250+ TOP MCQs on Advanced Problems Involving Parameters and Answers

Network Theory Multiple Choice Questions 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

Answer: B
Clarification: 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

Answer: A
Clarification: 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

Answer: B
Clarification: 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

Answer: A
Clarification: 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

Answer: D
Clarification: 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

Answer: C
Clarification: 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

Answer: D
Clarification: 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

Answer: C
Clarification: 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

Answer: B
Clarification: 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, Z_S = 2 kΩ and Z_L = 400 Ω. The value of the parameter Z_in is ______________
A. 250 Ω
B. 333.33 Ω
C. 650 Ω
D. 600 Ω

Answer: B
Clarification: Z_in = h_ie – (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, Z_S = 2 kΩ and Z_L = 400 Ω. The value of the parameter Z_out is ______________
A. 650 Ω
B. 500 Ω
C. 250 Ω
D. 600 Ω

Answer: A
Clarification: Z_out = (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})

Answer: C
Clarification: 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})

Answer: A
Clarification: 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

Answer: C
Clarification: 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

Answer: D
Clarification: 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.