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 RL 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
VOC = 3 × 10 I = 30 I
Or, VOC = 3 V
Now, ISC = 10 I = 1 A
Rth = 3/1 = 3 Ω
Vth = VOC = 3 V
RL = 3 Ω
Pmax = (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 RL for which maximum power is transferred to RL?

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

Answer: C
Clarification: Maximum power is transferred to RL when the load resistance equals the Thevenin resistance of the circuit.
RL = RTH = (frac{V_{OC}}{I_{SC}} )
Due to open-circuit, VOC = 100 V; ISC = I1 + I2
Applying KVL in lower loop, 100 – 8I1 = 0
Or, I1 = (frac{100}{8} = frac{25}{2})
And VX = -4I1 = -4 × (frac{25}{2}) = -50V
KVL in upper loop, 100 + VX – 4I2 = 0
I2 = (frac{100-50}{4} = frac{25}{2})
Hence, ISC = I1 + I2 = (frac{25}{2} + frac{25}{2}) = 25
RTH = (frac{V_{OC}}{I_{SC}} = frac{100}{25}) = 4 Ω
RL = RTH = 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 VNL = (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; VNL = 1V
∴ Power dissipated in RNL = 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 V1Ω = 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 – 0.1V1Ω + 0.5V – 10 – 0.5V1Ω + 0.2 – 0.2VOC = 2
Or, 0.8V – 0.6V1Ω = 12 + 0.2VOC
Or, 0.8 V – 0.2VOC = 12 +3=15 (Putting V1Ω = 5)
Again, (frac{V_{OC}-V}{5}) + 2 = 5
Or, 0.2VOC – 0.2V = 3
Again, RTH = {(10||2) + 1} + 5
= ((frac{20}{12+1})) + 5 = 7.67 Ω
Following the theorem of maximum power transfer, R = RTH = 7.67 Ω
And PMAX = (frac{V_{OC}^2}{4R} = frac{45^2}{4×7.67}) = 66 W.

9. In the circuit given below, the value of RL 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
VOC = 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
ISC = (frac{2.85}{100}) + 3 × 0.0142 = 0.07135 A
∴ RTH = (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
∴ PL = (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 Vm = 150 V and Im = 150 / 20 = 7 A
RMS value of the current, Irms = Im / 2 = 7/2 = 3.5 A
Average value of the current, Iavg = Im / π = 2.228 A
Form factor = Irms / Iavg = 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.

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