Network theory miscellaneous Difficult Questions and Answers | Page - 2

Network theory miscellaneous

The effective resistance between terminals A and B in the circuit shown below is—

1. R
1. R – 1
1. R/2
1. 6 R
  11

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The given circuit:

By converting star network into delta network

R_{eq. (AB)} = R_AB || (R_AC + R_BC)
Where,
R_AB = R || 3R = 3/4 R
R_AC = R || 3R = 3/4 R
R_BC = R || 3R = 3/4 R
Now,
R_{eq. (AB)} = 3R 3R + 3 R
4 4 4

= 3R 6 R = 3R/4 × 6R/4
4 4 3R/4 + 6R/4

= 18R² = R
16 × (9R/4) 2

Hence alternative (C) is the correct choice.

Correct Option: C

The given circuit:

By converting star network into delta network

R_{eq. (AB)} = R_AB || (R_AC + R_BC)
Where,
R_AB = R || 3R = 3/4 R
R_AC = R || 3R = 3/4 R
R_BC = R || 3R = 3/4 R
Now,
R_{eq. (AB)} = 3R 3R + 3 R
4 4 4

= 3R 6 R = 3R/4 × 6R/4
4 4 3R/4 + 6R/4

= 18R² = R
16 × (9R/4) 2

Hence alternative (C) is the correct choice.

The total power consumed in the circuit shown in the figure is—

1. 10W
1. 12W
1. 16W
1. 20W

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The given circuit:

Case 1: When current source is taken:

The net current in branch AB is 2A
Case 2: When voltage source is taken:

The net current in branch BC is 2/2 = 1A
Now, the total power consumed in the circuit:
= I²_ABR_AB + I²_BC × R_BC
= 2² × 2 + 1² × 2
= 10W.

Correct Option: A

The given circuit:

Case 1: When current source is taken:

The net current in branch AB is 2A
Case 2: When voltage source is taken:

The net current in branch BC is 2/2 = 1A
Now, the total power consumed in the circuit:
= I²_ABR_AB + I²_BC × R_BC
= 2² × 2 + 1² × 2
= 10W.

The time constant of the network shown in figure is—

1. 2RC
1. 3RC
1. RC/2
1. 2 RC
  3

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The given network:

The equivalent circuit for calculating time constant is shown below

R_eq = R || 2R = 2R
3

Time constant,
τ = R_eq C = 2R . C = 2 RC
3 3

Correct Option: D

The given network:

The equivalent circuit for calculating time constant is shown below

R_eq = R || 2R = 2R
3

Time constant,
τ = R_eq C = 2R . C = 2 RC
3 3

For the circuit shown below, the current I_x and power delivered by the independent source is—

1. 0·4A, 16W
1. – 0·4A, – 16W
1. 4A, 160W
1. – 4A, – 160W

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The given circuit

Apply KCL at node A
2I_x + V_A –0 = I_x
(400 + 200)Ω

or
I_x = - V_A
600
…(i)
From figure
I_x = 40 - V_A
500
…(ii)
From equation (i) and (ii)
– V_A = 40 – V_A
600 500

or
– 5V_A = 240 – 6V_A
or
V_A = 240V
and
I_x = – V_A = – 240 = – 0·4A
600 600

Power delivered by the independent source is
= 40 × I_x = 40 × (– 0·4) = – 16 W.

Correct Option: B

The given circuit

Apply KCL at node A
2I_x + V_A –0 = I_x
(400 + 200)Ω

or
I_x = - V_A
600
…(i)
From figure
I_x = 40 - V_A
500
…(ii)
From equation (i) and (ii)
– V_A = 40 – V_A
600 500

or
– 5V_A = 240 – 6V_A
or
V_A = 240V
and
I_x = – V_A = – 240 = – 0·4A
600 600

Power delivered by the independent source is
= 40 × I_x = 40 × (– 0·4) = – 16 W.

The circuit shown in figure below is equivalent to a load of—

1. 4 Ω
  3
1. 8 Ω
  3
1. 4Ω
1. 2Ω

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The given circuit

The load is given by ratio of voltage, V to the current, I
V = 4 (I – I₁) …(i)
and
V = 2I₁ + 2I …(ii)
From equation (i) and (ii):
4I – 4I₁ = 2I₁ + 2I
or
2I = 6I₁
or
I = 3I₁
or
I₁ = I/3
Now,
V = 4 I - I = 8I
3 3

or
V = 8 Ω
I 3

Hence alternative (B) is the correct choice.

Correct Option: B

The given circuit

The load is given by ratio of voltage, V to the current, I
V = 4 (I – I₁) …(i)
and
V = 2I₁ + 2I …(ii)
From equation (i) and (ii):
4I – 4I₁ = 2I₁ + 2I
or
2I = 6I₁
or
I = 3I₁
or
I₁ = I/3
Now,
V = 4 I - I = 8I
3 3

or
V = 8 Ω
I 3

Hence alternative (B) is the correct choice.