300+ REAL TIME Network Theory Objective Questions & Answers 2023

250+ TOP MCQs on Supernode Analysis and Answers

Network Theory Multiple Choice Questions on “Supernode Analysis”.

1. Consider the figure shown below. Find the voltage (V) at node 1.

A. 13
B. 14
C. 15
D. 16

Answer: B
Clarification: Applying Super Node Analysis, the combined equation of node 1 and node 2 is (V₁-V₃)/3+3+(V₂-V₃)/1-6+V₂/5=0. At node 3, (V₃-V₁)/3+(V₃-V₂)/1+V₃/2=0. Also V₁-V₂=10. On solving above equations, we get V₁ = 13.72V ≈ 14V.

2. Consider the figure shown below. Find the voltage (V) at node 2.

A. 3
B. 4
C. 5
D. 6

3. Consider the figure shown below. Find the voltage (V) at node 3.

A. 4.5
B. 5.5
C. 6.5
D. 7.5

Answer: A
Clarification: Applying Super Node Analysis, the combined equation of node 1 and node 2 is (V₁-V₃)/3+3+(V₂-V₃)/1-6+V₂/5=0. At node 3, (V₃-V₁)/3+(V₃-V₂)/1+V₃/2=0. Also V₁-V₂=10. On solving above equations, we get V₃ = 4.5V.

4. Consider the figure shown below. Find the power (W) delivered by the source 6A.

A. 20.3
B. 21.3
C. 22.3
D. 24.3

Answer: C
Clarification: The term power is defined as the product of voltage and current and the power delivered by the source (6A. = V₂x6 = 3.72×6 = 22.32W.

5. Find the voltage (V) at node 1 in the circuit shown below.

A. 18
B. 19
C. 20
D. 21

Answer: B
Clarification: The equation at node 1 is 10 = V₁/3+(V₁-V₂)/2. According to super Node analysis, (V₁-V₂)/2=V₂/1+(V₃-10)/5+V₃/2V₂-V₃=20. On solving, we get, V₁=19V.

6. Consider the figure shown below. Find the voltage (V) at node 2.

A. 11.5
B. 12
C. 12.5
D. 13

Answer: A
Clarification: The equation at node 1 is 10 = V₁/3+(V₁-V₂)/2
According to super Node analysis, (V₁-V₂)/2=V₂/1+(V₃-10)/5+V₃/2V₂-V₃=20. On solving, we get, V₂=11.5V.

7. Find the voltage (V) at node 3 in the figure shown below.

A. 18
B. 20
C. 22
D. 24

Answer: A
Clarification: At node 1, (V₁-40-V₃)/4+(V₁-V₂)/6-3-5=0. Applying Super Node Analysis at node 2 and 3, (V₂-V₁)/6+5+V₂/3+V₃/5+(V₃+40-V₁)/4=0. Also, V₃-V₂=20. On solving above equations, V₃ = 18.11V ≈ 18V.

8. Find the power absorbed by 5Ω resistor in the following figure.

A. 60
B. 65.5
C. 70.6
D. 75

Answer: B
Clarification: The current through 5Ω resistor = V₃/5=18.11/5=3.62A. The power absorbed by 5Ω resistor = (3.62)²)×5=65.52W.

9. Find the value of the voltage (V) in the equivalent voltage source of the current source shown below.

A. 20
B. 25
C. 30
D. 35

Answer: C
Clarification: The value of the voltage (V) in the equivalent voltage source of the current source the voltage across the terminals A and B is (6)(5) = 30V.

10. Find the value of the current (A. in the equivalent current source of the voltage source shown below.

A. 1
B. 2
C. 3
D. 4

Answer: B
Clarification: The value of the current (A. in the equivalent current source of the voltage source the short circuit current at the terminals A and B is I=60/30=2A.

250+ TOP MCQs on Advanced Problems on Millman’s Theorem and Maximum Power Transfer Theorem and Answers

Network Theory Questions & Answers for Exams on “Advanced Problems on Millman’s Theorem and Maximum Power Transfer Theorem”.

1. The basic elements of an electric circuit are _____________
A. R, L and C
B. Voltage
C. Current
D. Voltage as well as current

Answer: A
Clarification: The elements which show their behaviour only when excited are called as basic circuit elements. Here resistance, inductance and capacitance show their behaviour only when excited. Hence they are the basic elements of an electric circuit.

2. In the circuit given below, the value of the maximum power transferred through R_L is ___________

A. 0.75 W
B. 1.5 W
C. 2.25 W
D. 1.125 W

Answer: A
Clarification: I + 0.9 = 10 I
Or, I = 0.1 A
V_OC = 3 × 10 I = 30 I
Or, V_OC = 3 V
Now, I_SC = 10 I = 1 A
R_th = 3/1 = 3 Ω
V_th = V_OC = 3 V
R_L = 3 Ω
P_max = (frac{3^2}{4×3}) = 0.75 W.

3. The energy stored in the magnetic field at a solenoid 100 cm long and 10 cm diameter wound with 1000 turns of wire carrying an electric current of 10 A, is ___________
A. 1.49 J
B. 0.49 J
C. 0.1 J
D. 1 J

Answer: B
Clarification: L = (frac{N^2 μ_0 A}{l})
= (frac{10^6.4π.10^{-7}.frac{π}{4}.(100 X 10^{-4})}{1})
= (frac{π^2 X 10^{-3}}{1})
Energy = 0.5 LI²
= 0.49 J.

4. The resistance of a strip of copper of rectangular cross section is 2 Ω. A metal of resistivity twice that of a copper is coated on its upper surface to a thickness equal to that of copper strip. The resistance of composite strip will be _________
A. (frac{3}{4}) Ω
B. (frac{4}{3}) Ω
C. (frac{3}{2}) Ω
D. 6 Ω

Answer: B
Clarification: Given that copper and coated metal strip have resistance of 2 ohms respectively. These two strips are connected in parallel.
Hence, the resistance of the composite strip = (frac{2 X 4}{2 + 4})
= (frac{8}{6} = frac{4}{3}) Ω.

