Network Theory Questions and Answers – Advanced Problems Involving Parameters

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This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Advanced Problems Involving Parameters”.

1. For the circuit given below, the value of Transmission parameter A and C are ____________
network-theory-questions-answers-advanced-problems-involving-parameters-q1
a) A = -0.7692 + j0.3461 Ω, C = 0.03461 + j0.023 Ω
b) A = 0.7692 + j0.3461 Ω, C = 0.03461 + j0.023 Ω
c) A = -0.7692 – j0.3461 Ω, C = -0.03461 + j0.023 Ω
d) A = 0.7692 – j0.3461 Ω, C = 0.023 + j0.03461 Ω
View Answer

Answer: b
Explanation: V = [20 + (-j10) || (j15 − j20)] I1
V1 = \(\Big[20 + \frac{(-j10)(-j5)}{-j15}\Big]\) I1
= [20 – j\(\frac{10}{3}\)] I1
I0 = \(\left(\frac{-j10}{-j10-j5}\right)\) I1 = \(\frac{2}{3}\)I1
V2 = (-j20) I0 + 20I’0
= –\(\frac{j40}{3}I_1 + 20I_1 = (20 – \frac{j40}{3}) I_1 \)
∴ A = \(\frac{V_1}{V_2} = \frac{(20-\frac{j10}{3})I_1}{20-\frac{j40}{3}) I_1}\) = 0.7692 + j0.3461 Ω
∴ C = \(\frac{I_1}{V_2} = \frac{1}{20-j40/3}\) = 0.03461 + j0.023 Ω.
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2. For the circuit given below, the value of the Transmission parameter B and D are __________
network-theory-questions-answers-advanced-problems-involving-parameters-q1
a) D = 0.5385 + j0.6923 Ω, B = -6.923 + j25.385 Ω
b) D = 0.6923 + j0.5385 Ω, B = 6.923 + j25.385 Ω
c) D = -0.6923 + j0.5385 Ω, B= 25.385 + j6.923 Ω
d) D = -0.5385 + j0.6923 Ω, B = -6.923 + j25.385 Ω
View Answer

Answer: a
Explanation: Z1 = \(\frac{(-j15)(-j10)}{-j15-j10-j20}\) = j10
Z2 = \(\frac{(-j10)(-j20)}{-j15}\) = \(-\frac{j40}{3}\)
Z3 = \(\frac{(j15)(-j20)}{-j15}\) = j20
-I2 = \(\frac{20-j40/3}{20-\frac{j40}{3}+j20}I_1 = \frac{3-j2}{3+j}\) I1
∴ D = \(\frac{-I_1}{I_2} = \frac{3+j}{3-j2}\) = 0.5385 + j0.6923 Ω
V1 = [j10 + 2(9+j7)] I1
= jI1 (24 – j18)
So, B = –\(\frac{V_1}{I_2} = \frac{-jI_1 (24-j18)}{-\frac{3-j2}{3+j} I_1}\)
= \(\frac{6}{13}\)(-15+j55)
∴ B = -6.923 + j25.385 Ω.

3. For the circuit given below, the value of the Transmission parameters A and C are _________________
network-theory-questions-answers-advanced-problems-involving-parameters-q3
a) A = 0, C = 1
b) A = 1, C = 0
c) A = Z, C = 1
d) A = 1, C = Z
View Answer

Answer: b
Explanation: V1 = V2
Or, A = \(\frac{V_1}{I_2}\) = 1
I1 = 0 or, C = \(\frac{I_1}{V_2}\) = 0.

4. For the circuit given below, the value of the Transmission parameters B and D are _________________
network-theory-questions-answers-advanced-problems-involving-parameters-q3
a) B = Z, D = 1
b) B = 1, D = Z
c) B = Z, D = Z
d) B = 1, D = 1
View Answer

Answer: a
Explanation: V1 = ZI1
And I2 = -I1
B = \(\frac{V_1}{I_2}\)
= \(\frac{-ZI_1}{-I_1}\) = Z
D = \(\frac{-I_1}{I_2}\) = 1.

