# Network Theory Questions and Answers – Image Parameters

This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Image Parameters”.

1. A network is said to be symmetrical if the relation between A and D is?
a) A = D
b) A = 2 D
c) A = 3 D
d) A = 4 D

Explanation: We know V1=AV2-BI2 and I1=CV2-DI2. If the network is symmetrical, then the relation between A and D is A = D.

2. The relation between Z11 and Z12 if the network is symmetrical is?
a) Z11 = 2 Z12
b) Z11 = Z12
c) Z11 = 3 Z12
d) Z11 = 4 Z12

Explanation: For a network to be symmetrical A=D. So the relation between Z11 and Z12 for the network is symmetrical is Z11 = Z12.

3. The relation between Z12 and Z11 and B and C parameters if the network is symmetrical is?
a) Z11 = Z12 = B/C
b) Z11 = Z12 = C/B
c) Z11 = Z12 = √(B/C)
d) Z11 = Z12 = √(C/B)

Explanation: For symmetrical network, A=D. On substituting this we get the relation between Z12 and Z11 and B and C parameters if the network is symmetrical is Z11 = Z12 = √(B/C).

4. Determine the transmission parameter A in the circuit shown below.

a) 3/4
b) 4/3
c) 5/6
d) 6/5

Explanation: We know V1=AV2-BI2 and I1=CV2-DI2. A=(V1/V2) |I2=0. On solving we get the transmission parameter A as A = 6/5.

5. Determine the transmission parameter B in the circuit shown below.

a) 17/5
b) 5/17
c) 13/5
d) 5/13

Explanation: The transmission parameter B is -V1/I2 |V2=0. On short cicuiting the port 2, from the circuit we get -I2 = (5/17)V1 => -V1/I2 = 17/5. On substituting we get B = 17/5.
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6. Determine the transmission parameter C in the circuit shown below.

a) 2/5
b) 1/5
c) 4/5
d) 3/5

Explanation: The transmission parameter C is I1/V2 |I2=0. This parameter is obtained by open circuiting the port 2. So we get V2 = 5I1 => I1/V2 = 1/5. On substituting, we get C = 1/5.

7. Determine the transmission parameter D in the circuit shown below.

a) 3/5
b) 4/5
c) 7/5
d) 2/5

Explanation: The transmission parameter D is -I1/I2 |V2=0. This is obtained by short circuiting the port 2. We get I1 = (7/17)V1 and -I2 = (5/17)V1. On solving, we get -I1/I2 = 7/5. On substituting we get D = 7/5.

8. The value of Z11 in the circuit shown below is?

a) 1.8
b) 2.8
c) 3.8
d) 4.8

Explanation: The relation between Z11 and ABCD parameters is Z11=√(AB/CD). We know A = 6/5, B = 17/5, C = 1/5, D = 7/5. On substituting, Z11 = √((6/5×17/5)/(1/5×7/5)) = 3.8Ω.

9. The value of Z12 in the circuit shown below is?

a) 1.1
b) 2.2
c) 3.3
d) 4.4

Explanation: The relation between Z12 and ABCD parameters is Z12=√(BD/AC). We got B = 17/5, D = 7/5, A = 6/5, C = 1/5. On substituting Z12 = √(BD/AC) = √((17/5×7/5)/(6/5×1/5)) = 4.4Ω.

10. Determine the value of Ø in the circuit shown below.

a) 0.25
b) 0.5
c) 0.75
d) 1

Explanation: Ø is called image transfer constant and it is also used to describe reciprocal networks and this parameter is obtained from the voltage and current ratios. We know Ø = tanh-1⁡√(BC/AD) = tanh-1⁡√(17/42) = 0.75.

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