5. In the circuit shown below what is the value of R_L for which maximum power is transferred to R_L?

A. 2.4 Ω
B. (frac{8}{3}) Ω
C. 4 Ω
D. 6 Ω

Answer: C
Clarification: Maximum power is transferred to R_L when the load resistance equals the Thevenin resistance of the circuit.
R_L = R_TH = (frac{V_{OC}}{I_{SC}} )
Due to open-circuit, V_OC = 100 V; I_SC = I₁ + I₂
Applying KVL in lower loop, 100 – 8I₁ = 0
Or, I₁ = (frac{100}{8} = frac{25}{2})
And V_X = -4I₁ = -4 × (frac{25}{2}) = -50V
KVL in upper loop, 100 + V_X – 4I₂ = 0
I₂ = (frac{100-50}{4} = frac{25}{2})
Hence, I_SC = I₁ + I₂ = (frac{25}{2} + frac{25}{2}) = 25
R_TH = (frac{V_{OC}}{I_{SC}} = frac{100}{25}) = 4 Ω
R_L = R_TH = 4 Ω.

6. A 3 V DC supply with an internal resistance of 2 Ω supplies a passive non-linear resistance characterized by the relation V_NL = (I_{NL}^{2}). The power dissipated in the resistance is ___________
A. 1 W
B. 1.5 W
C. 2.5 W
D. 3 W

Answer: A
Clarification: 3 = 2I + I²
∴ I = 1 A; V_NL = 1V
∴ Power dissipated in R_NL = 1 × 1 = 1 W.

7. The two windings of a transformer have an inductor of 2 H each. If mutual inductor between them is also 2 H, then which of the following is correct?
A. Turns ratio of the transformer is also 2
B. Transformer is an ideal transformer
C. It is a perfect transformer
D. It is a perfect as well as an ideal transformer

Answer: C
Clarification: We know that, K = (frac{M}{sqrt{L_1 L_2}})
= (frac{2}{sqrt{2 X 2}}) = 1
Hence, it is a perfect transformer.

8. In the circuit given below, what is the amount of maximum power transfer to R?

A. 56 W
B. 76 W
C. 60 W
D. 66 W

Answer: D
Clarification: Drop across V_1Ω = 5 × 1 = 5V
Also, (frac{V-V_{1Ω}}{10} + frac{V-20-V_{1Ω}}{2} + frac{V-V_{OC}}{5}) = 2
Or, 0.1 V – 0.1V_1Ω + 0.5V – 10 – 0.5V_1Ω + 0.2 – 0.2V_OC = 2
Or, 0.8V – 0.6V_1Ω = 12 + 0.2V_OC
Or, 0.8 V – 0.2V_OC = 12 +3=15 (Putting V_1Ω = 5)
Again, (frac{V_{OC}-V}{5}) + 2 = 5
Or, 0.2V_OC – 0.2V = 3
Again, R_TH = {(10||2) + 1} + 5
= ((frac{20}{12+1})) + 5 = 7.67 Ω
Following the theorem of maximum power transfer, R = R_TH = 7.67 Ω
And P_MAX = (frac{V_{OC}^2}{4R} = frac{45^2}{4×7.67}) = 66 W.

9. In the circuit given below, the value of R_L for which it absorbs maximum power is ___________

A. (frac{400}{3}) Ω
B. (frac{2}{9}) kΩ
C. 350.38 Ω
D. (frac{4}{9}) kΩ

Answer: C
Clarification: 5 = 200I – 50 × 2I
Or, I = (frac{5}{100}) = 0.05 A
V_OC = 100 × 3I + 200 × I = 25 V
V₁ = (frac{frac{5}{50}}{frac{1}{50} + frac{1}{200} + frac{1}{100}} )
= (frac{0.1}{0.02+0.005+0.01} )
= 2.85 V
I = (frac{2.85}{100}) = 0.0142 A = 14.2 mA
I_SC = (frac{2.85}{100}) + 3 × 0.0142 = 0.07135 A
∴ R_TH = (frac{V_{OC}}{I_{SC}} = frac{25}{0.07135}) = 350.38 Ω.

10. The form factor of sinusoidal alternating electric current is ___________
A. 0
B. 1
C. 1.11
D. 1.15

Answer: C
Clarification: We know that for alternating electric current form factor is defined as the ratio of rms value and average value of alternating current.
Now the rms value of alternating electric current = 0.07 × maximum value of alternating current.
Average value of alternating electric current = 0.637 × maximum value of alternating current.
∴ Form factor = (frac{0.707}{0.637}) = 1.11.

11. The average power delivered to the 6 Ω load in the circuit of figure below is ___________

A. 8 W
B. 76.68 W
C. 625 kW
D. 2.50 kW

Answer: B
Clarification: I₂ = (frac{V_2}{6}), I₁ = (frac{I_2}{5} = frac{V_2}{30})
V₁ = 5V₂
50 = 400(I₁ – 0.04V₂) + V₁
Or, V₂ = 21.45 V
∴ P_L = (frac{V_2^2}{6} )
= (frac{21.45^2}{6} )
= (frac{460.1025}{6}) = 76.68 W.

12. The rms value of the sine wave is 100 A. Its peak value is ____________
A. 70.7 A
B. 141.4 A
C. 150 A
D. 282.8 A

Answer: B
Clarification: We know that for sinusoidal alternating electric current the peak factor or amplitude factor can be expressed the ratio of maximum or peak value and rms value of alternating current.
So the peak value = rms value of alternating electric current × peak factor of alternating electric current = 100 × 1.414 = 141.4 A.

13. Potential of earth is – 50 V. If the potential difference between anode and cathode (eartheD. is measured as 150 V, actual voltage on anode is __________
A. 0 V
B. 100 V
C. 200 V
D. 250 V

Answer: C
Clarification: Given that, potential difference between anode and cathode (eartheD. is measured as 150 V and potential of earth is – 50 V.
So, actual voltage on anode, V = 150 – (- 50)
= 150 + 50
= 200 V.

