Network Theory Questions and Answers – Relation between Transmission Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters

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This set of Network Theory Puzzles focuses 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 = \(\frac{A}{B}\)
View Answer

Answer: a
Explanation: 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 \(\left(\frac{A}{B}V_2 – \frac{1}{B}V_1\right) = \frac{D}{B}V_1 + \left(C-\frac{A}{B}\right) V_2\) …………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y11 = \(\frac{D}{B}\)
y12 = \(\frac{C-A}{B}\)
y21 = – \(\frac{1}{B}\)
y22 = \(\frac{A}{B}\).
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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 = \(\frac{A}{B}\)
View Answer

Answer: b
Explanation: 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 \(\left(\frac{A}{B}V_2 – \frac{1}{B}V_1\right) = \frac{D}{B}V_1 + \left(C-\frac{A}{B}\right) V_2\) …………… (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 = \(\frac{A}{B}\)
View Answer

Answer: c
Explanation: 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 \(\left(\frac{A}{B}V_2 – \frac{1}{B}V_1\right) = \frac{D}{B}V_1 + \left(C-\frac{A}{B}\right) V_2\) …………… (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 = \(\frac{A}{B}\)
View Answer

Answer: d
Explanation: 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 \(\left(\frac{A}{B}V_2 – \frac{1}{B}V_1\right) = \frac{D}{B}V_1 + \left(C-\frac{A}{B}\right) V_2\) …………… (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 = \(\frac{D}{C}\)
View Answer

Answer: a
Explanation: 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 = \(A \left(\frac{1}{C}I_1 + \frac{D}{C}I_2\right) – BI_2 = \frac{A}{C}I_1 + \left(\frac{AD}{C} – B\right) I_2\) ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z11 = \(\frac{A}{C}\)
z12 = \(\frac{AD}{C – B}\)
z21 = \(\frac{1}{C}\)
z22 = \(\frac{D}{C}\).
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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 = \(\frac{D}{C}\)
View Answer

Answer: b
Explanation: 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 = \(A \left(\frac{1}{C}I_1 + \frac{D}{C}I_2\right) – BI_2 = \frac{A}{C}I_1 + \left(\frac{AD}{C} – B\right) I_2\) ………….. (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 = \(\frac{D}{C}\)
View Answer

Answer: c
Explanation: 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 = \(A \left(\frac{1}{C}I_1 + \frac{D}{C}I_2\right) – BI_2 = \frac{A}{C}I_1 + \left(\frac{AD}{C} – B\right) I_2\) ………….. (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 = \(\frac{D}{C}\)
View Answer

Answer: d
Explanation: 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 = \(A \left(\frac{1}{C}I_1 + \frac{D}{C}I_2\right) – BI_2 = \frac{A}{C}I_1 + \left(\frac{AD}{C} – B\right) I_2\) ………….. (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 = \(\frac{D’}{B’}\)
View Answer

Answer: a
Explanation: 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’ \(\left(- \frac{1}{B’} V_2 + \frac{A’}{B’} V_1\right) = \left(C’ – \frac{D’ A’}{B’}\right) V_1 + \frac{D’}{B’} V_2\) ………… (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 = \(\frac{D’}{B’}\).
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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 = \(\frac{D’}{B’}\)
View Answer

Answer: b
Explanation: 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’ \(\left(- \frac{1}{B’} V_2 + \frac{A’}{B’} V_1\right) = \left(C’ – \frac{D’ A’}{B’}\right) V_1 + \frac{D’}{B’} V_2\) ………… (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 = \(\frac{D’}{B’}\)
View Answer

Answer: c
Explanation: 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’ \(\left(- \frac{1}{B’} V_2 + \frac{A’}{B’} V_1\right) = \left(C’ – \frac{D’ A’}{B’}\right) V_1 + \frac{D’}{B’} V_2\) ………… (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 = \(\frac{D’}{B’}\)
View Answer

Answer: d
Explanation: 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’ \(\left(- \frac{1}{B’} V_2 + \frac{A’}{B’} V_1\right) = \left(C’ – \frac{D’ A’}{B’}\right) V_1 + \frac{D’}{B’} V_2\) ………… (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 = \(\frac{A’}{C’}\)
View Answer

Answer: a
Explanation: 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’ \(\left(\frac{D’}{C’} I_1 + \frac{1}{C’} I_2\right) – B’I_1 = \left(\frac{A’ D’}{C’} – B’\right) I_1 + \frac{A’}{C’} I_2\) ………… (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 = \(\frac{A’}{C’}\).
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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 = \(\frac{A’}{C’}\)
View Answer

Answer: b
Explanation: 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’ \(\left(\frac{D’}{C’} I_1 + \frac{1}{C’} I_2\right) – B’I_1 = \left(\frac{A’ D’}{C’} – B’\right) I_1 + \frac{A’}{C’} I_2\) ………… (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 = \(\frac{A’}{C’}\)
View Answer

Answer: d
Explanation: 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’ \(\left(\frac{D’}{C’} I_1 + \frac{1}{C’} I_2\right) – B’I_1 = \left(\frac{A’ D’}{C’} – B’\right) I_1 + \frac{A’}{C’} I_2\) ………… (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 = \(\frac{A’}{C’}\).

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn