This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Inverse Hybrid (g) Parameter”.

1. For the circuit given below, the value of the Inverse hybrid parameter g_{11} is ___________

a) 0.067 Ω

b) 0.025 Ω

c) 0.3 Ω

d) 0.25 Ω

View Answer

Explanation: Inverse Hybrid parameter g

_{11}is given by, g

_{11}= \(\frac{I_1}{V_1}\), when I

_{2}= 0.

Therefore short circuiting the terminal Y-Y’, we get,

V

_{1}= I

_{1}((10||10) + 10)

= I

_{1}\(\left(\left(\frac{10×10}{10+10}\right)+10\right)\)

= 15I

_{1}

∴ \(\frac{I_1}{V_1} = \frac{1}{15}\) = 0.067 Ω

Hence g

_{11}= 15 Ω.

2. For the circuit given below, the value of the Inverse hybrid parameter g_{21} is ___________

a) 0.6 Ω

b) 0.5 Ω

c) 0.3 Ω

d) 0.2 Ω

View Answer

Explanation: Inverse Hybrid parameter g

_{21}is given by, g

_{21}= \(\frac{V_2}{V_1}\), when I

_{2}= 0.

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

V

_{1}= I

_{1}(10 + 10)

V

_{2}= I

_{1}10

∴ \(\frac{V_2}{V_1} = \frac{I_1 10}{I_1 20}\) = 0.5

Hence g

_{21}= 0.5 Ω.

3. For the 2 port network as shown below, the Z-matrix is ___________

a) [Z_{1}; Z_{1} + Z_{2}; Z_{1} + Z_{2}; Z_{3}]

b) [Z_{1}; Z_{1}; Z_{1} + Z_{2}; Z_{2}]

c) [Z_{1}; Z_{2}; Z_{2}; Z_{1} + Z_{2}]

d) [Z_{1}; Z_{1}; Z_{1}; Z_{1} + Z_{2}]

View Answer

Explanation: z

_{11}= \(\frac{V_1}{I_1}\), when I

_{2}= 0

z

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

_{1}= 0

z

_{12}= \(\frac{V_1}{I_2}\), when I

_{1}= 0

z

_{21}= \(\frac{V_2}{I_1}\), when I

_{2}= 0

Now, in the given circuit putting I

_{1}= 0, we get,

z

_{12}= Z

_{1}and z

_{22}= Z

_{1}+ Z

_{2}

And putting I

_{2}= 0, we get,

z

_{21}= Z

_{1}and z

_{11}= Z

_{1}.

4. Which one of the following parameters does not exist for the two-port network in the circuit given below?

a) h

b) Y

c) Z

d) g

View Answer

Explanation: Y-parameter = \(\frac{1}{Z}\)[1; -1; -1; 1]

And from the definition of the Y parameters, ∆Y = 0. Therefore the Y-parameter exists.

Since ∆Y = 0, so by property of reciprocity, ∆h = 0 and ∆g = 0.

Hence both hybrid and inverse hybrid parameters exist.

But the Z-parameters cannot exist here because if one terminal is opened the circuit will become invalid.

∴ Z- parameters do not exists.

5. In the circuit given below, the value of the Inverse hybrid parameter g_{11} is _________

a) 10 Ω

b) 0.133 Ω

c) 5 Ω

d) 2.5 Ω

View Answer

Explanation: Inverse Hybrid parameter g

_{11}is given by, g

_{11}= \(\frac{I_1}{V_1}\), when I

_{2}= 0.

Therefore short circuiting the terminal Y-Y’, we get,

V

_{1}= I

_{1}((5 || 5) + 5)

= I

_{1}\(\left(\left(\frac{5×5}{5+5}\right)+5\right)\)

= 7.5I

_{1}

∴ \(\frac{I_1}{V_1} = \frac{1}{7.5}\) = 0.133

Hence g

_{11}= 7.5 Ω.

6. In the circuit given below, the value of the Inverse hybrid parameter g_{21} is _________

a) 10 Ω

b) 0.5 Ω

c) 5 Ω

d) 2.5 Ω

View Answer

Explanation: Inverse Hybrid parameter g

_{21}is given by, g

_{21}= \(\frac{V_2}{V_1}\), when I

_{2}= 0.

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

V

_{1}= I

_{1}(5 + 5)

V

_{2}= I

_{1}5

∴ \(\frac{V_2}{V_1} = \frac{I_1 5}{I_1 10}\) = 0.5

Hence g

_{21}= 0.5 Ω.

7. 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: The network is non reciprocal because Y

_{12}≠ Y

_{21}and Y

_{12}is also negative which means either energy storing or providing device is available. So network is active. Therefore the network is Non- reciprocal and active.

8. 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].

9. For the circuit given below, the value of the Inverse hybrid parameter g_{22} is ___________

a) 7.5 Ω

b) 5 Ω

c) 6.25 Ω

d) 3 Ω

View Answer

Explanation: Inverse Hybrid parameter g

_{22}is given by, g

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

_{1}= 0.

