Network theory miscellaneous Difficult Questions and Answers | Page - 1

Network theory miscellaneous

For the circuit shown below, the V₁ = ?

1. V₁ = 40V, V₂ = 15V
1. V₁ = – 40V, V₂ = – 15V
1. V₁ = 15V, V₂ = 40V
1. V₁ = – 15V, V₂ = – 40V

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The given circuit

Applying KCL at node A
V₁ - 20 + V₁ - 0.5V₁ - V₂ = 2
20 5

or
V₁ 1 + 1 - V₂ = 3
20 10 5

or
3V₁ - V₂ = 3
20 5

or
3V₁ – 4V₂ = 60 . . . . . . …(i)
KCL at node B
V₂ - V₁ + 0.5V₁ + V₂ + V₂ - 40 = 4
5 2 10

V₂ 1 + 1 + 1 - V₁ = 8
5 2 10 10

or
V₂ 2 + 5 + 1 - V₁ = 8
10 8

or
8V₁ – V₁ = 80 …. . . (ii)
From equation (i) and (ii)
V₁ = 40V and V₂ = 15V
Hence alternative (A) is the correct choice.

Correct Option: A

The given circuit

Applying KCL at node A
V₁ - 20 + V₁ - 0.5V₁ - V₂ = 2
20 5

or
V₁ 1 + 1 - V₂ = 3
20 10 5

or
3V₁ - V₂ = 3
20 5

or
3V₁ – 4V₂ = 60 . . . . . . …(i)
KCL at node B
V₂ - V₁ + 0.5V₁ + V₂ + V₂ - 40 = 4
5 2 10

V₂ 1 + 1 + 1 - V₁ = 8
5 2 10 10

or
V₂ 2 + 5 + 1 - V₁ = 8
10 8

or
8V₁ – V₁ = 80 …. . . (ii)
From equation (i) and (ii)
V₁ = 40V and V₂ = 15V
Hence alternative (A) 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 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 circuit shown in figure 1 is replaced by that in figure 2. If current ‘I’ remains the same, then R₀ will be—

1. Zero
1. R
1. 2R
1. 4R

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The given circuit:

In order to make current-I same in both figures:
1 = 1 + R₀
1/32R + 1/16R + 1/8R 1/4R + 1/2R + 1/R

or
32R = 4R + R₀
7 7

or
R₀ = 32R - 4R = 28R
7 7 7

or
R₀ = 4R

Correct Option: D

The given circuit:

In order to make current-I same in both figures:
1 = 1 + R₀
1/32R + 1/16R + 1/8R 1/4R + 1/2R + 1/R

or
32R = 4R + R₀
7 7

or
R₀ = 32R - 4R = 28R
7 7 7

or
R₀ = 4R

For the circuit shown below, the current delivered by 24V source—

1. 1·142A
1. – 1·142A
1. 2·200A
1. – 2·284A

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The given circuit:

Applying KCL at node A, we get:
V_A - 24 + V_A - 0 + V_A - 36 = 2
5 2 10

or
V_A 1 + 1 + 1 = 2 + 24 + 36
5 2 10 5 10

or
V_A 2 + 5 + 1 = 20 + 24 × 2 + 36
10 10

or
V_A = 104 = 13
8

The current delivered by 24V source,
I = 24 + 13 = 2.2A
5

Hence alternative (C) is the correct choice.

Correct Option: C

The given circuit:

Applying KCL at node A, we get:
V_A - 24 + V_A - 0 + V_A - 36 = 2
5 2 10

or
V_A 1 + 1 + 1 = 2 + 24 + 36
5 2 10 5 10

or
V_A 2 + 5 + 1 = 20 + 24 × 2 + 36
10 10

or
V_A = 104 = 13
8

The current delivered by 24V source,
I = 24 + 13 = 2.2A
5

Hence alternative (C) is the correct choice.