Network Theory Questions and Answers – Advanced Problems on Two Port Network – 1

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This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Advanced Problems on Two Port Network – 1”.

1. For the circuit given below, the value of the z12 parameter is ___________
network-theory-questions-answers-advanced-problems-two-port-network-1-q1
a) z12 = 1 Ω
b) z12 = 4 Ω
c) z12 = 1.667 Ω
d) z12 = 2.33 Ω
View Answer

Answer: a
Explanation: z11 = \(\frac{V_1}{I_1}\) = 1 + 6 || (4+2) = 4Ω
I0 = \(\frac{1}{2} I_1 \)
V2 = 2I0 = I1
z21 = \(\frac{V_2}{I_1}\) = 1Ω
z22 = \(\frac{V_2}{I_2}\) = 2 || (4+6) = 1.667Ω
So, I’0 = \(\frac{2}{2+10}I_2 = \frac{1}{6}I_2 \)
V1 = 6I’0 = I2
z12 = \(\frac{V_1}{I_2}\) = 1Ω
Hence, [z] = [4:1; 1:1.667] Ω.
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2. For the network of figure, z11 is equal to ___________
network-theory-questions-answers-advanced-problems-two-port-network-1-q2
a) \(\frac{5}{3}\) Ω
b) \(\frac{3}{2}\) Ω
c) 2 Ω
d) \(\frac{2}{3}\) Ω
View Answer

Answer: a
Explanation: From the figure, we can infer that,
Z11 = 1 + \(\frac{1 X 2}{3}\)
= 1 + \(\frac{2}{3}\)
= \(\frac{5}{3}\) Ω.

3. For the circuit given below, the value of z11 parameter is ____________
network-theory-questions-answers-advanced-problems-two-port-network-1-q3
a) z11 = 4 + j6 Ω
b) z11 = j6 Ω
c) z11 = -j6 Ω
d) z11 = -j6 + 4 Ω
View Answer

Answer: a
Explanation: z12 = j6 = z21
z11 – z12 = 4
Or, z11 = z12 + 4 = 4 + j6 Ω
And z22 – z12 = -j10
Or, z22 = z12 + -j10 = -j4 Ω
∴ [z] = [4+j6:j6; j6:-j4] Ω.

4. In a series RLC circuit excited by a voltage 3e-t u (t), the resistance is equal to 1 Ω and capacitance = 2 F. For the circuit, the values of I (0+) and I (∞), are ____________
a) 0 and 1.5 A
b) 1.5 A and 3 A
c) 3 A and 0
d) 3 A and 1.5 A
View Answer

Answer: c
Explanation: I(s) = \(\frac{6}{s+1} – \frac{3}{s+0.5}\)
Or, I(t) = 6 e-t – 3 e-0.5t
Putting, t = 0, we get, I(0) = 3A
Putting t = ∞, we get, I (∞) = 0.

5. For the circuit given below, the value of z12 parameter is ____________
network-theory-questions-answers-advanced-problems-two-port-network-1-q5
a) Z12 = 20 Ω
b) Z12 = 25 Ω
c) Z12 = 30 Ω
d) z12 = 24 Ω
View Answer

Answer: a
Explanation: z11 = \(\frac{V_1}{I_1} = \frac{(20+5)I_1}{I_1}\) = 25Ω
V0 = \(\frac{20}{25}\)V1 = 20 I1
-V0 – 4I2 + V2 = 0
Or, V2 = V0 + 4I1 = 20I1 + 4I1 = 24 I1
Or, z21 = \(\frac{V_2}{I_1}\) = 24 Ω
V2 = (10+20) I2 = 30 I2
Or, z22 = \(\frac{V_2}{I_1}\) = 30 Ω
V1 = 20I2
Or, z12 = \(\frac{V_1}{I_2}\) = 20 Ω
∴ [z] = [25:20; 24:30] Ω.
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6. For the circuit given below, the value of the z22 parameter is ___________
network-theory-questions-answers-advanced-problems-two-port-network-1-q1
a) z22 = 1 Ω
b) z22 = 4 Ω
c) z22 = 1.667 Ω
d) z22 = 2.33 Ω
View Answer

