Network Elements and the Concept of Circuit Easy Questions and Answers

Network Elements and the Concept of Circuit

Voltage across the terminal a and b—

1. 3V
1. –3V
1. 5V
1. 5.5V

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Given circuit

KCL at node a (Assume potential at point b = 0V)
V_a – V_b + V_a – 5 = 3
2 2

or V_a – 0 + V_a – 5 = 3
2 2

or V_a 1 + 1
2 2

= 3 + 5
2

or V_a = 5.5 V
Therefore V_ab = 5.5 V

Correct Option: D

Given circuit

KCL at node a (Assume potential at point b = 0V)
V_a – V_b + V_a – 5 = 3
2 2

or V_a – 0 + V_a – 5 = 3
2 2

or V_a 1 + 1
2 2

= 3 + 5
2

or V_a = 5.5 V
Therefore V_ab = 5.5 V

Dependent source—

1. Delivers 80W
1. Delivers 45W
1. Absorbs 80W
1. Absorbs 75W

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The given circuit can be redrawn as shown below

KCL at point A
V_A – 0 + V_A – 15 = 3
5 5

or V_A 1 + 1 = 3 + 3
5 5

or V_A = 6 × 5 = 15 V
2

Now, Power = V_A × V₁ = 15 × 15 = 45 W
5 5

Positive sign of power indicates that it delivers 75W power.

Correct Option: B

The given circuit can be redrawn as shown below

KCL at point A
V_A – 0 + V_A – 15 = 3
5 5

or V_A 1 + 1 = 3 + 3
5 5

or V_A = 6 × 5 = 15 V
2

Now, Power = V_A × V₁ = 15 × 15 = 45 W
5 5

Positive sign of power indicates that it delivers 75W power.

The incidence matrix of a graph is as given below—

The number of possible tree are—

1. 11
1. 14
1. 16
1. 8

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Since the incidence matrix is complete because sum of every column is zero. Therefore, first make reduced incidence matrix, A_r
A_r = –1 0 1 1 –1 0
1 –1 0 0 0 –1

Number of tree = det
[A_r . A_r ^T]

= det 4 –1
–1 3

= 12 – 1 = 11
Hence alternative (A) is the correct choice.

Correct Option: A

Since the incidence matrix is complete because sum of every column is zero. Therefore, first make reduced incidence matrix, A_r
A_r = –1 0 1 1 –1 0
1 –1 0 0 0 –1

Number of tree = det
[A_r . A_r ^T]

= det 4 –1
–1 3

= 12 – 1 = 11
Hence alternative (A) is the correct choice.

The average power by the dependent source is—

1. Supplied by 59.4 W
1. Absorbed by 59.4 W
1. Supplied by 118.8 W
1. Absorbed by 118.8 W

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The given circuit can be redrawn by applying source transformation

KCL at node A
I_x + V_A = 1.6 I_x ..............(i)
8

Also, V_A – 10 90° = I_x
(5 + j2)

or V_A = I_x (5 + j2) + 10 90º ..........................(ii)
from equations (i) and (ii)
I_x + I_x (5 + j2) + 10 90° = 1.6 I_x
8

or 8I_x + I_x (5 + j2) – 12.8 I_x = – 10 90º
or I_x (0.2 + j2) = – 10 90º
or I_x = –10 90° = –10 90°
0.2 + j2 2.01 84.23°

or I_x = – 4.975 5.77°
V_A = (1.6 I_x – I_x) × 8
or V_A = .6 I_x × 8
or V_A = 4.8 I_x
or V_A = – 23.88 V
Now, average power = 1 V_A. I_x
2

= 1 (– 23.88) × (– 4.975)
2

= 59.40 W
Positive power means supplied by the dependent source.