14. An alternating voltage V = 150 sin(314)t is supplied to a device which offers a resistance of 20 Ω in forward direction of electric current while preventing the flow of electric current in reverse direction. The form factor is ___________
A. 0.57
B. 0.318
C. 1.414
D. 1.57

Answer: D
Clarification: From the voltage equation, we can get V_m = 150 V and I_m = 150 / 20 = 7 A
RMS value of the current, I_rms = I_m / 2 = 7/2 = 3.5 A
Average value of the current, I_avg = I_m / π = 2.228 A
Form factor = I_rms / I_avg = 3.5 / 2.228 = 1.57.

15. A coil of 300 turns is wound on a non-magnetic core having a mean circumference of 300 mm and a cross sectional area of 300 mm². The inductor of the coil corresponding to a magnetizing electric current of 3 A will be?
A. 37.68 μH
B. 47.68 μH
C. 113.04 μH
D. 120.58 μH

Answer: C
Clarification: Inductance of the coil, L = (frac{μ_0 n^2 A}{l})
= (frac{4π X 10^{-7} X 300 X 300 X 300 X 10^{-6}}{300 X 10^{-3}})
= 113.04 μH.

250+ TOP MCQs on Problems of Parallel Resonance Involving Quality Factor and Answers

Network Theory Questions & Answers for Exams on “Problems of Parallel Resonance Involving Quality Factor”.

1. For the parallel resonant circuit shown below, the value of the resonant frequency is _________

A. 1.25 MHz
B. 2.5 MHz
C. 5 MHz
D. 1.5 MHz

Answer: A
Clarification: Resonant Frequency, F_R = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(100×10^{-6})(162.11×10^{-12})}})
= (frac{1}{6.28sqrt{1.621×10^{-14}}})
= (frac{1}{799×10^{-9}}) = 1.25 MHz.

2. For the parallel resonant circuit given below, the value of the inductive and capacitive reactance is _________

A. X_L = 654.289 Ω; X_C = 458.216 Ω
B. X_L = 985.457 Ω; X_C = 875.245 Ω
C. X_L = 785.394 Ω; X_C = 785.417 Ω
D. X_L = 125.354 Ω; X_C = 657.215 Ω

Answer: C
Clarification: Resonant Frequency, f = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(100×10^{-6})(162.11×10^{-12})}})
= (frac{1}{6.28sqrt{1.621×10^{-14}}})
= (frac{1}{799×10^{-9}}) = 1.25 MHz.
Inductive Reactance, X_L = 2πfL = (6.28) (1.25×10⁶)(100×10^-6)
= 785.394 Ω
Capacitive Reactance, X_C = (frac{1}{2πfC} = frac{1}{(6.28)(1.25×10^6)(162.11×10^{-12})})
= (frac{1}{1.273×10^{-3}}) = 785.417 Ω.

3. For the parallel resonant circuit given below, the current through the capacitor and inductor are _________

A. I_C = 10.892 mA; I_L = 12.732 mA
B. I_C = 12.732 mA; I_L = 10.892 mA
C. I_C = 10.892 mA; I_L = 10.892 mA
D. I_C = 12.732 mA; I_L = 12.732 mA

Answer: D
Clarification: Resonant Frequency, f = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(100×10^{-6})(162.11×10^{-12})}})
= (frac{1}{6.28sqrt{1.621×10^{-14}}})
= (frac{1}{799×10^{-9}}) = 1.25 MHz.
Inductive Reactance, X_L = 2πfL = (6.28) (1.25×10⁶)(100×10^-6)
= 785.394 Ω
Capacitive Reactance, X_C = (frac{1}{2πfC} = frac{1}{(6.28)(1.25×10^6)(162.11×10^{-12})})
= (frac{1}{1.273×10^{-3}}) = 785.417 Ω.
I_C = (frac{V_A}{X_C} )
= (frac{10}{785.417} ) = 12.732 mA
I_L = (frac{V_A}{X_L} )
= (frac{10}{785.394} ) = 12.732 mA.

4. For the parallel resonant circuit given below, the value of the equivalent impedance of the circuit is ________

A. 56.48 kΩ
B. 78.58 kΩ
C. 89.12 kΩ
D. 26.35 kΩ

Answer: B
Clarification: f = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(100×10^{-6})(162.11×10^{-12})}})
= (frac{1}{6.28sqrt{1.621×10^{-14}}})
= (frac{1}{799×10^{-9}}) = 1.25 MHz.
Inductive Reactance, X_L = 2πfL = (6.28) (1.25×10⁶)(100×10^-6)
= 785.394 Ω
Q = (frac{X_L}{R} = frac{785.394}{7.85}) = 100.05
∴ Z_EQ = QX_L = (100.05)(785.394) = 78.58 kΩ.

5. For the parallel resonant circuit, the bandwidth of the circuit is ____________

A. ±6.25 kHz
B. ±8.56 kHz
C. ±10.35 kHz
D. ±6.37 kHz

Answer: A
Clarification: Resonant Frequency, f = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(100×10^{-6})(162.11×10^{-12})}})
= (frac{1}{6.28sqrt{1.621×10^{-14}}})
= (frac{1}{799×10^{-9}}) = 1.25 MHz.
Inductive Reactance, X_L = 2πfL = (6.28) (1.25×10⁶)(100×10^-6)
= 785.394 Ω
Q = (frac{X_L}{R} = frac{785.394}{7.85}) = 100.05
∴ ∆F = (frac{f}{Q} = frac{1.25 × 10^6}{100.05}) = 12.5 kHz
Hence, bandwidth = (frac{∆F}{2}) = 6.25 kHz.

