This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Advanced Problems on Network Theorems – 1”.

1. The temperature coefficient of a metal as the temperature increases will ____________

a) Decreases

b) Increases

c) Remains unchanged

d) Increases and remains same

View Answer

Explanation: We know that the temperature coefficient is,

Given by, α = \(\frac{α_0}{1 + α_0 t}\)

Since temperature is present at the denominator, so with increase in temperature t, the denominator increases and hence the fraction decreases.

So, temperature coefficient decreases.

2. Given a wire of resistance R Ω. The resistance of a wire of the same material and same weight and double the diameter is ___________

a) 0.5 R

b) 0.25 R

c) 0.125 R

d) 0.0625 R

View Answer

Explanation: Since diameter is double, area of cross-section is four times and length is one-fourth.

It can be verified by the following equation,

R

_{2}= \(\frac{\frac{ρl}{4}}{4A}\)

= \(\frac{ρl}{16 A} = \frac{R}{16}\).

3. The star equivalent resistance of 3 resistors having each resistance = 5 Ω is ____________

a) 1.5 Ω

b) 1.67 Ω

c) 3 Ω

d) 4.5 Ω

View Answer

Explanation: We know that for star connection, R

_{EQ}= \(\frac{R X R}{R+R+R}\)

Given R = 5 Ω

So, R

_{EQ}= \(\frac{5 X 5}{5+5+5}\)

= \(\frac{25}{15}\) = 1.67 Ω.

4. The charge associated with a bulb rated as 20 W, 200 V and used for 10 minutes is ____________

a) 36 C

b) 60 C

c) 72 C

d) 50 C

View Answer

Explanation: Charge Q= It

Given I = \(\frac{20}{200}\) = 0.1 A, t = 10 X 60 sec = 600 sec

So, Q = 0.1 X 600 = 60 C.

5. For a series RL circuit having L = 5 H, current = 1 A (at an instant). The energy stored in magnetic field is ___________

a) 3.6 J

b) 2.5 J

c) 1.5 J

d) 3 J

View Answer

Explanation: We know that, Energy, E = 0.5 LI

^{2}

Or, E = 0.5 X 5 X 1

^{2}= 2.5 J.

6. For a practical voltage source, which of the following is correct?

a) Cannot be less than source voltage

b) Cannot be higher than source voltage

c) Is always less than source voltage

d) Is always equal to source voltage

View Answer

Explanation: A practical voltage source has some resistance. Because of this resistance, some amount of voltage drop occurs across this resistance. Hence, the terminal voltage cannot be higher than source voltage. However, if current is zero, then terminal voltage and source voltage are equal.

7. Consider an electric motor having resistance of 10 Ω, 20 Ω and 30 Ω respectively. The percentage of energy dissipated by 10 Ω earthing plate is ____________

a) More than 50% of total energy

b) Less than 50% of total energy

c) Depends on the materials of the three plates

d) May be more or less than 50% of total energy

View Answer

Explanation: The parallel combination of 30 Ω and 20 Ω is 12 Ω. Since 12 Ω and 10 Ω are in parallel, the 10 ohm plate draws more than 50% current and dissipates more than 50% energy.

8. Consider a resistive network circuit, having 3 sources of 18 W, 50 W and 98 W respectively and a resistance R. When all source act together the maximum and minimum power is ____________

a) 98 W, 18 W

b) 166 W, 18 W

c) 450 W, 2 W

d) 166 W, 2 W

View Answer

Explanation: Let us suppose R = 1 Ω

Then, I

_{1}= 32 A, I

_{2}= 52 A and I

_{3}= 72 A

Or, (I

_{1}+ I

_{2}+ I

_{3})

^{2}R = (152)

^{2}R = 450 Ω.

9. A current waveform is of the shape of a right angled triangle with period of t = 1 sec. Given a resistance R = 1 Ω. The average power is __________

a) 1 W

b) 0.5 W

c) 0.333 W

d) 0.111 W

View Answer

Explanation: We know that RMS current is,

I = 1 \( \int_0^1 (1t)^2 \,dt\)

= \( \int_0^1 t^2 \,dt\)

= \(\frac{1}{3}\) A

Now, power P = I

^{2}R

= \(\frac{1}{9}\) X 1

= 0.111 W.

10. Given two voltages, V_{1} = sin (ωt + 60°) and V_{2} = cos (ωt). Which of the following is correct?

a) V_{1} is leading V_{2} by 15°

b) V_{1} is leading V_{2} by 30°

c) V_{2} is leading V_{1} by 60°

d) V_{2} is leading V_{1} by 30°

View Answer

Explanation: Given that, V

_{1}= sin (ωt + 30°) and V

_{2}= cos (ωt)

Now, V

_{2}can be written as,

V

_{2}= sin (ωt + 90°).

Hence, V

_{2}is leading V

_{1}by (90 – 30) = 60°.

11. Given two mutually coupled coils have a total inductance of 1500 mH, the self-inductance of each coils if the coefficient of coupling is 0.2 is ____________

a) 325 mH

b) 255 mH

c) 625 mH

d) 550 mH

View Answer

Explanation: We know that, M = k\(\sqrt{L_1 L_2}\)

Given that, L

_{EQ}= 1500 mH and k = 0.2

Again, total inductance = L

_{1}+ L

_{2}+ 2M

Or, 2L + 2kL = 1500 mH

Or, L (2 + 2 X 0.2) = 1500

Or, L = 625 mH.

12. For a series RL circuit, the impedance Z = 10 Ω at a frequency of 50 Hz. At 100 Hz the impedance is ___________

a) 10 Ω

b) 20 Ω

c) 1 Ω

d) More than 1 Ω but less than 10 Ω 0 Ω

View Answer

Explanation: We know that impedance, Z = \( \sqrt{R^2 + (\frac{1}{ω^2+c^2})^2}\)

Since frequency is doubled, so \(\frac{1}{ω^2+c^2}\) becomes one-fourth but R2 remains the same. Thus the impedance cannot be exactly measured but we can infer that the resistance is more than 1 Ω but less than 10 Ω.

13. Consider the self-inductances of two coils as 12 H and 5 H. 50 % of one flux links the other. The mutual inductance is ___________

a) 30 H

b) 24 H

c) 9 H

d) 4.5 H

View Answer

Explanation: We know that, M = k\(\sqrt{L_1 L_2}\)

Given that, L

_{1}= 12 H, L

_{2}= 5 H and k = 0.5

So, M = 0.5\(\sqrt{12 X 5}\)

= 0.5\(\sqrt{60}\)

= 0.5 X 7.75 = 3.875 H.

14. In the circuit given below the maximum power that can be transferred from the source voltage is __________

a) 1 W

b) 10 W

c) 0.25 W

d) 0.5 W

View Answer

Explanation: For maximum power transfer to the load resistor R

_{L}, R

_{L}must be equal to 100Ω.

∴ Maximum power = \(\frac{V^2}{4R_L}\)

= \(\frac{10^2}{4×100} = \frac{100}{400}\) = 0.25 W.

15. In the circuit given below, the value of RL for maximum power transfer is ___________

a) 2.4 Ω

b) 2.6 Ω

c) 2.8 Ω

d) 3.0 Ω

View Answer

Explanation: Using Y-∆ transformation,

R

_{AB}= (9+9 || 6) || (9||6)

= (18 || 6) || (9 || 6)

=\(\left(\frac{18×6}{18+6}\right) || \left(\frac{9×6}{9+6}\right)\)

= 4.5 || 3.6

= \(\frac{4.5×3.6}{4.5+3.6}\) = 2.8 Ω.

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