Answer: c
Explanation: z11 = \(\frac{V_1}{I_1}\) = 1 + 6 || (4+2) = 4Ω
I0 = \(\frac{1}{2} I_1 \)
V2 = 2I0 = I1
z21 = \(\frac{V_2}{I_1}\) = 1Ω
z22 = \(\frac{V_2}{I_2}\) = 2 || (4+6) = 1.667Ω
So, I’0 = \(\frac{2}{2+10}I_2 = \frac{1}{6}I_2 \)
V1 = 6I’0 = I2
z12 = \(\frac{V_1}{I_2}\) = 1Ω
Hence, [z] = [4:1; 1:1.667] Ω.

7. For the circuit given below, the value of z22 parameter is ____________
network-theory-questions-answers-advanced-problems-two-port-network-1-q7
a) z22 = 0.0667 Ω
b) z22 = 2.773 Ω
c) z22 = 1.667 Ω
d) z22 = 0.999 Ω
View Answer

Answer: b
Explanation: z11 = \(\frac{V_1}{I_1}\) = 2 + 1 || [2+1 || (2+1)]
z11 = 2 + 1 || (2 + \(\frac{3}{4}\)) = 2 + \(\frac{1×\frac{11}{4}}{1+\frac{11}{4}} = 2 + \frac{11}{15}\) = 2.733
I0 = \(\frac{1}{1+3}\) I’0 = \(\frac{1}{4}\) I’0
And I’0 = 1 + \(\frac{11}{4}\) I1 = \(\frac{4}{15}\) I1
Or, I0 = \(\frac{1}{4} × \frac{4}{5} I_1 = \frac{1}{15} I_1\)
Or, V2 = I0 = \(\frac{1}{15} I_1\)
z21 = \(\frac{V_2}{I_1} = \frac{1}{15}\) = z12 = 0.0667
z22 = \(\frac{V_2}{I_2}\) = 2+1 || (2+1||3) = z11 = 2.733
∴ [z] = [2.733:0.0667; 0.0667:2.733] Ω.

8. For the circuit given below, the value of z22 parameter is ____________
network-theory-questions-answers-advanced-problems-two-port-network-1-q3
a) z22 = 4 + j6 Ω
b) z22 = j6 Ω
c) z22 = -j4 Ω
d) z22 = -j6 + 4 Ω
View Answer

Answer: c
Explanation: z12 = j6 = z21
z11 – z12 = 4
Or, z11 = z12 + 4 = 4 + j6 Ω
And z22 – z12 = -j10
Or, z22 = z12 + -j10 = -j4 Ω
∴ [z] = [4+j6:j6; j6:-j4] Ω.

9. For the circuit given below, the value of z22 parameter is ____________
network-theory-questions-answers-advanced-problems-two-port-network-1-q9
a) z22 = 1.775 + j5.739 Ω
b) z22 = 1.775 – j4.26 Ω
c) z22 = -1.775 – j5.739 Ω
d) z22 = 1.775 + j4.26 Ω
View Answer

Answer: c
Explanation: z1 = \(\frac{12(j10)}{12+j10-j5} = \frac{j120}{12+j5}\)
z2 = \(\frac{j60}{12+j5}\)
z3 = \(\frac{50}{12+j5}\)
z12 = z21 = z2 = \(\frac{(-j60)(12-j5)}{144+25}\) = -1.775 – j4.26
z11 = z1 + z12 = \(\frac{(j120)(12-j5)}{144+25}\) + z12 = 1.775 + j4.26
z22 = z3 + z21 = \(\frac{(50)(12-j5)}{144+25}\) + z21 = 1.7758 – j5.739
∴ [z] = [1.775 + j4.26; -1.775 – j4.26; -1.775 – j4.26; 1.775 – j5.739] Ω.
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10. For the circuit given below, the value of z22 parameter is ____________
network-theory-questions-answers-advanced-problems-two-port-network-1-q5
a) z22 = 20 Ω
b) z22 = 25 Ω
c) z22 = 30 Ω
d) z22 = 24 Ω
View Answer