6. For the parallel resonant circuit given below, the cut-off frequencies are ____________

A. ∆f₁ = 2.389 MHz; ∆f₂ = 441.124 MHz
B. ∆f₁ = 1.256 MHz; ∆f₂ = 1.244 MHz
C. ∆f₁ = 5.658 MHz; ∆f₂ = 6.282 MHz
D. ∆f₁ = 3.656 MHz; ∆f₂ = 8.596 MHz

Answer: B
Clarification: Resonant Frequency, f = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(100×10^{-6})(162.11×10^{-12})}})
= (frac{1}{6.28sqrt{1.621×10^{-14}}})
= (frac{1}{799×10^{-9}}) = 1.25 MHz.
Inductive Reactance, X_L = 2πfL = (6.28) (1.25×10⁶)(100×10^-6)
= 785.394 Ω
Q = (frac{X_L}{R} = frac{785.394}{7.85}) = 100.05
∴ ∆F = (frac{f}{Q} = frac{1.25 × 10^6}{100.05}) = 12.5 kHz
Hence, bandwidth = (frac{∆F}{2}) = 6.25 kHz
∴ ∆f₁ = f + (frac{∆F}{2}) = 1.25 MHz + 6.25 kHz = 1.256 MHz
∴ ∆f₁ = f – (frac{∆F}{2}) = 1.25 MHz – 6.25 kHz = 1.244 MHz.

7. For the series resonant circuit shown below, the value of the resonant frequency is _________

A. 10.262 kHz
B. 44.631 kHz
C. 50.288 kHz
D. 73.412 kHz

Answer: D
Clarification: Resonant Frequency, F_R = (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(4.7×10^{-3})(0.001×10^{-6})}})
= (frac{1}{6.28sqrt{4.7×10^{-12}}})
= (frac{1}{1.362×10^{-5}}) = 73.412 kHz.

8. For the series resonant circuit given below, the value of the inductive and capacitive reactance is _________

A. X_L = 5.826 kΩ; X_C = 5.826 kΩ
B. X_L = 2.168 kΩ; X_C = 2.168 kΩ
C. X_L = 6.282 kΩ; X_C = 6.282 kΩ
D. X_L = 10.682 kΩ; X_C = 10.682 kΩ

9. For the series resonant circuit given below, the value of the equivalent impedance of the circuit is ________

A. 55 Ω
B. 47 Ω
C. 64 Ω
D. 10 Ω

Answer: B
Clarification: Resonant Frequency, (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(4.7×10^{-3})(0.001×10^{-6})}})
= (frac{1}{6.28sqrt{4.7×10^{-12}}})
= (frac{1}{1.362×10^{-5}}) = 73.412 kHz
Inductive Reactance, X_L = 2πfL = (6.28) (73.142 × 10³)(4.7 × 10^-6)
= 2.168 kΩ
Capacitive Reactance, X_C = (frac{1}{2πfC} = frac{1}{(6.28)(73.142×10^3)(0.001×10^{-6})})
= (frac{1}{4.613×10^{-4}}) = 2.168 kΩ
We see that, X_C = X_L are equal, along with being 180° out of phase.
Hence the net reactance is zero and the total impedance equal to the resistor.
∴ Z_EQ = R = 47 Ω.

10. For the series resonant circuit given below, the value of the total current flowing through the circuit is ____________

A. 7.521 mA
B. 6.327 mA
C. 2.168 mA
D. 9.136 mA

Answer: A
Clarification: Resonant Frequency, (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(4.7×10^{-3})(0.001×10^{-6})}})
= (frac{1}{6.28sqrt{4.7×10^{-12}}})
= (frac{1}{1.362×10^{-5}}) = 73.412 kHz
Inductive Reactance, X_L = 2πfL = (6.28) (73.142 × 10³)(4.7 × 10^-6)
= 2.168 kΩ
Capacitive Reactance, X_C = (frac{1}{2πfC} = frac{1}{(6.28)(73.142×10^3)(0.001×10^{-6})})
= (frac{1}{4.613×10^{-4}}) = 2.168 kΩ
Z_EQ = R = 47 Ω
I_T = (frac{V_{in}}{Z_{EQ}} = frac{V_{in}}{R} = frac{0.3535}{47}) = 7.521 mA.

11. For the series circuit given below, the value of the voltage across the capacitor and inductor are _____________

A. V_C = 16.306 V; V_L = 16.306 V
B. V_C = 11.268 V; V_L = 11.268 V
C. V_C = 16.306 V; V_L = 16.306 V
D. V_C = 14.441 V; V_L = 14.441 V

Answer: C
Clarification: Resonant Frequency, (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(4.7×10^{-3})(0.001×10^{-6})}})
= (frac{1}{6.28sqrt{4.7×10^{-12}}})
= (frac{1}{1.362×10^{-5}}) = 73.412 kHz
Inductive Reactance, X_L = 2πfL = (6.28) (73.142 × 10³)(4.7 × 10^-6)
= 2.168 kΩ
Capacitive Reactance, X_C = (frac{1}{2πfC} = frac{1}{(6.28)(73.142×10^3)(0.001×10^{-6})})
= (frac{1}{4.613×10^{-4}}) = 2.168 kΩ
Z_EQ = R = 47 Ω
I_T = (frac{V_{in}}{Z_{EQ}} = frac{V_{in}}{R} = frac{0.3535}{47}) = 7.521 mA
∴ Voltage across the capacitor, V_C = X_CI_T = (2.168 kΩ)(7.521 mA. = 16.306 V
∴ Voltage across the inductor, V_L = X_LI_T = (2.168 kΩ)(7.521 mA. = 16.306 V.

12. For the series resonant circuit given below, the value of the quality factor is ___________

A. 35.156
B. 56.118
C. 50.294
D. 46.128

Answer: D
Clarification: Resonant Frequency, (frac{1}{2πsqrt{LC}} )
= (frac{1}{6.28sqrt{(4.7×10^{-3})(0.001×10^{-6})}})
= (frac{1}{6.28sqrt{4.7×10^{-12}}})
= (frac{1}{1.362×10^{-5}}) = 73.412 kHz
Inductive Reactance, X_L = 2πfL = (6.28) (73.142 × 10³)(4.7 × 10^-6)
= 2.168 kΩ
Quality factor Q = (frac{X_L}{R} = frac{2.168 kΩ}{47}) = 46.128.

