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 ____________

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

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.

2. For the circuit given below, the value of the Transmission parameter B and D are __________

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

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 _________________

a) A = 0, C = 1
b) A = 1, C = 0
c) A = Z, C = 1
d) A = 1, C = Z

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 _________________

a) B = Z, D = 1
b) B = 1, D = Z
c) B = Z, D = Z
d) B = 1, D = 1

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 _______________

a) A = 1, C = 0
b) A = 0, C = 1
c) A = Y, C = 1
d) A = 1, C = Y

Explanation: V1 = V2
∴ A = $$\frac{V_1}{V_2}$$ = 1
And V1 = ZI1
∴ C = $$\frac{I_1}{V_2} = \frac{1}{Z}$$ = Y.

6. For the circuit given below, the value of the Transmission parameters B and D are ________________

a) B = Y, D = 1
b) B = 1, D = 0
c) B = 0, D = 1
d) B = 0, D = Y

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 ______________

a) 0.3299
b) 0.8942
c) 1.6
d) 0.2941

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 ______________

a) 0.3299
b) 0.8942
c) -1.6
d) 0.2941

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

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.

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

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

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 _________________

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 = $$\frac{R_2}{R_1-R_2}$$

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 _______________

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 – $$\frac{R_1 R_2}{R_1+R_2}$$

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 = $$\frac{V_2}{I_2} = R_3 + \frac{R_1 R_2}{R_1+R_2}$$.

14. For the circuit given below, the value of g11 and g21 are _________________

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

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 ________________

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

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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