Answer: c
Explanation: z11 = \(\frac{V_1}{I_1} = \frac{(20+5)I_1}{I_1}\) = 25Ω
V0 = \(\frac{20}{25}\)V1 = 20 I1
-V0 – 4I2 + V2 = 0
Or, V2 = V0 + 4I1 = 20I1 + 4I1 = 24 I1
Or, z21 = \(\frac{V_2}{I_1}\) = 24 Ω
V2 = (10+20) I2 = 30 I2
Or, z22 = \(\frac{V_2}{I_1}\) = 30 Ω
V1 = 20I2
Or, z12 = \(\frac{V_1}{I_2}\) = 20 Ω
∴ [z] = [25:20; 24:30] Ω.

11. A capacitor of 220 V, 50 Hz is needed for AC supply. The peak voltage rating of the capacitor is ____________
a) 220 V
b) 460 V
c) 440 V
d) 230 V
View Answer

Answer: c
Explanation: We know that,
Peak voltage rating = 2 (rms voltage rating)
Given that the RMS voltage rating = 220 V
So, the Peak Voltage Rating = 2 X 220 V
= 440 V.

12. In the circuit given below, the value of the hybrid parameter h21 is _________
network-theory-questions-answers-advanced-problems-two-port-network-1-q12
a) 10 Ω
b) 0.5 Ω
c) 5 Ω
d) 2.5 Ω
View Answer

Answer: b
Explanation: Hybrid parameter h21 is given by, h21 = \(\frac{I_2}{I_1}\), when V2 = 0.
Therefore short circuiting the terminal Y-Y’, and applying Kirchhoff’s law, we get,
-5 I2 – (I2 – I1)5 = 0
Or, -I2 = I2 – I1
Or, -2I2 = -I1
∴ \(\frac{I_2}{I_1} = \frac{1}{2}\)
Hence h21 = 0.5 Ω.

13. If a two port network is passive, then we have, with the usual notation, the relationship as _________
a) h21 = h12
b) h12 = -h21
c) h11 = h22
d) h11 h22 – h12 h22 = 1
View Answer

Answer: d
Explanation: We know that, I1 = y11 V1 + y12 V2 ……… (1)
I2 = y21 V1 + y22 V2 ………. (2)
And, V1 = h11 I1 + h12 V2 ………. (3)
I2 = h21 I1 + h22 V2 ……….. (4)
Now, (3) and (4) can be rewritten as,
I1 = \(\frac{V_1}{h_{11}} – \frac{h_{12} V_2}{h_{11}}\) ………. (5)
And I2 = \(\frac{h_21 V_1}{h_11} + \left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\right) V_2\) ………. (6)
Therefore using the above 6 equations in representing the hybrid parameters in terms of the Y parameters and applying ∆Y = 0, we get,
h11 h22 – h12 h22 = 1 [hence proved].
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14. In two-port networks the parameter h22 is called _________
a) Short circuit input impedance
b) Short circuit current gain
c) Open circuit reverse voltage gain
d) Open circuit output admittance
View Answer

Answer: d
Explanation: We know that, h22 = \(\frac{I_2}{V_2}\), when I1 = 0.
Since the current in the first loop is 0 when the ratio of the current and voltage in second loop is measured, therefore the parameter h12 is called as Open circuit output admittance.

15. The short-circuit admittance matrix of a two port network is as follows.
[0; -0.5; 0.5; 0] Then the 2 port network is ____________
a) Non-reciprocal and passive
b) Non-reciprocal and active
c) Reciprocal and passive
d) Reciprocal and active
View Answer

Answer: b
Explanation: So, network is non reciprocal because Y12 ≠ Y21 and Y12 are also negative which means either energy storing or providing device is available. So the network is active.
Therefore the network is Non- reciprocal and active.

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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