Network Elements and the Concept of Circuit Easy Questions and Answers

Network Elements and the Concept of Circuit

For the circuit shown below. Power is delivered by—

1. Dependent source of 192 W
1. Dependent source of 368 W
1. Independent source of 192 W
1. Independent source of 436 W

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The given circuit → Circuit correction

KCL at node A.
V_A – 40 + V_A – 0 = 0 ...................(i)
500 600

V_A = 2i_x ........................(ii)
Again from given circuit
i_x = 40 – V_A
500

V_A 1 + 1 = 40
500 600 500

or V_A 1100 = 40
500 × 600 500

or V_A = 40 × 600
1100

or V_A = 21.82 V
l_x = V_A = 21.82 = 10.90 A
2 2

Since lx is positive. So power is delivered by independent source i.e. P_delivered = 40 × l_x
or = 40 × 10.90 = 436 W

Correct Option: D

The given circuit → Circuit correction

KCL at node A.
V_A – 40 + V_A – 0 = 0 ...................(i)
500 600

V_A = 2i_x ........................(ii)
Again from given circuit
i_x = 40 – V_A
500

V_A 1 + 1 = 40
500 600 500

or V_A 1100 = 40
500 × 600 500

or V_A = 40 × 600
1100

or V_A = 21.82 V
l_x = V_A = 21.82 = 10.90 A
2 2

Since lx is positive. So power is delivered by independent source i.e. P_delivered = 40 × l_x
or = 40 × 10.90 = 436 W

For the circuit shown below. The dependent source—

1. Supplies 40 W
1. Absorbs 50 W
1. Supplies 50 W
1. Absorbs 40 W

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Given, l_x = 5A
Voltage at dependent source terminal is
= – 2 i_x
= – 2 × 5 = – 10V
P = V × i_x
or P = 10 × 5 = 50 W
Since current is entering in the independent source. It means power is absorbed by the dependent source.

Correct Option: B

Given, l_x = 5A
Voltage at dependent source terminal is
= – 2 i_x
= – 2 × 5 = – 10V
P = V × i_x
or P = 10 × 5 = 50 W
Since current is entering in the independent source. It means power is absorbed by the dependent source.

Calculate Z parameter for the circuit given below—

1. 14 8
  13 13
  216 314
  117 117
1. 13 13
  14 8
  117 117
  216 314
1. 12 9
  13 13
  116 114
  117 117
1. None of these

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The given circuit can be redrawn as shown below

Applying KVL in loop 1.
V₁ = 2 (I₁ – I) ................(i)
KVL in loop 2
– 3I + 2V₁ – (I₂ + I) 4 + 2 (I₁ – I) = 0
or 9I = 2V₁ – 4 I₂ + 2I₁
or I = 2V₁ + 2I₁ – 4I₂ ...................(ii)
9

From equations (i) and (ii)
V₁ = 2I₁ – 2 (2V₁ + 2I₁ – 4I₂)
9

or 9 V₁ 1 + 4 = 18I₁ – 4I₁ + 8I₂
9

or 13V₁ = 14 I₁ + 8I₂
or V₁ = 14 I₁ + 8 I₂ ............(iii)
13 13

We know that
V₁ = Z₁₁ I₁ + Z₁₂ I₂ ..........(iv)
(for Z-parameter)
Comparing equations (iii) and (iv), we get
Z₁₁ = 14 , Z₁₂ = 8
13 13

Applying KVL in loop 3.
V₂ = 4 (I₂ + I)
or V₂ = 4 I₂ + 4 2V₁ + 2I₁ – 4I₂
9

[from equation (ii)]
or V₂ = 4 I₂ + 8 V₁ + 8 I₁ –16 I₂
9 9 9

or V₂ = 4 I₂ + 8 14 I₁ + 8 I₂ + 8 I₁ –16 I₂
9 13 13 9 9

or V₂ = I₂ 4 + 64 – 16 + I₁ 112 + 8
117 9 117 9

or V₂ = I₂ (466 + 64 – 208) + I₁ 112 + 104
117 117

or V₂ = 314 I₂ + 216 I₁ .......(v)
117 117

We know that
V₂ = Z₂₁ I₁ + Z₂₂ I₂ (for Z-parameters) ..............(vi)
Comparing equations (v) and (vi), we get
Z₂₁ = 216 , Z₂₂ = 314
117 117

Hence alternative (A) is the correct choice.

Correct Option: B

The given circuit can be redrawn as shown below

Applying KVL in loop 1.
V₁ = 2 (I₁ – I) ................(i)
KVL in loop 2
– 3I + 2V₁ – (I₂ + I) 4 + 2 (I₁ – I) = 0
or 9I = 2V₁ – 4 I₂ + 2I₁
or I = 2V₁ + 2I₁ – 4I₂ ...................(ii)
9

From equations (i) and (ii)
V₁ = 2I₁ – 2 (2V₁ + 2I₁ – 4I₂)
9

or 9 V₁ 1 + 4 = 18I₁ – 4I₁ + 8I₂
9

or 13V₁ = 14 I₁ + 8I₂
or V₁ = 14 I₁ + 8 I₂ ............(iii)
13 13

We know that
V₁ = Z₁₁ I₁ + Z₁₂ I₂ ..........(iv)
(for Z-parameter)
Comparing equations (iii) and (iv), we get
Z₁₁ = 14 , Z₁₂ = 8
13 13