5. For the circuit given below, the value of the Transmission parameters A and C are _______________
network-theory-questions-answers-advanced-problems-involving-parameters-q5
a) A = 1, C = 0
b) A = 0, C = 1
c) A = Y, C = 1
d) A = 1, C = Y
View Answer

Answer: d
Explanation: V1 = V2
∴ A = \(\frac{V_1}{V_2}\) =1
And V1 = ZI1
∴ C = \(\frac{I_1}{V_2} = \frac{1}{Z}\) = Y.
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6. For the circuit given below, the value of the Transmission parameters B and D are ________________
network-theory-questions-answers-advanced-problems-involving-parameters-q5
a) B = Y, D = 1
b) B = 1, D = 0
c) B = 0, D = 1
d) B = 0, D = Y
View Answer

Answer: c
Explanation: V1 = V2 = 0
And I2 = -I1
∴ B = \(\frac{V_1}{I_2}\) = 0
∴ D = \(\frac{-I_1}{I_2}\) = 1.

7. For the circuit given below, the values of the h parameter is given as follows h = [16, 3; -2, 0.01]. The value of the ratio \(\frac{V_2}{V_1}\) is ______________
network-theory-questions-answers-advanced-problems-involving-parameters-q8
a) 0.3299
b) 0.8942
c) 1.6
d) 0.2941
View Answer

Answer: d
Explanation: Replacing the given 2-port circuit by an equivalent circuit and applying nodal analysis, we get,
V2 = (20) (2I1) = 40 I1
Or, -10 + 20I1 + 3V2 = 0
Or, 10 = 20I1 + (3) (40I1) = 140I1
∴ I1 = \(\frac{1}{14}\) and V2 = \(\frac{40}{14}\)
So, V1 = 16I1 + 3V2 = \(\frac{136}{14}\)
And I2 = (\(\frac{100}{125}\)) (2I1) = \(\frac{-8}{70}\)
∴ \(\frac{V_2}{V_1} = \frac{40}{136}\) = 0.2941.

8. For the circuit given below, the values of the h parameter is given as follows h = [16, 3; -2, 0.01]. The value of the ratio \(\frac{I_2}{I_1}\) is ______________
network-theory-questions-answers-advanced-problems-involving-parameters-q8
a) 0.3299
b) 0.8942
c) -1.6
d) 0.2941
View Answer

Answer: c
Explanation: Replacing the given 2-port circuit by an equivalent circuit and applying nodal analysis, we get,
V2 = (20) (2I1) = 40 I1
Or, -10 + 20I1 + 3V2 = 0
Or, 10 = 20I1 + (3) (40I1) = 140I1
∴ I1 = \(\frac{1}{14}\) and V2 = \(\frac{40}{14}\)
So, V1 = 16I1 + 3V2 = \(\frac{136}{14}\)
And I2 = (\(\frac{100}{125}\)) (2I1) = \(\frac{-8}{70}\)
∴ \(\frac{I_2}{I_1}\) = -1.6.

9. If for a circuit the value of the h parameter is given as h = [8, 2/3; -2/3, 4/9]. Then the value of the voltage source V is _________________
a) 2.38 V
b) 1.19 V
c) 1.6 V
d) 3.2 V
View Answer

Answer: b
Explanation: 8I1 + \(\frac{2}{3V_2}\) = 10
V2 = \(\frac{2}{3}\)I1 (5||\(\frac{9}{4}\))
= \(\frac{2}{3}\)I1 \((\frac{45}{29})= \frac{30}{29}I_1\)
I1 = \(\frac{29}{30}\)V2
(8)(\(\frac{29}{30}\)) V2 + \(\frac{2}{3}\)V2 = 10
V2 = \(\frac{300}{252}\) = 1.19 V.
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10. For a 2-port network, the value of the h parameter is as h=[600, 0.04; 30, 2×10-3]. Given that, ZS = 2 kΩ and ZL = 400 Ω. The value of the parameter Zin is ______________
a) 250 Ω
b) 333.33 Ω
c) 650 Ω
d) 600 Ω
View Answer

Answer: b
Explanation: Zin = hie – \(\frac{h_{re} h_{fe} R_L}{1 + h_{oe} R_L}\)
= h11 – \(\frac{h_{12} h_{21} R_L}{1+h_{22} R_L}\)
= 600 – \(\frac{0.04×30×400}{1+2×10^{-3}×400}\) = 333.33 Ω.

11. For a 2-port network, the value of the h parameter is as h=[600, 0.04; 30, 2×10-3]. Given that, ZS = 2 kΩ and ZL = 400 Ω. The value of the parameter Zout is ______________
a) 650 Ω
b) 500 Ω
c) 250 Ω
d) 600 Ω
View Answer

Answer: a
Explanation: Zout = \(\frac{R_s+h_{ie}}{(R_s+h_{ie}) h_{oe}-h_{re} h_{fe}}\)
= \(\frac{R_s+h_{11}}{(R_s+h_{11}) h_{22}-h_{21} h_{12}}\)
= \(\frac{2000+600}{2600×2×10^{-3}-30×0.04}\) = 650 Ω.

