250+ TOP MCQs on Relation between Transmission Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters and Answers

Network Theory Puzzles on “Relation between Transmission Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters”.

1. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₁₁ in terms of Transmission parameters can be expressed as ________
A. y₁₁ = (frac{D}{B})
B. y₁₁ = (frac{C-A}{B})
C. y₁₁ = – (frac{1}{B})
D. y₁₁ = (frac{A}{B})

Answer: A
Clarification: We know that, V₁ = AV₂ – BI₂ ……… (1)
I₁ = CV₂ – DI₂ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₂ = (frac{A}{B}V_2 – frac{1}{B}V_1) …………. (5)
And I₁ = CV₂ – D (left(frac{A}{B}V_2 – frac{1}{B}V_1right) = frac{D}{B}V_1 + left(C-frac{A}{B}right) V_2) …………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = (frac{D}{B})
y₁₂ = (frac{C-A}{B})
y₂₁ = – (frac{1}{B})
y₂₂ = (frac{A}{B}).

2. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₁₂ in terms of Transmission parameters can be expressed as ________
A. y₁₂ = (frac{D}{B})
B. y₁₂ = (frac{C-A}{B})
C. y₁₂ = – (frac{1}{B})
D. y₁₂ = (frac{A}{B})

Answer: B
Clarification: We know that, V₁ = AV₂ – BI₂ ……… (1)
I₁ = CV₂ – DI₂ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₂ = (frac{A}{B}V_2 – frac{1}{B}V_1) …………. (5)
And I₁ = CV₂ – D (left(frac{A}{B}V_2 – frac{1}{B}V_1right) = frac{D}{B}V_1 + left(C-frac{A}{B}right) V_2) …………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = (frac{D}{B})
y₁₂ = (frac{C-A}{B})
y₂₁ = – (frac{1}{B})
y₂₂ = (frac{A}{B}).

3. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₂₁ in terms of Transmission parameters can be expressed as ________
A. Y₂₁ = (frac{D}{B})
B. Y₂₁ = (frac{C-A}{B})
C. Y₂₁ = – (frac{1}{B})
D. Y₂₁ = (frac{A}{B})

Answer: C
Clarification: We know that, V₁ = AV₂ – BI₂ ……… (1)
I₁ = CV₂ – DI₂ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₂ = (frac{A}{B}V_2 – frac{1}{B}V_1) …………. (5)
And I₁ = CV₂ – D (left(frac{A}{B}V_2 – frac{1}{B}V_1right) = frac{D}{B}V_1 + left(C-frac{A}{B}right) V_2) …………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = (frac{D}{B})
y₁₂ = (frac{C-A}{B})
y₂₁ = – (frac{1}{B})
y₂₂ = (frac{A}{B}).

4. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₂₂ in terms of Transmission parameters can be expressed as ________
A. y₂₂ = (frac{D}{B})
B. y₂₂ = (frac{C-A}{B})
C. y₂₂ = – (frac{1}{B})
D. y₂₂ = (frac{A}{B})

Answer: D
Clarification: We know that, V₁ = AV₂ – BI₂ ……… (1)
I₁ = CV₂ – DI₂ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₂ = (frac{A}{B}V_2 – frac{1}{B}V_1) …………. (5)
And I₁ = CV₂ – D (left(frac{A}{B}V_2 – frac{1}{B}V_1right) = frac{D}{B}V_1 + left(C-frac{A}{B}right) V_2) …………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = (frac{D}{B})
y₁₂ = (frac{C-A}{B})
y₂₁ = – (frac{1}{B})
y₂₂ = (frac{A}{B}).

5. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₁₁ in terms of Transmission parameters can be expressed as ____________
A. z₁₁ = (frac{A}{C})
B. z₁₁ = (frac{AD}{C – B})
C. z₁₁ = (frac{1}{C})
D. z₁₁ = (frac{D}{C})

Answer: A
Clarification: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₁ = AV₂ – BI₂ ……… (3)
I₁ = CV₂ – DI₂ …………… (4)
Rewriting (3) and (4), we get,
V₂ = (frac{1}{C}I_1 + frac{D}{C}I_2) …………… (5)
And V₁ = (A left(frac{1}{C}I_1 + frac{D}{C}I_2right) – BI_2 = frac{A}{C}I_1 + left(frac{AD}{C} – Bright) I_2) ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = (frac{A}{C})
z₁₂ = (frac{AD}{C – B})
z₂₁ = (frac{1}{C})
z₂₂ = (frac{D}{C}).

6. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₁₂ in terms of Transmission parameters can be expressed as ____________
A. z₁₂ = (frac{A}{C})
B. z₁₂ = (frac{AD}{C – B})
C. z₁₂ = (frac{1}{C})
D. z₁₂ = (frac{D}{C})

Answer: B
Clarification: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₁ = AV₂ – BI₂ ……… (3)
I₁ = CV₂ – DI₂ …………… (4)
Rewriting (3) and (4), we get,
V₂ = (frac{1}{C}I_1 + frac{D}{C}I_2) …………… (5)
And V₁ = (A left(frac{1}{C}I_1 + frac{D}{C}I_2right) – BI_2 = frac{A}{C}I_1 + left(frac{AD}{C} – Bright) I_2) ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = (frac{A}{C})
z₁₂ = (frac{AD}{C – B})
z₂₁ = (frac{1}{C})
z₂₂ = (frac{D}{C}).

7. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₂₁ in terms of Transmission parameters can be expressed as ____________
A. z₂₁ = (frac{A}{C})
B. z₂₁ = (frac{AD}{C – B})
C. z₂₁ = (frac{1}{C})
D. z₂₁ = (frac{D}{C})

Answer: C
Clarification: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₁ = AV₂ – BI₂ ……… (3)
I₁ = CV₂ – DI₂ …………… (4)
Rewriting (3) and (4), we get,
V₂ = (frac{1}{C}I_1 + frac{D}{C}I_2) …………… (5)
And V₁ = (A left(frac{1}{C}I_1 + frac{D}{C}I_2right) – BI_2 = frac{A}{C}I_1 + left(frac{AD}{C} – Bright) I_2) ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = (frac{A}{C})
z₁₂ = (frac{AD}{C – B})
z₂₁ = (frac{1}{C})
z₂₂ = (frac{D}{C}).

8. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₂₂ in terms of Transmission parameters can be expressed as ____________
A. z₂₂ = (frac{A}{C})
B. z₂₂ = (frac{AD}{C – B})
C. z₂₂ = (frac{1}{C})
D. z₂₂ = (frac{D}{C})

Answer: D
Clarification: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₁ = AV₂ – BI₂ ……… (3)
I₁ = CV₂ – DI₂ …………… (4)
Rewriting (3) and (4), we get,
V₂ = (frac{1}{C}I_1 + frac{D}{C}I_2) …………… (5)
And V₁ = (A left(frac{1}{C}I_1 + frac{D}{C}I_2right) – BI_2 = frac{A}{C}I_1 + left(frac{AD}{C} – Bright) I_2) ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = (frac{A}{C})
z₁₂ = (frac{AD}{C – B})
z₂₁ = (frac{1}{C})
z₂₂ = (frac{D}{C}).

9. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₁₁ in terms of Inverse Transmission parameters can be expressed as ________
A. y₁₁ = (frac{A’}{B’})
B. y₁₁ = – (frac{1}{B’})
C. y₁₁ = (left(C’ – frac{D’ A’}{B’}right))
D. y₁₁ = (frac{D’}{B’})

Answer: A
Clarification: We know that, V₂ = A’V₁ – B’I₁ ……… (1)
I₂ = C’V₁ – D’I₁ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₁ = – (frac{1}{B’} V_2 + frac{A’}{B’} V_1) …………. (5)
And I₂ = C’V₁ – D’ (left(- frac{1}{B’} V_2 + frac{A’}{B’} V_1right) = left(C’ – frac{D’ A’}{B’}right) V_1 + frac{D’}{B’} V_2) ………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = (frac{A’}{B’})
y₁₂ = – (frac{1}{B’})
y₂₁ = (left(C’ – frac{D’ A’}{B’}right))
y₂₂ = (frac{D’}{B’}).

10. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₁₂ in terms of Inverse Transmission parameters can be expressed as ________
A. y₁₂ = (frac{A’}{B’})
B. y₁₂ = – (frac{1}{B’})
C. y₁₂ = (left(C’ – frac{D’ A’}{B’}right))
D. y₁₂ = (frac{D’}{B’})

Answer: B
Clarification: We know that, V₂ = A’V₁ – B’I₁ ……… (1)
I₂ = C’V₁ – D’I₁ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₁ = – (frac{1}{B’} V_2 + frac{A’}{B’} V_1) …………. (5)
And I₂ = C’V₁ – D’ (left(- frac{1}{B’} V_2 + frac{A’}{B’} V_1right) = left(C’ – frac{D’ A’}{B’}right) V_1 + frac{D’}{B’} V_2) ………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = (frac{A’}{B’})
y₁₂ = – (frac{1}{B’})
y₂₁ = (left(C’ – frac{D’ A’}{B’}right))
y₂₂ = (frac{D’}{B’}).

11. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₂₁ in terms of Inverse Transmission parameters can be expressed as ________
A. y₂₁ = (frac{A’}{B’})
B. y₂₁ = – (frac{1}{B’})
C. y₂₁ = (left(C’ – frac{D’ A’}{B’}right))
D. y₂₁ = (frac{D’}{B’})

Answer: C
Clarification: We know that, V₂ = A’V₁ – B’I₁ ……… (1)
I₂ = C’V₁ – D’I₁ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₁ = – (frac{1}{B’} V_2 + frac{A’}{B’} V_1) …………. (5)
And I₂ = C’V₁ – D’ (left(- frac{1}{B’} V_2 + frac{A’}{B’} V_1right) = left(C’ – frac{D’ A’}{B’}right) V_1 + frac{D’}{B’} V_2) ………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = (frac{A’}{B’})
y₁₂ = – (frac{1}{B’})
y₂₁ = (left(C’ – frac{D’ A’}{B’}right))
y₂₂ = (frac{D’}{B’}).

12. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₂₂ in terms of Inverse Transmission parameters can be expressed as ________
A. y₂₂ = (frac{A’}{B’})
B. y₂₂ = – (frac{1}{B’})
C. y₂₂ = (left(C’ – frac{D’ A’}{B’}right))
D. y₂₂ = (frac{D’}{B’})

Answer: D
Clarification: We know that, V₂ = A’V₁ – B’I₁ ……… (1)
I₂ = C’V₁ – D’I₁ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₁ = – (frac{1}{B’} V_2 + frac{A’}{B’} V_1) …………. (5)
And I₂ = C’V₁ – D’ (left(- frac{1}{B’} V_2 + frac{A’}{B’} V_1right) = left(C’ – frac{D’ A’}{B’}right) V_1 + frac{D’}{B’} V_2) ………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = (frac{A’}{B’})
y₁₂ = – (frac{1}{B’})
y₂₁ = (left(C’ – frac{D’ A’}{B’}right))
y₂₂ = (frac{D’}{B’}).

13. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₁₁ in terms of Transmission parameters can be expressed as ____________
A. z₁₁ = (frac{D’}{C’})
B. z₁₁ = (frac{1}{C’})
C. z₁₁ = (left(frac{A’ D’}{C’} – B’right))
D. z₁₁ = (frac{A’}{C’})

Answer: A
Clarification: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₂ = A’V₁ – B’I₁ ……… (3)
I₂ = C’V₁ – D’I₁ …………… (4)
Rewriting (3) and (4), we get,
V₂ = A’ (left(frac{D’}{C’} I_1 + frac{1}{C’} I_2right) – B’I_1 = left(frac{A’ D’}{C’} – B’right) I_1 + frac{A’}{C’} I_2) ………… (5)
And V₁ = (frac{D’}{C’} I_1 + frac{1}{C’} I_2) ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = (frac{D’}{C’})
z₁₂ = (frac{1}{C’})
z₂₁ = (left(frac{A’ D’}{C’} – B’right))
z₂₂ = (frac{A’}{C’}).

14. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₁₂ in terms of Transmission parameters can be expressed as ____________
A. z₁₂ = (frac{D’}{C’})
B. z₁₂ = (frac{1}{C’})
C. z₁₂ = (left(frac{A’ D’}{C’} – B’right))
D. z₁₂ = (frac{A’}{C’})

Answer: B
Clarification: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₂ = A’V₁ – B’I₁ ……… (3)
I₂ = C’V₁ – D’I₁ …………… (4)
Rewriting (3) and (4), we get,
V₂ = A’ (left(frac{D’}{C’} I_1 + frac{1}{C’} I_2right) – B’I_1 = left(frac{A’ D’}{C’} – B’right) I_1 + frac{A’}{C’} I_2) ………… (5)
And V₁ = (frac{D’}{C’} I_1 + frac{1}{C’} I_2) ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = (frac{D’}{C’})
z₁₂ = (frac{1}{C’})
z₂₁ = (left(frac{A’ D’}{C’} – B’right))
z₂₂ = (frac{A’}{C’}).

15. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₂₂ in terms of Transmission parameters can be expressed as ____________
A. z₂₂ = (frac{D’}{C’})
B. z₂₂ = (frac{1}{C’})
C. z₂₂ = (left(frac{A’ D’}{C’} – B’right))
D. z₂₂ = (frac{A’}{C’})

Answer: D
Clarification: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₂ = A’V₁ – B’I₁ ……… (3)
I₂ = C’V₁ – D’I₁ …………… (4)
Rewriting (3) and (4), we get,
V₂ = A’ (left(frac{D’}{C’} I_1 + frac{1}{C’} I_2right) – B’I_1 = left(frac{A’ D’}{C’} – B’right) I_1 + frac{A’}{C’} I_2) ………… (5)
And V₁ = (frac{D’}{C’} I_1 + frac{1}{C’} I_2) ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = (frac{D’}{C’})
z₁₂ = (frac{1}{C’})
z₂₁ = (left(frac{A’ D’}{C’} – B’right))
z₂₂ = (frac{A’}{C’}).