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 z_{12} parameter is ___________

a) z_{12} = 1 Ω

b) z_{12} = 4 Ω

c) z_{12} = 1.667 Ω

d) z_{12} = 2.33 Ω

View Answer

Explanation: z

_{11}= \(\frac{V_1}{I_1}\) = 1 + 6 || (4+2) = 4Ω

I

_{0}= \(\frac{1}{2} I_1 \)

V

_{2}= 2I

_{0}= I

_{1}

z

_{21}= \(\frac{V_2}{I_1}\) = 1Ω

z

_{22}= \(\frac{V_2}{I_2}\) = 2 || (4+6) = 1.667Ω

So, I’

_{0}= \(\frac{2}{2+10}I_2 = \frac{1}{6}I_2 \)

V

_{1}= 6I’

_{0}= I

_{2}

z

_{12}= \(\frac{V_1}{I_2}\) = 1Ω

Hence, [z] = [4:1; 1:1.667] Ω.

2. For the network of figure, z_{11} is equal to ___________

a) \(\frac{5}{3}\) Ω

b) \(\frac{3}{2}\) Ω

c) 2 Ω

d) \(\frac{2}{3}\) Ω

View Answer

Explanation: From the figure, we can infer that,

Z

_{11}= 1 + \(\frac{1 X 2}{3}\)

= 1 + \(\frac{2}{3}\)

= \(\frac{5}{3}\) Ω.

3. For the circuit given below, the value of z_{11} parameter is ____________

a) z_{11} = 4 + j6 Ω

b) z_{11} = j6 Ω

c) z_{11} = -j6 Ω

d) z_{11} = -j6 + 4 Ω

View Answer

Explanation: z

_{12}= j6 = z

_{21}

z

_{11}– z

_{12}= 4

Or, z

_{11}= z

_{12}+ 4 = 4 + j6 Ω

And z

_{22}– z

_{12}= -j10

Or, z

_{22}= z

_{12}+ -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

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 z_{12} parameter is ____________

a) Z_{12} = 20 Ω

b) Z_{12} = 25 Ω

c) Z_{12} = 30 Ω

d) z_{12} = 24 Ω

View Answer

Explanation: z

_{11}= \(\frac{V_1}{I_1} = \frac{(20+5)I_1}{I_1}\) = 25Ω

V

_{0}= \(\frac{20}{25}\)V

_{1}= 20 I

_{1}

-V

_{0}– 4I

_{2}+ V

_{2}= 0

Or, V

_{2}= V

_{0}+ 4I

_{1}= 20I

_{1}+ 4I

_{1}= 24 I

_{1}

Or, z

_{21}= \(\frac{V_2}{I_1}\) = 24 Ω

V

_{2}= (10+20) I

_{2}= 30 I

_{2}

Or, z

_{22}= \(\frac{V_2}{I_1}\) = 30 Ω

V

_{1}= 20I

_{2}

Or, z

_{12}= \(\frac{V_1}{I_2}\) = 20 Ω

∴ [z] = [25:20; 24:30] Ω.

6. For the circuit given below, the value of the z_{22} parameter is ___________

a) z_{22} = 1 Ω

b) z_{22} = 4 Ω

c) z_{22} = 1.667 Ω

d) z_{22} = 2.33 Ω

View Answer

Explanation: z

_{11}= \(\frac{V_1}{I_1}\) = 1 + 6 || (4+2) = 4Ω

I

_{0}= \(\frac{1}{2} I_1 \)

V

_{2}= 2I

_{0}= I

_{1}

z

_{21}= \(\frac{V_2}{I_1}\) = 1Ω

z

_{22}= \(\frac{V_2}{I_2}\) = 2 || (4+6) = 1.667Ω

So, I’

_{0}= \(\frac{2}{2+10}I_2 = \frac{1}{6}I_2 \)

V

_{1}= 6I’

_{0}= I

_{2}

z

_{12}= \(\frac{V_1}{I_2}\) = 1Ω

Hence, [z] = [4:1; 1:1.667] Ω.

7. For the circuit given below, the value of z_{22} parameter is ____________

a) z_{22} = 0.0667 Ω

b) z_{22} = 2.773 Ω

c) z_{22} = 1.667 Ω

d) z_{22} = 0.999 Ω

View Answer

Explanation: z

_{11}= \(\frac{V_1}{I_1}\) = 2 + 1 || [2+1 || (2+1)]

z

_{11}= 2 + 1 || (2 + \(\frac{3}{4}\)) = 2 + \(\frac{1×\frac{11}{4}}{1+\frac{11}{4}} = 2 + \frac{11}{15}\) = 2.733

I

_{0}= \(\frac{1}{1+3}\) I’

_{0}= \(\frac{1}{4}\) I’

_{0}

And I’

_{0}= 1 + \(\frac{11}{4}\) I

_{1}= \(\frac{4}{15}\) I

_{1}

Or, I

_{0}= \(\frac{1}{4} × \frac{4}{5} I_1 = \frac{1}{15} I_1\)

Or, V

_{2}= I

_{0}= \(\frac{1}{15} I_1\)

z

_{21}= \(\frac{V_2}{I_1} = \frac{1}{15}\) = z

_{12}= 0.0667

z

_{22}= \(\frac{V_2}{I_2}\) = 2+1 || (2+1||3) = z

_{11}= 2.733

∴ [z] = [2.733:0.0667; 0.0667:2.733] Ω.