12. For the circuit given below, the value of the g11 and g21 are _________________
network-theory-questions-answers-advanced-problems-involving-parameters-q12
a) g11 = –\(\frac{1}{R_1+R_2}\), g21 = \(\frac{R_2}{R_1+R_2}\)
b) g11 = \(\frac{1}{R_1-R_2}\), g21 = –\(\frac{R_2}{R_1+R_2}\)
c) g11 = \(\frac{1}{R_1+R_2}\), g21 = \(\frac{R_2}{R_1+R_2}\)
d) g11 = \(\frac{1}{R_1-R_2}\), g21 = \(\frac{R_2}{R_1-R_2}\)
View Answer

Answer: c
Explanation: I1 = \(\frac{V_1}{R_1+R_2}\)
Or, g11 = \(\frac{I_1}{V_1} = \frac{1}{R_1+R_2}\)
By voltage division, V2 = \(\frac{R_2}{R_1+R_2}\)V1
Or, g21 = \(\frac{V_2}{V_1} = \frac{R_2}{R_1+R_2}\).

13. For the circuit given below, the value of the g12 and g22 are _______________
network-theory-questions-answers-advanced-problems-involving-parameters-q12
a) g12 = –\(\frac{R_2}{R_1+R_2}\), g22 = R3 + \(\frac{R_1 R_2}{R_1+R_2}\)
b) g12 = \(\frac{R_2}{R_1+R_2}\), g22 = -R3 + \(\frac{R_1 R_2}{R_1+R_2}\)
c) g12 = –\(\frac{R_2}{R_1+R_2}\), g22 = R3 – \(\frac{R_1 R_2}{R_1+R_2}\)
d) g12 = \(\frac{R_2}{R_1+R_2}\), g22 = -R3 – \(\frac{R_1 R_2}{R_1+R_2}\)
View Answer

Answer: a
Explanation: I1 = –\(\frac{R_2}{R_1+R_2}\)I2
Or, g12 = \(\frac{I_1}{I_2} = -\frac{R_2}{R_1+R_2}\)
Also, I2 (R3 + R1 //R2)
= I2 \((R_3 + \frac{R_1 R_2}{R_1+R_2})\)
∴ g22 = \(\frac{V_2}{I_2} = R_3 + \frac{R_1 R_2}{R_1+R_2}\).
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14. For the circuit given below, the value of g11 and g21 are _________________
network-theory-questions-answers-advanced-problems-involving-parameters-q14
a) g11 = 0.0667 – j0.0333 Ω, g21 = 0.8 + j0.4 Ω
b) g11 = -0.0667 – j0.0333 Ω, g21 = -0.8 – j0.4 Ω
c) g11 = 0.0667 + j0.0333 Ω, g21 = 0.8 + j0.4 Ω
d) g11 = -0.0667 + j0.0333 Ω, g21 = 0.8 – j0.4 Ω
View Answer

Answer: c
Explanation: V1 = (12-j6) I1
Or, g11 = \(\frac{I_1}{V_1} = \frac{1}{12-j6}\) = 0.0667 + j0.0333 Ω
g21 = \(\frac{V_2}{V_1} = \frac{12I_1}{(12-j6) I_1}\)
= \(\frac{2}{2-j}\) = 0.8 + j0.4 Ω.

15. For the circuit given below, the value of g12 and g22 are ________________
network-theory-questions-answers-advanced-problems-involving-parameters-q14
a) g12 = 0.8 + j0.4 Ω, g22 = 2.4 + j5.2 Ω
b) g12 = -0.8 + j0.4 Ω, g22 = -2.4 – j5.2 Ω
c) g12 = 0.8 – j0.4 Ω, g22 = 2.4 – j5.2 Ω
d) g12 = -0.8 – j0.4 Ω, g22 = 2.4 + j5.2 Ω
View Answer

Answer: d
Explanation: I1 = \(\frac{-12}{12-j6}\)I2
Or, g12 = \(\frac{I_1}{I_2}\) = -g21 = -0.8 – j0.4 Ω
V2 = (j10 + 12 || -j6) I2
g22 = \(\frac{V_2}{I_2}\) = 2.4 + j5.2 Ω.

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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