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

1. A voltage waveform V (t) = 12t^{2} is applied across a 1 H inductor for t≥0, with initial current through it being 0. The current through the inductor for t greater than 0 is ___________

a) 12t

b) 24t

c) 12t^{3}

d) 4t^{3}

View Answer

Explanation: We know that, I = \(\frac{1}{L} \int_0^t V \,dt\)

Here, V (t) = 12t

^{2}

∴ I = \(\int_0^t 12t^2 \,dt\)

= \(\big[\frac{12t^3}{3}\big]_0^t\) = 4t

^{3}A.

2. In the circuit given below, Z_{1} = 10∠-60°, Z_{2} = 10∠60°, Z_{3} = 50∠53.13°. The Thevenin impedance as seen from A-B is _____________

a) 56.5∠45°

b) 60∠30°

c) 70∠30°

d) 34.4∠65°

View Answer

Explanation: Z

_{TH}= Z

_{A-B}= Z

_{1}|| Z

_{2}+ Z

_{3}

= \(\frac{Z_1 × Z_2}{Z_1 + Z_2} + Z_3\)

= \(\frac{10∠-60° × 10∠60°}{10∠-60° + 10∠60°}\) + 50∠53.13°

= 56.66∠45°.

3. In the circuit given below, the value of Z which would cause parallel resonance at 500 Hz is ___________

a) 125 mH

b) 34.20 μF

c) 2 μF

d) 0.05 μF

View Answer

Explanation: At resonance the circuit should have unity power factor

Hence, Z should be capacitive.

Now, \(\frac{1}{jLω} + \frac{1}{1/jCω}\) = 0

Or, \(\frac{-1}{jLω}\) + jCω = 0

∴ C = \(\frac{1}{L × ω^2} = \frac{1}{2 × (2π × 500)^2}\)

= 0.05 μF.

4. In the circuit given below, the current source is 1∠0° A, R = 1 Ω, the impedances are Z_{C} = -j Ω and Z_{L} = 2j Ω. The Thevenin equivalent looking across A-B is __________

a) \(\sqrt{2}\)∠0 V, (1+2j) Ω

b) 2∠45° V, (1-2j) Ω

c) 2∠45° V, (1+j) Ω

d) \(\sqrt{2}\)∠45° V, (1+j) Ω

View Answer

Explanation: Z

_{TH}= 1 + (2j-j) = (1 + j) Ω

∴ V

_{TH}= 1(1+j) = \(\sqrt{2}\)∠45° V.

5. A periodic rectangular signal X (t) has the waveform as shown below. The frequency of the fifth harmonic of its spectrum is ______________

a) 40 Hz

b) 200 Hz

c) 250 Hz

d) 1250 Hz

View Answer

Explanation: Periodic time = 4 ms = 4 × 10

^{-3}

Fundamental frequency = \(\frac{10^3}{4}\) = 250 Hz

∴ Frequency of the fifth harmonic = 250 × 5 = 1250 Hz.

6. A voltage having the waveform of a sine curve is applied across a capacitor. When the frequency of the voltage is increased, what happens to the current through the capacitor?

a) Increases

b) Decreases

c) Remains same

d) Is zero

View Answer

Explanation: The current through the capacitor is given by,

I

_{C}= ωCV cos (ωt + 90°).

As the frequency is increased, I

_{C}also increases.

7. An inductive coil connected to a 200 V, 50 Hz AC supply with 10 A current flowing through it dissipates 1000 W. Which of the following will have least value?

a) Resistance R

b) Reactance X_{L}

c) Impedance Z

d) Reactance X_{C}

View Answer

Explanation: Power dissipated is given by,

I

^{2}R = 1000 W

Hence, R = \(\frac{1000}{100}\)

Or, R = 10 Ω

Impedance Z = \(\frac{V}{I} = \frac{200}{10}\)

Or, Z = 20 Ω

Reactance X

_{L}= \(\sqrt{Z^2 – R^2}\)

= \(\sqrt{20^2 – 10^2}\)

∴ X

_{L}= 17.3 Ω.

8. An inductive coil connected to a 200 V, 50 Hz AC supply with 10 A current flowing through it dissipates 1000 W. The value of the impedance Z and Reactance X_{L} are respectively ____________

a) 10 Ω and 15 Ω

b) 20 Ω and 17.3 Ω

c) 17.3 Ω and 20 Ω

d) 15 Ω and 10 Ω

View Answer

Explanation: Power dissipated is given by,

I

^{2}R = 1000 W

Hence, R = \(\frac{1000}{100}\)

Or, R = 10 Ω

Impedance Z = \(\frac{V}{I} = \frac{200}{10}\)

Or, Z = 20 Ω

Reactance X

_{L}= \(\sqrt{Z^2 – R^2}\)

= \(\sqrt{20^2 – 10^2}\)

∴ X

_{L}= 17.3 Ω.

