250+ TOP MCQs on Problems Involving Standard Time Functions and Answers

Network Theory Multiple Choice Questions on “Problems Involving Standard Time Functions”.

1. A voltage waveform V (t) = 12t² 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. 12t³
D. 4t³

Answer: D
Clarification: We know that, I = (frac{1}{L} int_0^t V ,dt)
Here, V (t) = 12t²
∴ I = (int_0^t 12t^2 ,dt)
= (big[frac{12t^3}{3}big]_0^t) = 4t³ A.

2. In the circuit given below, Z₁ = 10∠-60°, Z₂ = 10∠60°, Z₃ = 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°

Answer: A
Clarification: Z_TH = Z_A-B = Z₁ || Z₂ + Z₃
= (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

Answer: D
Clarification: 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

Answer: D
Clarification: 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

Answer: D
Clarification: 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

Answer: A
Clarification: 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

Answer: A
Clarification: Power dissipated is given by,
I²R = 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

Answer: B
Clarification: Power dissipated is given by,
I²R = 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

Answer: A
Clarification: 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

Answer: C
Clarification: 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 = V² + 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

Answer: B
Clarification: 7I = V² + 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

Answer: C
Clarification: 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{6sqrt{2}}{sqrt{2}})^2right)}) = 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

Answer: C
Clarification: 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 = 10³ 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

Answer: A
Clarification: 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

Answer: A
Clarification: 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.