Network theory miscellaneous Moderate Questions and Answers | Page - 14

Network theory miscellaneous

For the circuit shown below, the potential difference between points P and Q is—

1. 12V
1. 10V
1. – 6V
1. 8V

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The given circuit:

KCL at node
V_P - V_R + V_P - 0 + 2 = 0
2 8

or
V_P - 10 + V_P + 2 = 0
2 8

or
V_P 1 + 1 = 3
2 8

or
V_P = 3 × 8 = 24 V
5 5

KCL at node Q
V_Q - V_R + V_Q - 0 = 2
4 6

or V_Q = V_Q = 2 + 10
4 6 4

or
V_Q 1 + 1 = 8 + 10
4 6 4

or
V_Q = 18 × 24 = 54
4 10 5

Now,
V_Q -V_Q 24 - 54 = -30 = -6V
5 5 5

Correct Option: C

The given circuit:

KCL at node
V_P - V_R + V_P - 0 + 2 = 0
2 8

or
V_P - 10 + V_P + 2 = 0
2 8

or
V_P 1 + 1 = 3
2 8

or
V_P = 3 × 8 = 24 V
5 5

KCL at node Q
V_Q - V_R + V_Q - 0 = 2
4 6

or V_Q = V_Q = 2 + 10
4 6 4

or
V_Q 1 + 1 = 8 + 10
4 6 4

or
V_Q = 18 × 24 = 54
4 10 5

Now,
V_Q -V_Q 24 - 54 = -30 = -6V
5 5 5

A dc circuit shown below has a voltage V, a current source I and several resistors. A particular resistor R dissipates a power of 4W when V alone is active. The same resistor dissipates a power of 9W when I alone is active. The power dissipated by R when both sources are active will be—

1. 1W
1. 5W
1. 13W
1. 25W

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The given circuit:

Resistive network Applying superposition theorem (since here one energy source is acting at a time)
● When only voltage source is active:
V²/R = 4W = I²R
Let R = 1Ω
Assume I₁ = I = 4 = 2A
● When only current source is active
I²R = 9W
or
I² = 9W
Assume
I₂2 = I² = 9
or
I₂ = 3A
Since current flowing in both the case follow the same direction. Therefore, net current will add up.
So,
I = I₁ + I₂ = 2 + 3 = 5A
Now, the power dissipated by R when both sources are active will be:
P=I₂R = 52.1 = 25W
Hence alternative (D) is the correct choice.

Correct Option: D

The given circuit:

Resistive network Applying superposition theorem (since here one energy source is acting at a time)
● When only voltage source is active:
V²/R = 4W = I²R
Let R = 1Ω
Assume I₁ = I = 4 = 2A
● When only current source is active
I²R = 9W
or
I² = 9W
Assume
I₂2 = I² = 9
or
I₂ = 3A
Since current flowing in both the case follow the same direction. Therefore, net current will add up.
So,
I = I₁ + I₂ = 2 + 3 = 5A
Now, the power dissipated by R when both sources are active will be:
P=I₂R = 52.1 = 25W
Hence alternative (D) is the correct choice.

What are the source voltage and source resistance, respectively for the Thevenin’s equivalent circuit as seen from the terminals indicated in the circuit given above?

1. 2·0 V, 24Ω
1. 20V, 48Ω
1. 20V, 4·8Ω
1. 20V, 12Ω

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The given circuit:

Voltage at point A, V_A = 40 × 100 = 40V
40 + 60

Voltage at point B, V_B = 60 × 100 = 60V
40 + 60

V_{th>/sub> = V_BA = 60 – 40 = 20V
Equivalent circuit for calculating R_th is shown below:

From figue, R_th = R_AB = (60||40) + (40 ||60)
= 24 + 24 = 48Ω.}

Correct Option: B

The given circuit:

Voltage at point A, V_A = 40 × 100 = 40V
40 + 60

Voltage at point B, V_B = 60 × 100 = 60V
40 + 60

V_{th>/sub> = V_BA = 60 – 40 = 20V
Equivalent circuit for calculating R_th is shown below:

From figue, R_th = R_AB = (60||40) + (40 ||60)
= 24 + 24 = 48Ω.}

A network contains only independent current sources and resistors. If the values of all resistors are doubled, the values of node voltage—

1. will become half
1. will remain unchanged
1. will become double
1. cannot be determined unless the circuit configuration and the values of the resistors are known

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Let I₁, I₂, I₃, … independent current sources are connected in parallel with resistors R₁, R₂, R₃,…

V_A + V_A + V_A + .........= l₁+l₂+l₃+.....
R₁ R₂ R₃

or
V_A 1 + 1 + 1 + .........= l₁+l₂+l₃+.....
R₁ R₂ R₃

or
V_A = l₁ + l₂ + l₃ +.....
1 + 1 + 1
R₁ R₂ R₃

According to question, if the values of all resistors are doubled then
V_A = l₁ + l₂ + l₃ +.....
1 + 1 + 1
2R₁ 2R₂ 2R₃

or
V_A = 2 × l₁ + l₂ + l₃ +.....
1 + 1 + 1 +...............
R₁ R₂ R₃

Therefore the value of node voltage will become double.
Hence alternative (C) is the correct choice.

Correct Option: C

Let I₁, I₂, I₃, … independent current sources are connected in parallel with resistors R₁, R₂, R₃,…

V_A + V_A + V_A + .........= l₁+l₂+l₃+.....
R₁ R₂ R₃

or
V_A 1 + 1 + 1 + .........= l₁+l₂+l₃+.....
R₁ R₂ R₃

or
V_A = l₁ + l₂ + l₃ +.....
1 + 1 + 1
R₁ R₂ R₃

According to question, if the values of all resistors are doubled then
V_A = l₁ + l₂ + l₃ +.....
1 + 1 + 1
2R₁ 2R₂ 2R₃

or
V_A = 2 × l₁ + l₂ + l₃ +.....
1 + 1 + 1 +...............
R₁ R₂ R₃

Therefore the value of node voltage will become double.
Hence alternative (C) is the correct choice.

Consider the circuit shown below. What is the power delivered by the 24V source?

1. 96W
1. 144W
1. 192W
1. 288W

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The given circuit:

From figure we conclude that the magnitude of current dependent source is 2 times that of 6Ω branch. It means resistance of dependent sources is half of the 6Ω, i.e., 3Ω
I = V = 24 = 24 = 12A
R_eq 3|| 6 2

Now, the power delivered by 24V source

= VI = 24 × 12 = 288W.
Alternative Method:
From figure, I = I_R + 2I_R = 3I_R
Applying KVL in the input loop
24 = 6I_R
or
I_R = 4A
and
I = 3I_R = 3 × 4 = 12A
Now, power delivered by 24V source is:
VI = 24 × 12 = 288W.

Correct Option: D

The given circuit:

From figure we conclude that the magnitude of current dependent source is 2 times that of 6Ω branch. It means resistance of dependent sources is half of the 6Ω, i.e., 3Ω
I = V = 24 = 24 = 12A
R_eq 3|| 6 2

Now, the power delivered by 24V source

= VI = 24 × 12 = 288W.
Alternative Method:
From figure, I = I_R + 2I_R = 3I_R
Applying KVL in the input loop
24 = 6I_R
or
I_R = 4A
and
I = 3I_R = 3 × 4 = 12A
Now, power delivered by 24V source is:
VI = 24 × 12 = 288W.