8. For the circuit given below, the value of z_{22} parameter is ____________

a) z_{22} = 4 + j6 Ω

b) z_{22} = j6 Ω

c) z_{22} = -j4 Ω

d) z_{22} = -j6 + 4 Ω

View Answer

Explanation: z

_{12}= j6 = z

_{21}

z

_{11}– z

_{12}= 4

Or, z

_{11}= z

_{12}+ 4 = 4 + j6 Ω

And z

_{22}– z

_{12}= -j10

Or, z

_{22}= z

_{12}+ -j10 = -j4 Ω

∴ [z] = [4+j6:j6; j6:-j4] Ω.

9. For the circuit given below, the value of z_{22} parameter is ____________

a) z_{22} = 1.775 + j5.739 Ω

b) z_{22} = 1.775 – j4.26 Ω

c) z_{22} = -1.775 – j5.739 Ω

d) z_{22} = 1.775 + j4.26 Ω

View Answer

Explanation: z

_{1}= \(\frac{12(j10)}{12+j10-j5} = \frac{j120}{12+j5}\)

z

_{2}= \(\frac{j60}{12+j5}\)

z

_{3}= \(\frac{50}{12+j5}\)

z

_{12}= z

_{21}= z

_{2}= \(\frac{(-j60)(12-j5)}{144+25}\) = -1.775 – j4.26

z

_{11}= z

_{1}+ z

_{12}= \(\frac{(j120)(12-j5)}{144+25}\) + z

_{12}= 1.775 + j4.26

z

_{22}= z

_{3}+ z

_{21}= \(\frac{(50)(12-j5)}{144+25}\) + z

_{21}= 1.7758 – j5.739

∴ [z] = [1.775 + j4.26; -1.775 – j4.26; -1.775 – j4.26; 1.775 – j5.739] Ω.

10. For the circuit given below, the value of z_{22} parameter is ____________

a) z_{22} = 20 Ω

b) z_{22} = 25 Ω

c) z_{22} = 30 Ω

d) z_{22} = 24 Ω

View Answer

Explanation: z

_{11}= \(\frac{V_1}{I_1} = \frac{(20+5)I_1}{I_1}\) = 25Ω

V

_{0}= \(\frac{20}{25}\)V

_{1}= 20 I

_{1}

-V

_{0}– 4I

_{2}+ V

_{2}= 0

Or, V

_{2}= V

_{0}+ 4I

_{1}= 20I

_{1}+ 4I

_{1}= 24 I

_{1}

Or, z

_{21}= \(\frac{V_2}{I_1}\) = 24 Ω

V

_{2}= (10+20) I

_{2}= 30 I

_{2}

Or, z

_{22}= \(\frac{V_2}{I_1}\) = 30 Ω

V

_{1}= 20I

_{2}

Or, z

_{12}= \(\frac{V_1}{I_2}\) = 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

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 h_{21} is _________

a) 10 Ω

b) 0.5 Ω

c) 5 Ω

d) 2.5 Ω

View Answer

Explanation: Hybrid parameter h

_{21}is given by, h

_{21}= \(\frac{I_2}{I_1}\), when V

_{2}= 0.

Therefore short circuiting the terminal Y-Y’, and applying Kirchhoff’s law, we get,

-5 I

_{2}– (I

_{2}– I

_{1})5 = 0

Or, -I

_{2}= I

_{2}– I

_{1}

Or, -2I

_{2}= -I

_{1}

∴ \(\frac{I_2}{I_1} = \frac{1}{2}\)

Hence h

_{21}= 0.5 Ω.

13. If a two port network is passive, then we have, with the usual notation, the relationship as _________

a) h_{21} = h_{12}

b) h_{12} = -h_{21 }

c) h_{11} = h_{22}

d) h_{11} h_{22} – h_{12} h_{22} = 1

View Answer

Explanation: We know that, I

_{1}= y

_{11}V

_{1}+ y

_{12}V

_{2}……… (1)

I

_{2}= y

_{21}V

_{1}+ y

_{22}V

_{2}………. (2)

And, V

_{1}= h

_{11}I

_{1}+ h

_{12}V

_{2}………. (3)

I

_{2}= h

_{21}I

_{1}+ h

_{22}V

_{2}……….. (4)

Now, (3) and (4) can be rewritten as,

I

_{1}= \(\frac{V_1}{h_{11}} – \frac{h_{12} V_2}{h_{11}}\) ………. (5)

And I

_{2}= \(\frac{h_21 V_1}{h_11} + \left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\right) V_2\) ………. (6)

Therefore using the above 6 equations in representing the hybrid parameters in terms of the Y parameters and applying ∆Y = 0, we get,

h

_{11}h

_{22}– h

_{12}h

_{22}= 1 [hence proved].

14. In two-port networks the parameter h_{22} 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

Explanation: We know that, h

_{22}= \(\frac{I_2}{V_2}\), when I

_{1}= 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 h

_{12}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

Explanation: So, network is non reciprocal because Y

_{12}≠ Y

_{21}and Y

_{12}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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