A) 4 V (B) −4 V (C) 6 V (D) 2 V 5. For the circuit shown in the figure, the current ‘I’ is given by

6A

(B) 80 V, 48 V, 32 V (D) 20 V, 12 V, 8 V

1Ω e at

+ v (t ) −

e bt

(A) eat − ebt (B) eat + ebt at bt (C) a e − b e (D) a eat + b ebt 12. The rms value of the voltage defined by v(t) = 5 + 5 sin (314t + p 6) is

+ 3V –

(D) zero

(A) 5 V

(B) 2.5 V

(C) 6.12 V

(D) 10 V

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1.6 || Part Networks 3.8 III • Unit 1 • Networks 13. If R1 = R2 = R4 = R and R3 = 1.1R in the bridge circuit shown in the figure, then the reading in the ideal voltmeter connected between a and b is

16. + V1

+

+

a

10 V

R4

−

V −

b

R2

If V2 = 1 V in the abovementioned network, the value of V1 will be (A) 2.5 V (B) 4 V (C) 5 V (D) 8 V

(A)

(B) 0.138 V (D) 1 V

Z

(C) 3 3Z

(B) 3Z

3

8Ω

2V

+ −

E=?

25 A

30 Ω

5A

The current I in the circuit is (A) 20 A (B) −20 A (C) 16.2 A (D) −16.12 A 15.

a

− V1

+ 4V

a

b

R ab

=

9

1.5

b

c

R ac R bc

Previous Years’ Questions 1. If R1 = R2 = R4 = R and R3 = 1.1 R in the bridge circuit shown in figure, then the reading in the ideal voltmeter connected between a and b is [2005]

a

R4

+ R2

v

1.5V

(A) 0.6 V

R3

(B) 1.8

(C) 0.9 V

(D) 0 V

(A) 0.238 V (B) 0.138 V (C) −0.238 V (D) 1 V 2. A fully charged mobile phone with a 12 V battery is good for a 10−min talk time. Assume that, during the talk time, the battery delivers a constant current of 2 A and its voltage drops linearly from 12 V to 10 V, as shown in the figure. How much energy does the battery deliver during this talk time?

[2009] v(t)

b

−

2Ω

3Ω

+ −

10 V

c

The value of Rab, Rac, and Rbc are (A) 30 W, 15 W, 5 W (B) 5 W, 30 W, 15 W (C) 5 W, 30 W, 15 W (D) 15 W, 5 W, 30 W

R1

+ 5V −

(A) −16 V (B) 4 V (C) −6 V (D) 16 V 19. What is the current through resistor 2 W in the circuit given?

0.2 µF

3

Z 3

0V + 1V

−

I 12 Ω

(D)

18. In the given circuit, the value of the voltage source E is V2

14.

2 Ω V2 = 1 V −

17. If each branch of a Delta circuit has impedance 3Z , then each branch of the equivalent Wye circuit has impedance

R3

(A) 0.238 V (C) −0.238 V

+

2Ω

2Ω

− R1

2Ω

2Ω

12V 10V 0

10min

t

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