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
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 = \(\frac{A}{B}\).
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
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
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
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
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 = \(\frac{D}{C}\).
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
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
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
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
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 = \(\frac{D’}{B’}\).
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
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
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
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
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 = \(\frac{A’}{C’}\).
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
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
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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