Therefore short circuiting the terminal X-X’ we get,

-5 I

_{2}– 5 I

_{1}+ V

_{2}= 0

-5 I

_{1}– 5(I

_{1}– I

_{2}) = 0

Or, 2I

_{1}= I

_{2}

Putting the above equation in the first equation, we get,

-7.5 I

_{2}= -V

_{2}

Or, \(\frac{V_2}{I_2}\) = 7.5

Hence g

_{22}= 7.5 Ω.

10. In two-port networks the parameter g_{11} is called _________

a) Short circuit input impedance

b) Short circuit current ratio

c) Open circuit voltage ratio

d) Open circuit input admittance

View Answer

Explanation: We know that, g

_{11}= \(\frac{I_1}{V_1}\), when I

_{2}= 0.

Since the second voltage terminal is short circuited when the ratio of the current and voltage is measured, therefore the parameter g

_{11}is called as Open circuit input admittance.

11. In two-port networks the parameter g_{21} is called _________

a) Short circuit input impedance

b) Short circuit current ratio

c) Open circuit voltage ratio

d) Open circuit input admittance

View Answer

Explanation: We know that, g

_{21}= \(\frac{V_2}{V_1}\), when I

_{2}= 0.

Since the second output terminal is short circuited when the ratio of the two voltages is measured, therefore the parameter g

_{21}is called as Open circuit voltage ratio.

12. In two-port networks the parameter g_{12} 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, g

_{12}= \(\frac{I_1}{I_2}\), when V

_{1}= 0.

Since the primary terminal is short circuited and the ratio of the two currents is measured, therefore the parameter g

_{12}is called as Short circuit current ratio.

13. In two-port networks the parameter g_{22} is called _________

a) Short circuit input impedance

b) Short circuit current ratio

c) Open circuit voltage ratio

d) Open circuit input admittance

View Answer

Explanation: We know that, g

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

_{1}= 0.

Since the primary voltage terminal is short circuited and the ratio of the voltage and current in second loop is measured, therefore the parameter g

_{22}is called as Short circuit current ratio.

14. For a T-network if the Short circuit admittance parameters are given as y_{11}, y_{21}, y_{12}, y_{22}, then y_{12} in terms of Inverse Hybrid parameters can be expressed as ________

a) y_{12} = \(\left(g_{11} – \frac{g_{12} g_{21}}{g_{22}}\right)\)

b) y_{12} = \(\frac{g_{12}}{g_{22}} \)

c) y_{12} = –\(\frac{g_{21}}{g_{22}} \)

d) y_{12} = \(\frac{1}{g_{22}}\)

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, I

_{1}= g

_{11}V

_{1}+ g

_{12}I

_{2}………. (3)

V

_{2}= g

_{21}V

_{1}+ g

_{22}I

_{2}……….. (4)

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

I

_{1}= \(\left(g_{11} – \frac{g_{12} g_{21}}{g_{22}}\right)V_1 + \frac{g_{12}}{g_{22}} V_2\)………. (5)

And I

_{2}= –\(\frac{g_{21} V_1}{g_{22}} + \frac{V_2}{g_{22}}\)………. (6)

∴ Comparing (1), (2) and (5), (6), we get,

y

_{11}= \(\left(g_{11} – \frac{g_{12} g_{21}}{g_{22}}\right)\)

y

_{12}= \(\frac{g_{12}}{g_{22}} \)

y

_{21}= –\(\frac{g_{21}}{g_{22}} \)

y

_{22}= \(\frac{1}{g_{22}}\).

15. For a T-network if the Short circuit admittance parameters are given as y_{11}, y_{21}, y_{12}, y_{22}, then y_{21} in terms of Inverse Hybrid parameters can be expressed as ________

a) y_{21} = \(\left(g_{11} – \frac{g_{12} g_{21}}{g_{22}}\right)\)

b) y_{21} = \(\frac{g_{12}}{g_{22}} \)

c) y_{21} = –\(\frac{g_{21}}{g_{22}} \)

d) y_{21} = \(\frac{1}{g_{22}}\)

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, I

_{1}= g

_{11}V

_{1}+ g

_{12}I

_{2}………. (3)

V

_{2}= g

_{21}V

_{1}+ g

_{22}I

_{2}……….. (4)

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

I

_{1}= \(\left(g_{11} – \frac{g_{12} g_{21}}{g_{22}}\right)V_1 + \frac{g_{12}}{g_{22}} V_2\)………. (5)

And I

_{2}= –\(\frac{g_{21} V_1}{g_{22}} + \frac{V_2}{g_{22}}\)………. (6)

∴ Comparing (1), (2) and (5), (6), we get,

y

_{11}= \(\left(g_{11} – \frac{g_{12} g_{21}}{g_{22}}\right)\)

y

_{12}= \(\frac{g_{12}}{g_{22}} \)

y

_{21}= –\(\frac{g_{21}}{g_{22}} \)

y

_{22}= \(\frac{1}{g_{22}}\).

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