13. For a parallel RLC circuit, the incorrect statement among the following is _____________
A. The bandwidth of the circuit decreases if R is increased
B. The bandwidth of the circuit remains same if L is increased
C. At resonance, input impedance is a real quantity
D. At resonance, the magnitude of input impedance attains its minimum value

Answer: D
Clarification: BW = 1/RC
It is clear that the bandwidth of a parallel RLC circuit is independent of L and decreases if R is increased.
At resonance, imaginary part of input impedance is zero. Hence, at resonance input impedance is a real quantity.
In parallel RLC circuit, the admittance is minimum, at resonance. Hence the magnitude of input impedance attains its maximum value at resonance.

14. A circuit excited by voltage V has a resistance R which is in series with an inductor and capacitor, which are connected in parallel. The voltage across the resistor at the resonant frequency is ___________
A. 0
B. (frac{V}{2})
C. (frac{V}{3})
D. V

Answer: A
Clarification: Dynamic resistance of the tank circuit, Z_DY = L/(R_LC.
But given that R_L = 0
So, Z_DY = L/(0X_C) = ∞
Therefore current through circuit, I = (frac{V}{∞}) = 0
∴ V_D = 0.

15. For the circuit given below, the nature of the circuit is ____________

A. Inductive
B. Capacitive
C. Resistive
D. Both inductive as well as capacitive

Answer: C
Clarification: θ = 0° since X_L and X_C are cancelling, which means at resonance the circuit is purely resistive.

250+ TOP MCQs on DC Response of an R-C Circuit and Answers

Network Theory Multiple Choice Questions on “DC Response of an R-C Circuit”.

1. The current in the R-L circuit at a time t = 0⁺ is?
A. V/R
B. R/V
C. V
D. R

Answer: A
Clarification: The capacitor never allows sudden changes in voltage, it will act as a short circuit at t = 0⁺. So the current in the circuit at t = 0⁺ is V/R.

2. The expression of current in R-C circuit is?
A. i=(V/R)exp⁡(t/RC.
B. i=(V/R)exp⁡(-t/RC.
C. i=(V/R)-exp(⁡t/RC.
D. i=(V/R)-exp⁡(-t/RC.

Answer: B
Clarification: The particular solution of the current equation is zero. So the expression of current in R-C circuit is i=(V/R)exp⁡(-t/RC..

3. In an R-C circuit, when the switch is closed, the response ____________
A. do not vary with time
B. decays with time
C. rises with time
D. first increases and then decreases

Answer: B
Clarification: In a R-C circuit, when the switch is closed, the response decays with time that is the response V/R decreases with increase in time.

4. The time constant of an R-C circuit is?
A. RC
B. R/C
C. R
D. C

Answer: A
Clarification: The time constant of an R-C circuit is RC and it is denoted by τ and the value of τ in dc response of R-C circuit is RC sec.

5. After how many time constants, the transient part reaches more than 99 percent of its final value?
A. 2
B. 3
C. 4
D. 5

Answer: D
Clarification: After five time constants, the transient part of the response reaches more than 99 percent of its final value.

6.A series R-C circuit consists of resistor of 10 and capacitor of 0.1F as shown in the figure. A constant voltage of 20V is applied to the circuit at t = 0. What is the current in the circuit at t = 0?

A. 1
B. 2
C. 3
D. 4

Answer: B
Clarification: At t = 0, switch S is closed. Since the capacitor does not allow sudden changes in voltage, the current in the circuit is i = V/R = 20/10 = 2A. At t = 0, i = 2A.

7. The expression of current obtained from the circuit in terms of differentiation from the circuit shown below?

A. di/dt+i=1
B. di/dt+i=2
C. di/dt+i=3
D. di/dt+i=0

Answer: D
Clarification: By applying Kirchhoff’s law, we get

Differentiating with respect to t, we get 10 di/dt+i/0.1=0 => di/dt+i=0.

8. The current equation in the circuit shown below is?

A. i=2(e^-2t)A
B. i=2(e^2t)A
C. i=2(-e^-2t)A
D. i=2(-e^2t)A

Answer: A
Clarification: At t = 0, switch S is closed. Since the capacitor does not allow sudden changes in voltage, the current in the circuit is i = V/R = 20/10 = 2A. At t = 0, i = 2A. The current equation is i=2(e^-2t)A.

9. The expression of voltage across resistor in the circuit shown below is?

A. V_R = 20(e^t)V
B. V_R = 20(-e^-t)V
C. V_R = 20(-e^t)V
D. V_R = 20(e^-t)V

Answer: D
Clarification: The expression of voltage across resistor in the circuit is V_R = iR =(2(e^-t))×10=20(e^-t)V.

10. Determine the voltage across the capacitor in the circuit shown below is?

A. V_C = 60(1-e^-t)V
B. V_C = 60(1+e^t)V
C. V_C = 60(1-e^t)V
D. V_C = 60(1+e^-t)V

Answer: A
Clarification: The expression of voltage across capacitor in the circuit V_C = V(1-e^-t/RC) = 20(1-e^-t)V.

250+ TOP MCQs on The Concept of Complex Frequency and Answers

Network Theory Multiple Choice Questions on “The Concept of Complex Frequency”.

1. The solution of differential equations for networks is of the form?
A. i(t)=K_n e^{⁡(s_n t)}
B. i(t)=K_n e⁡^{(-s_n t)}
C. i(t)=-K_n e⁡^{(-s_n t)}
D. i(t)=-K_n e⁡^{(s_n t)}
Answer: A
Clarification: The solution of differential equations for networks is of the form
i(t)=K_n e^{⁡(s_n t)} where S_n is a complex number which is a root of the characteristic equation.