Applying KVL in loop 3.
V₂ = 4 (I₂ + I)
or V₂ = 4 I₂ + 4 2V₁ + 2I₁ – 4I₂
9

[from equation (ii)]
or V₂ = 4 I₂ + 8 V₁ + 8 I₁ –16 I₂
9 9 9

or V₂ = 4 I₂ + 8 14 I₁ + 8 I₂ + 8 I₁ –16 I₂
9 13 13 9 9

or V₂ = I₂ 4 + 64 – 16 + I₁ 112 + 8
117 9 117 9

or V₂ = I₂ (466 + 64 – 208) + I₁ 112 + 104
117 117

or V₂ = 314 I₂ + 216 I₁ .......(v)
117 117

We know that
V₂ = Z₂₁ I₁ + Z₂₂ I₂ (for Z-parameters) ..............(vi)
Comparing equations (v) and (vi), we get
Z₂₁ = 216 , Z₂₂ = 314
117 117

Hence alternative (A) is the correct choice.

Calculate the Y parameter for the given network—

1. 2.5 3.5
  1.5 4.5
1. 2.5 3.5
  0.5 3.5
1. 1.5 3.0
  1.0 4.5
1. None of these

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Given network

Y-parameter for network

Standard equations of Y-parameters are
I'₁ = Y'₁₁ V₁ + Y'₁₂ V₂ ........(i)
I'₂ = Y'₂₁ V₁ + Y'₂₂ V₂ .......(ii)
I'₁ = – I'₂ (from above circuit) ...........(iii)
Y'₁₁ = I₁
V₁ _{v₂ = 0}

From above circuit V₁ = 2 I'₁ or I'₁ = 1 = 0.5 mho
V₁ 2

or Y'₁₁ = 0.5 mho
Y'₁₂ = I₁
V₂ _{v₁ = 0}

V₂ = I'₂ 2
or V₂ = – I'₁ 2
Y'₁₂ = I₁ = – 1 = – 0.5 mho
V₂ 2

Similarly
Y'₂₁ = I'₂
V₁ _{v₂ = 0}

or Y'₂₁ = – 0.5 mho and
Y'₂₂ = I₂
V₂ _{v₁ = 0}

or Y'₂₂ = 0.5 mho.
Now,
Y' = 0.5 –0.5 and
–0.5 0.5

The new Y parameter
Y = Y' + Y''
or Y = 0.5 –0.5 2 4
–0.5 0.5 1 3

or Y = 2.5 3.5 and
0.5 3.5

Correct Option: B

Given network

Y-parameter for network

Standard equations of Y-parameters are
I'₁ = Y'₁₁ V₁ + Y'₁₂ V₂ ........(i)
I'₂ = Y'₂₁ V₁ + Y'₂₂ V₂ .......(ii)
I'₁ = – I'₂ (from above circuit) ...........(iii)
Y'₁₁ = I₁
V₁ _{v₂ = 0}

From above circuit V₁ = 2 I'₁ or I'₁ = 1 = 0.5 mho
V₁ 2

or Y'₁₁ = 0.5 mho
Y'₁₂ = I₁
V₂ _{v₁ = 0}

V₂ = I'₂ 2
or V₂ = – I'₁ 2
Y'₁₂ = I₁ = – 1 = – 0.5 mho
V₂ 2

Similarly
Y'₂₁ = I'₂
V₁ _{v₂ = 0}

or Y'₂₁ = – 0.5 mho and
Y'₂₂ = I₂
V₂ _{v₁ = 0}

or Y'₂₂ = 0.5 mho.
Now,
Y' = 0.5 –0.5 and
–0.5 0.5

The new Y parameter
Y = Y' + Y''
or Y = 0.5 –0.5 2 4
–0.5 0.5 1 3

or Y = 2.5 3.5 and
0.5 3.5

For the circuit shown below. Calculate the value of load impedance when the load is pure resistive, under maximum power transfer condition—

1. (60 – j80)Ω
1. (60 + j80)Ω
1. 100Ω
1. None of these

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Under maximum power transfer condition, when load is pure resistive the load impedance is given by
R_L = √R² + (X_L or X_C)²
Here, R = 60Ω
X_C = 80Ω
Now, R_L = √60² + 80²
or R_L = 100Ω

Correct Option: C

Under maximum power transfer condition, when load is pure resistive the load impedance is given by
R_L = √R² + (X_L or X_C)²
Here, R = 60Ω
X_C = 80Ω
Now, R_L = √60² + 80²
or R_L = 100Ω

	V_A – 40	+	V_A – 0	= 0 ...................(i)
	500		600

i_x =	40 – V_A
	500

V_A		1	+	1		=	40
		500		600			500

or V_A			1100		=	40
			500 × 600			500

Network Elements and the Concept of Circuit

Network Elements and the Concept of Circuit

Network Theory

Correct Option: D

Correct Option: B

Correct Option: B

Correct Option: B

Correct Option: C