This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Problems Involving Dot Conventions”.

1. The current through an electrical conductor is 1A when the temperature of the conductor 0°C and 0.7 A when the temperature is 100°C. The current when the temperature of the conductor is 1200°C is ___________

a) 0.08 A

b) 0.16 A

c) 0.32 A

d) 0.64 A

View Answer

Explanation: \(\frac{1}{0.7} = \frac{R_O (1+αt)}{R_O}\)

= 1 + α100

∴α = 0.0043 per °C

Current at 1200 °C is given by, \(\frac{1}{I} = \frac{R_O (1+α1200)}{R_O}\)

= 1 + α1200

= 1 + 0.0043 × 1200 = 6.16

∴ I = \(\frac{1}{6.16}\) = 0.16 A.

2. In the circuit given below, the equivalent capacitance is ____________

a) 1.625 F

b) 1.583 F

c) 0.583 F

d) 0.615 F

View Answer

Explanation: C

_{CB}= \(\left(\frac{C_2 C_3}{C_2+ C_3}\right)\) + C

_{5}= 1.5 F

Now, C

_{AB}=\(\left(\frac{C_1 C_{CB}}{C_1+ C_{CB}}\right)\) + C

_{6}= 1.6 F

C

_{XY}= \(\frac{C_{AB} × C_4}{C_{AB} + C_4}\) = 0.615 F.

3. In the circuit given below, the equivalent capacitance is ______________

a) 3.5 μF

b) 1.2 μF

c) 2.4 μF

d) 4.05 μF

View Answer

Explanation: The 2.5 μF capacitor is in parallel with 1 μF capacitor and this combination is in series with 1.5 μF.

Hence, C

_{1}= \(\frac{1.5(2.5+1)}{1.5+2.5+1}\)

= \(\frac{5.25}{5}\) = 1.05

Now, C

_{1}is in parallel with the 3 μF capacitor.

∴ C

_{EQ}= 1.05 + 3 = 4.05 μF.

4. In the circuit given below, the voltage across A and B is _______________

a) 13.04 V

b) 17.84 V

c) 12 V

d) 10.96 V

View Answer

Explanation: Loop current I

_{1}= \(\frac{6}{10}\) = 0.6 A

I

_{2}= \(\frac{12}{14}\) = 0.86 A

V

_{AB}= (0.6) (4) + 12 + (0.86) (4)

= 2.4 + 12 + 3.44

= 17.84 V.

5. In the figure given below, the voltage source provides the circuit with a voltage V. The number of non-planar graph of independent loop equations is ______________

a) 8

b) 12

c) 7

d) 5

View Answer

Explanation: The total number of independent loop equations are given by L = B – N + 1 where,

L = number of loop equations

B = number of branches = 12

N = number of nodes = 8

∴ L = 12 – 8 + 1 = 5.

6. When a DC voltage is applied to an inductor, the current through it is found to build up in accordance with I = 20(1-e^{-50t}). After the lapse of 0.02 s, the voltage is equal to 2 V. The value of inductance is ________________

a) 2 mH

b) 5.43 mH

c) 1.54 mH

d) 0.74 mH

View Answer

Explanation: V

_{L}= L\(\frac{dI}{dt}\)

Where, I = 20(1-e

^{-50t})

Therefore, V

_{L}= L\(\frac{d 20(1-e^{-50t})}{dt}\)

= L × 20 × 50e

^{-50t}

At t = 0.02 s, V

_{L}= 2 V

∴ L = \(\frac{2}{20 × 50 × e^{-50×0.02}}\)

= 5.43 μH.

7. An air capacitor of capacitance 0.005 μF is connected to a direct voltage of 500 V. It is disconnected and then immersed in oil with a relative permittivity of 2.5. The energy after immersion is _____________

a) 275 μJ

b) 250 μJ

c) 225 μJ

d) 625 μJ

View Answer

Explanation: E = \(\frac{1}{2}\) CV

^{2}

Or, C = \(\frac{ε_0 ε_r A}{d} = ϵ_r (\frac{ε_0 A}{d})\)

= 2.5 × 0.005 × 10

^{-6}

∴ C

_{NEW}= 12.5 × 10

^{-9}F

Now, q = CV = 0.005 × 10

^{-6}× 500 = 2.5 × 10

^{-6}

V

_{NEW}= \(\frac{q} {C_{NEW}}\)

= \(\frac{2.5 × 10^{-6}}{12.5 × 10^{-9}}\) V

_{NEW}= 200

E = \(\frac{1}{2}\) CV

^{2}

= \(\frac{1}{2}\) × 12.5 × 10

^{-9}× (200)

^{2}

= 250 μJ.