2. The real part of the complex frequency is called?
A. radian frequency
B. neper frequency
C. sampling frequency
D. angular frequency
Answer: B
Clarification: The complex number consists of two parts, the real part and the imaginary part. The real part of the complex frequency is called neper frequency.

3. The imaginary part of the complex frequency is called?
A. angular frequency
B. sampling frequency
C. neper frequency
D. radian frequency
Answer: D
Clarification: The complex number consists of two parts, the real part of the complex frequency is called radian frequency. The radian frequency is expressed in radian/sec and is related to the frequency or the periodic time.

4. The ratio of transform voltage to the transform current is defined as _________ of the resistor.
A. transform voltage
B. transform current
C. transform impedance
D. transform admittance
Answer: C
Clarification: Transform impedance of the resistor is defined as the ratio of transform voltage to the transform current and is expressed as Z_R(s) = V_R(s)/I_R(s) = R.

5. The ratio of transform current to the transform voltage is defined as ________ of the resistor.
A. transform admittance
B. transform impedance
C. transform current
D. transform voltage
Answer: A
Clarification: Transform admittance of the resistor is defined as the ratio of transform current to the transform voltage and it is also defined as the reciprocal of transform impedance. Y_R(s) = I_R(s)/V_R(s) = G.

6. The transform impedance of the inductor is?
A. L
B. 1/L
C. sL
D. 1/sL
Answer: C
Clarification: Considering the sum of the transform voltage and the initial current voltage as V₁(s) we have the transform impedance of the inductor. The transform impedance of the inductor is Z_L(s) = V₁(s)/I_L(s) = sL.

7. The transform admittance of the inductor is?
A. 1/sL
B. sL
C. 1/L
D. L
Answer: A
Clarification: The transform admittance of the inductor is Y_L(s) = I₁(s)/V_L(s) = 1/sL where I₁(s) is the total transform current through the inductor L.

8. The equivalent transform circuit contains an admittance of value ____ and equivalent transform current source.
A. 1/L
B. 1/sL
C. L
D. sL
Answer: B
Clarification: The time domain representation of inductor L has initial current i_L(0⁺). The equivalent transform circuit contains an admittance of value 1/sL and equivalent transform current source.

9. The transform impedance of the capacitor is?
A. C
B. 1/C
C. sC
D. 1/sC
Answer: D
Clarification: The transform impedance of the capacitor is the ratio of the transform voltage V₁(s) to the transform current I_C(s) and is Z_C(s) = 1/Cs.

10. The transform admittance of the capacitor is?
A. 1/sC
B. sC
C. 1/C
D. C
Answer: B
Clarification: The transform admittance of the capacitor is the ratio of transform current I₁(s) to transform voltage V_C(s) and is Y_C(s) = sC.

250+ TOP MCQs on Relation between Transmission Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters and Answers

Network Theory Puzzles on “Relation between Transmission Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters”.

1. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₁₁ in terms of Transmission parameters can be expressed as ________
A. y₁₁ = (frac{D}{B})
B. y₁₁ = (frac{C-A}{B})
C. y₁₁ = – (frac{1}{B})
D. y₁₁ = (frac{A}{B})

Answer: A
Clarification: We know that, V₁ = AV₂ – BI₂ ……… (1)
I₁ = CV₂ – DI₂ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₂ = (frac{A}{B}V_2 – frac{1}{B}V_1) …………. (5)
And I₁ = CV₂ – D (left(frac{A}{B}V_2 – frac{1}{B}V_1right) = frac{D}{B}V_1 + left(C-frac{A}{B}right) V_2) …………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = (frac{D}{B})
y₁₂ = (frac{C-A}{B})
y₂₁ = – (frac{1}{B})
y₂₂ = (frac{A}{B}).

2. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₁₂ in terms of Transmission parameters can be expressed as ________
A. y₁₂ = (frac{D}{B})
B. y₁₂ = (frac{C-A}{B})
C. y₁₂ = – (frac{1}{B})
D. y₁₂ = (frac{A}{B})

Answer: B
Clarification: We know that, V₁ = AV₂ – BI₂ ……… (1)
I₁ = CV₂ – DI₂ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₂ = (frac{A}{B}V_2 – frac{1}{B}V_1) …………. (5)
And I₁ = CV₂ – D (left(frac{A}{B}V_2 – frac{1}{B}V_1right) = frac{D}{B}V_1 + left(C-frac{A}{B}right) V_2) …………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = (frac{D}{B})
y₁₂ = (frac{C-A}{B})
y₂₁ = – (frac{1}{B})
y₂₂ = (frac{A}{B}).

3. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₂₁ in terms of Transmission parameters can be expressed as ________
A. Y₂₁ = (frac{D}{B})
B. Y₂₁ = (frac{C-A}{B})
C. Y₂₁ = – (frac{1}{B})
D. Y₂₁ = (frac{A}{B})

Answer: C
Clarification: We know that, V₁ = AV₂ – BI₂ ……… (1)
I₁ = CV₂ – DI₂ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₂ = (frac{A}{B}V_2 – frac{1}{B}V_1) …………. (5)
And I₁ = CV₂ – D (left(frac{A}{B}V_2 – frac{1}{B}V_1right) = frac{D}{B}V_1 + left(C-frac{A}{B}right) V_2) …………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = (frac{D}{B})
y₁₂ = (frac{C-A}{B})
y₂₁ = – (frac{1}{B})
y₂₂ = (frac{A}{B}).

4. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₂₂ in terms of Transmission parameters can be expressed as ________
A. y₂₂ = (frac{D}{B})
B. y₂₂ = (frac{C-A}{B})
C. y₂₂ = – (frac{1}{B})
D. y₂₂ = (frac{A}{B})

Answer: D
Clarification: We know that, V₁ = AV₂ – BI₂ ……… (1)
I₁ = CV₂ – DI₂ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₂ = (frac{A}{B}V_2 – frac{1}{B}V_1) …………. (5)
And I₁ = CV₂ – D (left(frac{A}{B}V_2 – frac{1}{B}V_1right) = frac{D}{B}V_1 + left(C-frac{A}{B}right) V_2) …………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = (frac{D}{B})
y₁₂ = (frac{C-A}{B})
y₂₁ = – (frac{1}{B})
y₂₂ = (frac{A}{B}).

5. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₁₁ in terms of Transmission parameters can be expressed as ____________
A. z₁₁ = (frac{A}{C})
B. z₁₁ = (frac{AD}{C – B})
C. z₁₁ = (frac{1}{C})
D. z₁₁ = (frac{D}{C})

Answer: A
Clarification: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₁ = AV₂ – BI₂ ……… (3)
I₁ = CV₂ – DI₂ …………… (4)
Rewriting (3) and (4), we get,
V₂ = (frac{1}{C}I_1 + frac{D}{C}I_2) …………… (5)
And V₁ = (A left(frac{1}{C}I_1 + frac{D}{C}I_2right) – BI_2 = frac{A}{C}I_1 + left(frac{AD}{C} – Bright) I_2) ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = (frac{A}{C})
z₁₂ = (frac{AD}{C – B})
z₂₁ = (frac{1}{C})
z₂₂ = (frac{D}{C}).

6. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₁₂ in terms of Transmission parameters can be expressed as ____________
A. z₁₂ = (frac{A}{C})
B. z₁₂ = (frac{AD}{C – B})
C. z₁₂ = (frac{1}{C})
D. z₁₂ = (frac{D}{C})

Answer: B
Clarification: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₁ = AV₂ – BI₂ ……… (3)
I₁ = CV₂ – DI₂ …………… (4)
Rewriting (3) and (4), we get,
V₂ = (frac{1}{C}I_1 + frac{D}{C}I_2) …………… (5)
And V₁ = (A left(frac{1}{C}I_1 + frac{D}{C}I_2right) – BI_2 = frac{A}{C}I_1 + left(frac{AD}{C} – Bright) I_2) ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = (frac{A}{C})
z₁₂ = (frac{AD}{C – B})
z₂₁ = (frac{1}{C})
z₂₂ = (frac{D}{C}).

7. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₂₁ in terms of Transmission parameters can be expressed as ____________
A. z₂₁ = (frac{A}{C})
B. z₂₁ = (frac{AD}{C – B})
C. z₂₁ = (frac{1}{C})
D. z₂₁ = (frac{D}{C})

Answer: C
Clarification: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₁ = AV₂ – BI₂ ……… (3)
I₁ = CV₂ – DI₂ …………… (4)
Rewriting (3) and (4), we get,
V₂ = (frac{1}{C}I_1 + frac{D}{C}I_2) …………… (5)
And V₁ = (A left(frac{1}{C}I_1 + frac{D}{C}I_2right) – BI_2 = frac{A}{C}I_1 + left(frac{AD}{C} – Bright) I_2) ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = (frac{A}{C})
z₁₂ = (frac{AD}{C – B})
z₂₁ = (frac{1}{C})
z₂₂ = (frac{D}{C}).

8. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₂₂ in terms of Transmission parameters can be expressed as ____________
A. z₂₂ = (frac{A}{C})
B. z₂₂ = (frac{AD}{C – B})
C. z₂₂ = (frac{1}{C})
D. z₂₂ = (frac{D}{C})

Answer: D
Clarification: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₁ = AV₂ – BI₂ ……… (3)
I₁ = CV₂ – DI₂ …………… (4)
Rewriting (3) and (4), we get,
V₂ = (frac{1}{C}I_1 + frac{D}{C}I_2) …………… (5)
And V₁ = (A left(frac{1}{C}I_1 + frac{D}{C}I_2right) – BI_2 = frac{A}{C}I_1 + left(frac{AD}{C} – Bright) I_2) ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = (frac{A}{C})
z₁₂ = (frac{AD}{C – B})
z₂₁ = (frac{1}{C})
z₂₂ = (frac{D}{C}).

9. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₁₁ in terms of Inverse Transmission parameters can be expressed as ________
A. y₁₁ = (frac{A’}{B’})
B. y₁₁ = – (frac{1}{B’})
C. y₁₁ = (left(C’ – frac{D’ A’}{B’}right))
D. y₁₁ = (frac{D’}{B’})

Answer: A
Clarification: We know that, V₂ = A’V₁ – B’I₁ ……… (1)
I₂ = C’V₁ – D’I₁ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₁ = – (frac{1}{B’} V_2 + frac{A’}{B’} V_1) …………. (5)
And I₂ = C’V₁ – D’ (left(- frac{1}{B’} V_2 + frac{A’}{B’} V_1right) = left(C’ – frac{D’ A’}{B’}right) V_1 + frac{D’}{B’} V_2) ………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = (frac{A’}{B’})
y₁₂ = – (frac{1}{B’})
y₂₁ = (left(C’ – frac{D’ A’}{B’}right))
y₂₂ = (frac{D’}{B’}).

10. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₁₂ in terms of Inverse Transmission parameters can be expressed as ________
A. y₁₂ = (frac{A’}{B’})
B. y₁₂ = – (frac{1}{B’})
C. y₁₂ = (left(C’ – frac{D’ A’}{B’}right))
D. y₁₂ = (frac{D’}{B’})

Answer: B
Clarification: We know that, V₂ = A’V₁ – B’I₁ ……… (1)
I₂ = C’V₁ – D’I₁ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₁ = – (frac{1}{B’} V_2 + frac{A’}{B’} V_1) …………. (5)
And I₂ = C’V₁ – D’ (left(- frac{1}{B’} V_2 + frac{A’}{B’} V_1right) = left(C’ – frac{D’ A’}{B’}right) V_1 + frac{D’}{B’} V_2) ………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = (frac{A’}{B’})
y₁₂ = – (frac{1}{B’})
y₂₁ = (left(C’ – frac{D’ A’}{B’}right))
y₂₂ = (frac{D’}{B’}).

11. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₂₁ in terms of Inverse Transmission parameters can be expressed as ________
A. y₂₁ = (frac{A’}{B’})
B. y₂₁ = – (frac{1}{B’})
C. y₂₁ = (left(C’ – frac{D’ A’}{B’}right))
D. y₂₁ = (frac{D’}{B’})

Answer: C
Clarification: We know that, V₂ = A’V₁ – B’I₁ ……… (1)
I₂ = C’V₁ – D’I₁ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₁ = – (frac{1}{B’} V_2 + frac{A’}{B’} V_1) …………. (5)
And I₂ = C’V₁ – D’ (left(- frac{1}{B’} V_2 + frac{A’}{B’} V_1right) = left(C’ – frac{D’ A’}{B’}right) V_1 + frac{D’}{B’} V_2) ………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = (frac{A’}{B’})
y₁₂ = – (frac{1}{B’})
y₂₁ = (left(C’ – frac{D’ A’}{B’}right))
y₂₂ = (frac{D’}{B’}).

12. For a T shaped network, if the Short-circuit admittance parameters are y₁₁, y₁₂, y₂₁, y₂₂, then y₂₂ in terms of Inverse Transmission parameters can be expressed as ________
A. y₂₂ = (frac{A’}{B’})
B. y₂₂ = – (frac{1}{B’})
C. y₂₂ = (left(C’ – frac{D’ A’}{B’}right))
D. y₂₂ = (frac{D’}{B’})

Answer: D
Clarification: We know that, V₂ = A’V₁ – B’I₁ ……… (1)
I₂ = C’V₁ – D’I₁ …………… (2)
And, I₁ = y₁₁ V₁ + y₁₂ V₂ ……… (3)
I₂ = y₂₁ V₁ + y₂₂ V₂ ………. (4)
Now, (1) and (2) can be rewritten as, I₁ = – (frac{1}{B’} V_2 + frac{A’}{B’} V_1) …………. (5)
And I₂ = C’V₁ – D’ (left(- frac{1}{B’} V_2 + frac{A’}{B’} V_1right) = left(C’ – frac{D’ A’}{B’}right) V_1 + frac{D’}{B’} V_2) ………… (6)
Comparing equations (3), (4) and (5), (6), we get,
y₁₁ = (frac{A’}{B’})
y₁₂ = – (frac{1}{B’})
y₂₁ = (left(C’ – frac{D’ A’}{B’}right))
y₂₂ = (frac{D’}{B’}).

13. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₁₁ in terms of Transmission parameters can be expressed as ____________
A. z₁₁ = (frac{D’}{C’})
B. z₁₁ = (frac{1}{C’})
C. z₁₁ = (left(frac{A’ D’}{C’} – B’right))
D. z₁₁ = (frac{A’}{C’})

Answer: A
Clarification: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₂ = A’V₁ – B’I₁ ……… (3)
I₂ = C’V₁ – D’I₁ …………… (4)
Rewriting (3) and (4), we get,
V₂ = A’ (left(frac{D’}{C’} I_1 + frac{1}{C’} I_2right) – B’I_1 = left(frac{A’ D’}{C’} – B’right) I_1 + frac{A’}{C’} I_2) ………… (5)
And V₁ = (frac{D’}{C’} I_1 + frac{1}{C’} I_2) ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = (frac{D’}{C’})
z₁₂ = (frac{1}{C’})
z₂₁ = (left(frac{A’ D’}{C’} – B’right))
z₂₂ = (frac{A’}{C’}).

14. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₁₂ in terms of Transmission parameters can be expressed as ____________
A. z₁₂ = (frac{D’}{C’})
B. z₁₂ = (frac{1}{C’})
C. z₁₂ = (left(frac{A’ D’}{C’} – B’right))
D. z₁₂ = (frac{A’}{C’})

Answer: B
Clarification: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₂ = A’V₁ – B’I₁ ……… (3)
I₂ = C’V₁ – D’I₁ …………… (4)
Rewriting (3) and (4), we get,
V₂ = A’ (left(frac{D’}{C’} I_1 + frac{1}{C’} I_2right) – B’I_1 = left(frac{A’ D’}{C’} – B’right) I_1 + frac{A’}{C’} I_2) ………… (5)
And V₁ = (frac{D’}{C’} I_1 + frac{1}{C’} I_2) ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = (frac{D’}{C’})
z₁₂ = (frac{1}{C’})
z₂₁ = (left(frac{A’ D’}{C’} – B’right))
z₂₂ = (frac{A’}{C’}).

15. For a T-network if the Open circuit Impedance parameters are z₁₁, z₁₂, z₂₁, z₂₂, then z₂₂ in terms of Transmission parameters can be expressed as ____________
A. z₂₂ = (frac{D’}{C’})
B. z₂₂ = (frac{1}{C’})
C. z₂₂ = (left(frac{A’ D’}{C’} – B’right))
D. z₂₂ = (frac{A’}{C’})

Answer: D
Clarification: We know that, V₁ = z₁₁ I₁ + z₁₂ I₂ …………. (1)
V₂ = z₂₁ I₁ + z₂₂ I₂ ……………. (2)
And V₂ = A’V₁ – B’I₁ ……… (3)
I₂ = C’V₁ – D’I₁ …………… (4)
Rewriting (3) and (4), we get,
V₂ = A’ (left(frac{D’}{C’} I_1 + frac{1}{C’} I_2right) – B’I_1 = left(frac{A’ D’}{C’} – B’right) I_1 + frac{A’}{C’} I_2) ………… (5)
And V₁ = (frac{D’}{C’} I_1 + frac{1}{C’} I_2) ………….. (6)
Comparing (1), (2) and (5), (6), we get,
z₁₁ = (frac{D’}{C’})
z₁₂ = (frac{1}{C’})
z₂₁ = (left(frac{A’ D’}{C’} – B’right))
z₂₂ = (frac{A’}{C’}).