Solutions Manual for GATE Electronics and Communication Engineering 2020 by Trishna

, Networks

Chapter 1 N
 etwork Elements and
U
n
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




i
Chapter 5 Resonance1.93




t
1
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, 1.2 | Networks




Unit 01.indd 2
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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 alterna- (A) 132 J (B) 98 J
tive from the given choices. (C) 144 J (D) 168 J
1. The current in the 2 W resistor ‘I’ is 6. In the circuit of the given figure, the source current ‘I’ is
1Ω 1Ω 2 2 2
+ V I
10V 1Ω + I
2Ω 1A 4 4 4
–
–
8V 4
4
1 3 21 23
(A) A (B) A (C) A (D) A 8
8 8 8 8 (A) 2 A (B) 4 A (C) 1 A (D) A
3
2. The voltage eo in the figure is
2Ω 7. In the circuit shown in the figure, the current I is
3Ω + –
16 V IA o+
10 Ω eo I Vx 2Ω
8A 12 Ω 2 Vx
6Ω o– +
25 V 1A 1Ω
(A) 48 V (B) 24 V (C) 36 V (D) 28 V –
3. The driving point impedance of the infinite ladder net-
work shown in the figure is _____. 25 25
Given R1 = 3 W and R2 = 2 W (A) A (B) A (C) 2 A (D) 3 A
3 6
R1 R1 R1
A
8. In the circuit shown in the figure,
2Ω 2Ω
R2 R2 R2 A
B
2Ω 6Ω 6Ω
3 2  2
(A) Ω (B) Ω (C) 3 +  Ω (D) 21 2Ω
2 3  3 B
6Ω
4. The current wave formed in a pure resistor of 5 W is
shown in the figure. The power dissipated in the resistor is RAB =?
IA 21 5
(A) W (B) W (C) 10 W (D) 8 W
4 6
6
9. What should be the value of current ‘I’ to have zero
current flowing through AB?
1Ω 1V
A 1Ω B
– +
+ IA
0 2 4 6 t sec 1 A 1Ω
–
5V

(A) 20 W (B) 45 W (C) 60 W (D) 90 W
(A) 2 A (B) 4 A (C) −4 A (D) −2 A
5. Figure shows the waveform of the current passing
10. The two electrical sub-networks N1 and N2 are con-
through an inductor of resistance 1 W and inductance
nected through three resistors, as shown in the figure.
2 H. The energy absorbed by the inductor in the first
The voltage across the 2 W resistor is 8 V and the 3 W
four seconds is _____.
I
resistor is 6 V. The voltage across the 4 W resistor is
+ 8Ω –
2Ω
6A 4Ω
N1 N2

3Ω

2s 4s + 6Ω –
t in sec
(A) 24 V (B) −24 V (C) 8 V (D) −8 V




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