8. The resistance of copper motor winding at t=20°C is 3.42 Ω. After extended operation at full load, the motor windings measures 4.22 Ω. If the temperature coefficient is 0.0426 the, what is the rise in temperature?

a) 60°C

b) 45.2°C

c) 72.9°C

d) 10.16°C

View Answer

Explanation: Given that, R

_{1}= 3.42 Ω

T

_{1}= 20° and α = 0.0426

R

_{2}= 4.22 Ω

Now, \(\frac{R_1}{1 + αT_1} = \frac{R_2}{1 + αT_2}\)

Or, \(\frac{3.42}{1 + 0.0426 × 20} = \frac{4.22}{1 + 0.0426 T_2}\)

∴ Rise in temperature = T

_{2}– T

_{1}

= 30.16 – 20

= 10.16°C.

9. A capacitor of capacitance 50 μF is connected in parallel to another capacitor of capacitance 100 μF. They are connected across a time-varying voltage source. At a particular time, the current supplied by the source is 5 A. The magnitude of instantaneous current through the capacitor of capacitance 50 μF is _____________

a) 1.57 A

b) 1.87 A

c) 1.67 A

d) 2.83 A

View Answer

Explanation: As the capacitors are in parallel, then the voltage V is given by,

V = \(\frac{1}{C_1} \int I_1 \,dt \)

∴ I

_{1}= C

_{1}\(\frac{dV}{dt}\)

That is, \(\frac{I_1}{I_2} = \frac{C_1}{C_2} = \frac{50}{100} = \frac{1}{2}\) ………………….. (1)

Also, I

_{1}+ I

_{2}= 5 A ………………………….. (2)

Solving (1) and (2), we get, I

_{1}= 1.67 A.

10. A capacitor of capacitance 50 μF is connected in parallel to another capacitor of capacitance 100 μF. They are connected across a time-varying voltage source. At a particular time, the current supplied by the source is 5 A. The magnitude of instantaneous current through the capacitor of capacitance 100 μF is _____________

a) 2.33 A

b) 3.33 A

c) 1.33 A

d) 4.33 A

View Answer

Explanation: As the capacitors are in parallel, then the voltage V is given by,

V = \(\frac{1}{C_2} \int I_2 \,dt \)

∴ I

_{2}= C

_{2}\(\frac{dV}{dt}\)

That is, \(\frac{I_1}{I_2} = \frac{C_1}{C_2} = \frac{50}{100} = \frac{1}{2}\) ………………….. (1)

Also, I

_{1}+ I

_{2}= 5 A ………………………….. (2)

Solving (1) and (2), we get, I

_{2}= 3.33 A.

11. In the circuit given below, the resonant frequency is _______________

a) \(\frac{1}{2\sqrt{2} π}\) Hz

b) \(\frac{1}{2π}\) Hz

c) \(\frac{1}{4π}\) Hz

d) \(\frac{1}{\sqrt{2}2π}\) Hz

View Answer

Explanation: I

_{EQ}= L

_{1}+ L

_{2}+ 2M

L

_{EQ}= 1 + 2 + 2 × \(\frac{1}{2}\) = 4 H

∴ F

_{O}= \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{2π\sqrt{4 × 1}} \)

= \(\frac{1}{4π}\) Hz.

12. In a series resonant circuit, V_{C} = 300 V, V_{L} = 300 V and V_{R} = 100 V. What is the value of the source voltage?

a) Zero

b) 100 V

c) 350 V

d) 200 V

View Answer

Explanation: As V

_{C}and V

_{L}are equal, then X

_{C}is equal to X

_{L}and both the voltages are then cancelled out.

That is V

_{S}= V

_{R}

∴ V

_{S}= 100 V.

13. For the circuit given below, the value of the Q factor for the inductor is ______________

a) 4.74

b) 4.472

c) 4.358

d) 4.853

View Answer

Explanation: Q

_{IN}= \(\sqrt{\frac{L}{CR^2} – 1}\)

= \(\sqrt{\frac{1}{2 × 5^2 × 10^{-3} – 1}}\)

= \(\sqrt{\frac{1}{50 × 10^{-3} – 1}}\)

= \(\sqrt{19}\) = 4.358.

14. In the circuit given below, the value of the voltage source E is _______________

a) -65 V

b) 40 V

c) -60 V

d) 65 V

View Answer

Explanation: Going from 10 V to 0 V, we get,

50 + 10 + E + 5 = 0

Or, E + 65 = 0

Hence, E = -65 V.

15. In the circuit given below, bulb X uses 48 W when lit, bulb Y uses 22 W when lit and bulb Z uses 14.4 W when lit. The number of additional bulbs in parallel to this circuit, that would be required to below the fuse is _______________

a) 4

b) 5

c) 6

d) 7

View Answer

Explanation: I

_{X}= \(\frac{48}{12}\) = 4 A

I

_{Y}= \(\frac{22}{12}\) = 1.8 A

I

_{Z}= \(\frac{14.4}{12}\) = 1.2 A

Current required to below the fuse = 20 A

∴ Additional bulbs must draw current = 20 – (4 + 18 + 1.2)

= 20 – 7 = 13

∴ Number of additional bulbs required = \(\frac{13}{3}\) = 4.33

So, 4 additional bulbs are required.

**Sanfoundry Global Education & Learning Series – Network Theory.**

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