Network theory miscellaneous Moderate Questions and Answers | Page - 8

Network theory miscellaneous

For the circuit shown below, the power absorbed by 2Ω resistor is—

1. 50W
1. 8W
1. 18W
1. 32W

View Hint View Answer Discuss in Forum

The given circuit:

From above figure we conclude that the current passing through 2Ω resistor is 3A.
So,
Power absorbed = I²R = 3² × 2 = 18W.

Correct Option: C

The given circuit:

From above figure we conclude that the current passing through 2Ω resistor is 3A.
So,
Power absorbed = I²R = 3² × 2 = 18W.

The effective inductance of the circuit shown below—

1. 13H
1. 17H
1. 20H
1. None of these

View Hint View Answer Discuss in Forum

The given circuit:

L_eq = L₁ + L₂ + L₃ – 2M₁₂ + 2M₂₃
where
L₁ = 3H, L₂ = 2H, L₃ = 6H, M₁₂ = 2H, M₂₃ = 3H
Now,
L_eq = 3H + 2H + 6H – 2 × 2H + 2 × 3H
or
L_eq = 11H – 4H + 6H = 13H
Hence alternative (A) is the correct choice.

Correct Option: A

The given circuit:

L_eq = L₁ + L₂ + L₃ – 2M₁₂ + 2M₂₃
where
L₁ = 3H, L₂ = 2H, L₃ = 6H, M₁₂ = 2H, M₂₃ = 3H
Now,
L_eq = 3H + 2H + 6H – 2 × 2H + 2 × 3H
or
L_eq = 11H – 4H + 6H = 13H
Hence alternative (A) is the correct choice.

The circuit shown below, will act as an ideal current source with respect to terminals A and B, when frequency is—

1. Zero
1. 1 rad/sec
1. 4 rad/sec
1. 16 rad/sec

View Hint View Answer Discuss in Forum

= 1 rad/sec
LC

or
= 1
1 × 1
16

= 4 rad/sec.

Correct Option: C

= 1 rad/sec
LC

or
= 1
1 × 1
16

= 4 rad/sec.

For the circuit shown below, the power absorbed by 2Ω resistor and power delivered by 10V source will be—

1. 2W, 10W
1. – 2W, 10W
1. 2W, – 10W
1. – 2W, – 10W

View Hint View Answer Discuss in Forum

The given circuit:

From figure, 20 – 2I – 8 I – 10 = 0
or
I = 1A
Power absorbed by 2Ω resistor = I²R = 1² × 2 = 2W
Power delivered by 10V source = – VI = – 10 × 1 = – 10W
[Note: Here negative sign is taken because current is entering at the negative terminal of the 10V voltage source]
Hence alternative (C) is the correct choice.

Correct Option: C

The given circuit:

From figure, 20 – 2I – 8 I – 10 = 0
or
I = 1A
Power absorbed by 2Ω resistor = I²R = 1² × 2 = 2W
Power delivered by 10V source = – VI = – 10 × 1 = – 10W
[Note: Here negative sign is taken because current is entering at the negative terminal of the 10V voltage source]
Hence alternative (C) is the correct choice.

In the given figure, the Thevenin’s equivalent pair (voltage, impedance), as seen at the terminals P–Q, is given by—

1. (2V, 5Ω)
1. (2V, 7·5Ω)
1. (4V, 5Ω)
1. (4V, 7·5Ω)

View Hint View Answer Discuss in Forum

The given circuit:

R_th Calculation:

R_th = 10 || 10 = 5Ω
V_th Calculation:

I = 4V = 4V = 0.4 A
R_eq 10Ω

I₁ = 0.4 × 20 = 0.2A
20 + 20

V_th = I₁ × 10Ω = 0·2 × 10 = 2V
Hence alternative (A) is the correct choice.

Correct Option: A

The given circuit:

R_th Calculation:

R_th = 10 || 10 = 5Ω
V_th Calculation:

I = 4V = 4V = 0.4 A
R_eq 10Ω

I₁ = 0.4 × 20 = 0.2A
20 + 20

V_th = I₁ × 10Ω = 0·2 × 10 = 2V
Hence alternative (A) is the correct choice.