This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Short Circuit Admittance(Y) Parameters”.

1. In determining short circuit impedance parameters, among V_{1}, V_{2}, I_{1}, I_{2}, which of the following are dependent variables?

a) V_{1} and V_{2}

b) I_{1} and I_{2}

c) V_{1} and I_{2}

d) I_{1} and V_{2}

View Answer

Explanation: In determining short circuit impedance parameters, among V

_{1}, V

_{2}, I

_{1}, I

_{2}; I

_{1}and I

_{2}are dependent variables and V

_{1}, V

_{2}are independent variables i.e., dependent variables depend on independent variables.

2. In determining short circuit impedance parameters, among V_{1}, V_{2}, I_{1}, I_{2}, which of the following are independent variables?

a) I_{1} and V_{2}

b) V_{1} and I_{2}

c) I_{1} and I_{2}

d) V_{1} and V_{2}

View Answer

Explanation: In determining short circuit impedance parameters, among V

_{1}, V

_{2}, I

_{1}, I

_{2}; V

_{1}and V

_{2}are independent variables and I

_{1}, I

_{2}are dependent variables. Independent variables are the variables that do not depend on any other variable.

3. Which of the following expression is true in case of short circuit parameters?

a) I_{1} = Y_{11} V_{1} + Y_{12} V_{2}

b) I_{1} = Y_{11} I_{1} + Y_{12} V_{2}

c) V_{1} = Y_{11} I_{1} + Y_{12} V_{2}

d) V_{1} = Y_{11} V_{1} + Y_{12} V_{2}

View Answer

Explanation: The expression relating the short circuit parameters Y

_{11}, Y

_{12}and voltages V

_{1}, V

_{2}and current is I

_{1}, is I

_{1}= Y

_{11}V

_{1}+ Y

_{12}V

_{2}.

4. Which of the following expression is true in case of short circuit parameters?

a) I_{2} = Y_{21}I_{1} + Y_{22} I_{2}

b) V_{2} = Y_{21}I_{1} + Y_{22} V_{2}

c) I_{2} = Y_{21}V_{1} + Y_{22} V_{2}

d) I_{2} = Y_{21}V_{1} + Y_{22} I_{2}

View Answer

Explanation: The expression relating the voltages V

_{1}, V

_{2}and current is I

_{2}and short circuit parameters Y

_{11}, Y

_{12}is I

_{2}= Y

_{21}V

_{1}+ Y

_{22}V

_{2}.

5. The parameters Y_{11}, Y_{12}, Y_{21}, Y_{22} are called?

a) Open circuit impedance parameters

b) Short circuit admittance parameters

c) Inverse transmission parameters

d) Transmission parameters

View Answer

Explanation: The parameters Y

_{11}, Y

_{12}, Y

_{21}, Y

_{22}are called short circuit admittance parameters also called network functions as they are obtained by short circuiting port 1 or port 2.

6. Find the Y – parameter Y_{11} in the circuit shown below.

a) 2

b) 3/2

c) 1

d) 1/2

View Answer

Explanation: After short circuiting b-b’, the equation will be V

_{1}= (I

_{1}) x 2. We know Y

_{11}= I

_{1}/V

_{1}. From the equation we get I

_{1}/V

_{1}= 2. On substituting we get Y

_{11}= 2 mho.

7. Find the Y – parameter Y_{21} in the circuit shown in question 6.

a) -1/4

b) 1/4

c) 1/2

d) -1/2

View Answer

Explanation: After short circuiting b-b’, the equation will be -I

_{2}=I

_{1}× 2/4=I

_{1}/2 and -I

_{2}= V

_{1}/4 and on solving and substituting we get Y

_{21}=I

_{2}/V

_{1}=-1/4 mho.

8. Find the Y – parameter Y_{22} in the circuit shown in question 6.

a) 3/8

b) 5/8

c) 7/8

d) 9/8

View Answer

Explanation: On short circuiting a-a’,we get Z

_{eq}= 8/5 Ω. V

_{2}=I

_{2}× 8/5. We know Y

_{22}= I

_{2}/V

_{2}. We got I

_{2}/V

_{1}= 5/8. ON substituting we get Y

_{22}= 5/8 mho.

9. Find the Y – parameter Y_{12} in the circuit shown in question 6.

a) 1/2

b) -1/2

c) -1/4

d) 1/4

View Answer

Explanation: Short circuiting a-a’, -I

_{1}= 2/5 I

_{2}and I

_{2}= 5 V

_{2}/8. On solving -I

_{1}= 2/5×5/8 V

_{2}= V

_{2}/4. We know

Y

_{12}= I

_{1}/V

_{2}. We got I

_{1}/V

_{2}= -1/4. So the value of Y

_{12}will be -1/4 mho.

10. Which of the following equation is true in the circuit shown in question 6?

a) I_{1}=0.5(V_{1})+0.25(V_{2})

b) I_{1}=0.25(V_{1})+0.625(V_{2})

c) I_{1}=-0.25(V_{1})+0.625(V_{2})

d) I_{1}=0.5(V_{1})-0.25(V_{2})

View Answer

Explanation: We got the admittance parameters as Y

_{11}= 0.5, Y

_{12}= -0.25, Y

_{21}= -0.25, Y

_{22}= 0.625. So the equations in terms of admittance parameters is

I

_{1}=0.5(V

_{1})-0.25(V

_{2}) and I

_{2}=-0.25(V

_{1})+0.625(V

_{2}).

**Sanfoundry Global Education & Learning Series – Network Theory.**

To practice all areas of Network Theory, __here is complete set of 1000+ Multiple Choice Questions and Answers__.