Correct Option: A

The given circuit can be redrawn by applying source transformation

KCL at node A
I_x + V_A = 1.6 I_x ..............(i)
8

Also, V_A – 10 90° = I_x
(5 + j2)

or V_A = I_x (5 + j2) + 10 90º ..........................(ii)
from equations (i) and (ii)
I_x + I_x (5 + j2) + 10 90° = 1.6 I_x
8

or 8I_x + I_x (5 + j2) – 12.8 I_x = – 10 90º
or I_x (0.2 + j2) = – 10 90º
or I_x = –10 90° = –10 90°
0.2 + j2 2.01 84.23°

or I_x = – 4.975 5.77°
V_A = (1.6 I_x – I_x) × 8
or V_A = .6 I_x × 8
or V_A = 4.8 I_x
or V_A = – 23.88 V
Now, average power = 1 V_A. I_x
2

= 1 (– 23.88) × (– 4.975)
2

= 59.40 W
Positive power means supplied by the dependent source.

The power factor seen by the voltage source—

1. 0.8 lagging
1. 0.8 leading
1. 0.6 lagging
1. 0.6 leading

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KCL at node A
V_A – 10 cos 2t + V_A – 0 = 3 V_i ...................(i)
4 1 + 1 4
sC

Again from given circuit –
V₁ = V_A – 10 cos 2t ................(ii)
4 4

or V_i = 10 cos2t – V_A ...................(iii)
From equations (i) and (iii)
V_A – 10 cos 2t + V_A sC = 3 (10 cos 2t – V_A)
4 1 + sC 4

or V_A 1 + j + 3
4 3 + j 4

= 10 cos 2t 3 + 1
4 4

or V_A 1 + j2 = 10 cos 2t
3 + j2

V_A = 10 cos 2t 3 + 2j
3 + 4j

Current supplied by voltage source, say
I = 10 cos 2t – V_A
4

or I = 1 10 cos 2t – 10 cos 2t (3 + 2j)
4 3 + 4j

or I = 10 cos 2t 3 + 4j – 3 – 2j
4 3 + 4j

or I = 2.5 cos 2t 2j
3 + 4j

or I = 2.5 cos 2t 2 90°
5 53.13°

or I = cos 2t 36.86º
Since here current leads voltage, hence power factor will be leading and
cos φ = cos 36.86º = 0.8
Hence alternative (B) is the correct choice.

Correct Option: B

KCL at node A
V_A – 10 cos 2t + V_A – 0 = 3 V_i ...................(i)
4 1 + 1 4
sC

Again from given circuit –
V₁ = V_A – 10 cos 2t ................(ii)
4 4

or V_i = 10 cos2t – V_A ...................(iii)
From equations (i) and (iii)
V_A – 10 cos 2t + V_A sC = 3 (10 cos 2t – V_A)
4 1 + sC 4

or V_A 1 + j + 3
4 3 + j 4

= 10 cos 2t 3 + 1
4 4

or V_A 1 + j2 = 10 cos 2t
3 + j2

V_A = 10 cos 2t 3 + 2j
3 + 4j

Current supplied by voltage source, say
I = 10 cos 2t – V_A
4

or I = 1 10 cos 2t – 10 cos 2t (3 + 2j)
4 3 + 4j

or I = 10 cos 2t 3 + 4j – 3 – 2j
4 3 + 4j

or I = 2.5 cos 2t 2j
3 + 4j

or I = 2.5 cos 2t 2 90°
5 53.13°

or I = cos 2t 36.86º
Since here current leads voltage, hence power factor will be leading and
cos φ = cos 36.86º = 0.8
Hence alternative (B) is the correct choice.

V_A – 10 cos 2t	+	V_A – 0		=	3	V_i ...................(i)
4		1 +	1		4
			sC

	V_a – V_b	+	V_a – 5	= 3
	2		2

or	V_a – 0	+	V_a – 5	= 3
	2		2

or V_a		1	+	1
		2		2

	V_a – V_b	+	V_a – 5	= 3
	2		2

Network Elements and the Concept of Circuit

Network Elements and the Concept of Circuit

Network Theory

Correct Option: D

Correct Option: B

Correct Option: A

Correct Option: A

Correct Option: B