9. A relay coil having voltage of magnitude 210 V and frequency 50 Hz is connected to a 210 V, 50 Hz supply. If it has resistance of 50 Ω and an inductance of 0.2 H, the apparent power is _____________

a) 549.39 VA

b) 275.6 VA

c) 157 VA

d) 187 VA

View Answer

Explanation: Z= 50 + j (0.2) (2π) (50) = 50 + 62.8

S = \(\frac{|V|^2}{Z} = \frac{210^2}{50 – 62.8}\)

Apparent Power |S| = \(\frac{210^2}{\sqrt{50^2 + 62.8^2}}\)

= 549.39 VA.

10. In the circuit given below, the value of Z_{L} for maximum power to be transferred is _____________

a) R

b) R + jωL

c) R – jωL

d) jωL

View Answer

Explanation: The value of load for maximum power transfer is given by the complex conjugate of Z

_{AB}

Z

_{AB}= R + jX

_{L}

= R + jωL

∴ Z

_{L}for maximum power transfer is given by Z

_{L}= R – jωL.

11. A 10 V is connected across a load whose V-I characteristics is given by 7I = V^{2} + 2V. The internal resistance of the battery is of magnitude 1Ω. The current delivered by the battery is ____________

a) 6 A

b) 5 A

c) 7 A

d) 8 A

View Answer

Explanation: 7I = V

^{2}+ 2V …………………. (1)

Now, V = 10 – 1 ×I

Putting the value of V in eqt (1), we get,

&I = (10 – I) 2 + 2(10 – I)

Or, I = 100 + I2 – 20I + 20 – 2I

Or, I2 – 29I + 120 = 0…………………. (2)

∴ I = \(\frac{+29 ± \sqrt{29^2 – 4(120)}}{2} = \frac{29 ± 19}{2}\)

I = 5 A, 24 A

Now, I = 24 A is not possible because V will be negative from eqt (2)

∴ I = 5 A.

12. A current I given by I = – 8 + 6\(\sqrt{2}\) (sin (ωt + 30°)) A is passed through three meters. The respective readings (in ampere) will be?

a) 8, 6 and 10

b) 8, 6 and 8

c) – 8, 10 and 10

d) -8, 2 and 2

View Answer

Explanation: PMMC instrument reads only DC value and since it is a centre zero type, so it will give – 8 values.

So, rms = \(\sqrt{8^2 + \left(\frac{6\sqrt{2}}{\sqrt{2}})^2\right)}\) = 10 A

Moving iron also reads rms value, so its reading will also be 10 A.

13. A rectifier type AC voltmeter consists of a series resistance R, an ideal full-wave rectifier bridge and a PMMC instrument. A DC current of 1 mA produces a full-scale deflection of the instrument. The resistance of the instrument is 100 Ω. A voltage of 100 V (rms) is applied to the input terminals. The value of R required is?

a) 63.56 Ω

b) 89.83 Ω

c) 89.83 kΩ

d) 141.3 kΩ

View Answer

Explanation: V

_{OAverage}= 0.636 × \(\sqrt{2}\)V

_{rms}= 0.8993 V

_{rms}

The deflection with AC is 0.8993 times that with DC for the same value of voltage V

S

_{AC}= 0.8993 S

_{DC}

S

_{DC}of a rectifier type instrument is \(\frac{1}{I_{fs}}\) where I

_{fs}is the current required to produce full scale deflection, I

_{fs}= 1 mA; R

_{m}= 100 Ω; S

_{DC}= 10

^{3}Ω/V

S

_{AC}= 0.8993 × 1000 = 899.3 Ω/V. Resistance of multiplier R

_{S}= S

_{AC}V – R

_{m}– 2R

_{d}, where R

_{d}is the resistance of diode, for ideal diode R

_{d}= 0

∴ R

_{S}= 899.3 × 100 – 100 = 89.83 kΩ.

14. The discharge of a capacitor through a ballistic galvanometer produces a damped frequency of 0.125 Hz and successive swings of 120, 96 and 76.8 mm. The logarithmic decrement is?

a) 0.225

b) 0.565

c) 0.484

d) 0.7887

View Answer

Explanation: Logarithmic decrement, δ = ln \(\frac{x_1}{x_2} = ln \frac{120}{96}\) = 0.225

Now, δ is related to damping ratio K as, K = \(\frac{1}{\left(1 + (\frac{2π}{δ})^2)^{0.5}\right)}\)

∴ K = 0.0357.

15. 12, 1 H inductors are used as edge to form a cube. The inductance between two diagonally opposite corners of the cube is ____________

a) \(\frac{5}{6}\) H

b) \(\frac{10}{6}\) H

c) 2 H

d) \(\frac{3}{2}\) H

View Answer

Explanation: V

_{AB}= \(\frac{1}{3}L + \frac{1}{6} L + \frac{1}{3}L\)

V

_{AB}= IL \([\frac{5}{6}] \)

Now, \(\frac{V_{AB}}{I} = \frac{5L}{6}\)

Or, L

_{AB}= \(\frac{5L}{6}\)

Here, L = 1Ω

∴ L

_{AB}= \(\frac{5}{6}\) H.

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