Alternating Currents Chapter-Wise Test 18

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 30 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the rms current?

\( X_L = \omega L \), \( \omega = 2\pi \times 50 = 314 \, \text{rad/s} \).

\( L = 30 \times 10^{-3} \, \text{H} \).

\( X_L = 314 \times 0.03 = 9.42 \, \Omega \).

RMS current: \( I = \frac{V}{X_L} = \frac{220}{9.42} \approx 23.35 \, \text{A} \).

20 A
23.35 A
25 A
28 A
2

A \( 150 \, \text{V} \) (rms) AC source supplies a \( 75 \, \Omega \) resistor. What is the average power consumed?

RMS current: \( I = \frac{V}{R} = \frac{150}{75} = 2 \, \text{A} \).

Average power: \( P = I^2 R = 2^2 \times 75 = 4 \times 75 = 300 \, \text{W} \).

250 W
300 W
350 W
400 W
2

A \( 50 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the peak current?

\( X_L = \omega L \), \( \omega = 2\pi \times 50 = 314 \, \text{rad/s} \).

\( L = 50 \times 10^{-3} \, \text{H} \).

\( X_L = 314 \times 0.05 = 15.7 \, \Omega \).

RMS current: \( I = \frac{V}{X_L} = \frac{220}{15.7} \approx 14.01 \, \text{A} \).

Peak current: \( i_m = \sqrt{2} I = 1.414 \times 14.01 \approx 19.81 \, \text{A} \).

19 A
19.5 A
19.81 A
20 A
3

A \( 28 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the rms current?

\( X_C = \frac{1}{\omega C} \), \( \omega = 2\pi \times 50 = 314 \, \text{rad/s} \).

\( C = 28 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{314 \times 28 \times 10^{-6}} \approx 113.6 \, \Omega \).

RMS current: \( I = \frac{V}{X_C} = \frac{220}{113.6} \approx 1.936 \, \text{A} \).

1.8 A
1.9 A
2 A
1.936 A
4

A \( 50 \, \Omega \) resistor and \( 20 \, \mu\text{F} \) capacitor are in series with a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the impedance?

\( X_C = \frac{1}{\omega C} = \frac{1}{314 \times 20 \times 10^{-6}} \approx 159.2 \, \Omega \).

\( Z = \sqrt{R^2 + X_C^2} = \sqrt{50^2 + 159.2^2} = \sqrt{2500 + 25344.64} \approx 166.6 \, \Omega \).

160 Ω
165 Ω
170 Ω
166.6 Ω
4

A series LCR circuit with \( R = 40 \, \Omega \) is at resonance with a \( 200 \, \text{V} \) (rms) source. What is the rms current?

At resonance, \( Z = R = 40 \, \Omega \).

RMS current: \( I = \frac{V}{R} = \frac{200}{40} = 5 \, \text{A} \).

4.5 A
4.8 A
5 A
5.2 A
3

A series LCR circuit has \( R = 10 \, \Omega \), \( X_L = 20 \, \Omega \), \( X_C = 30 \, \Omega \). What is the phase angle?

\( \tan \phi = \frac{X_C - X_L}{R} = \frac{30 - 20}{10} = 1 \).

\( \phi = \tan^{-1}(1) = 45^\circ \).

Since \( X_C > X_L \), current leads voltage.

40°
42°
45°
50°
3

A transformer has \( N_p = 300 \), \( N_s = 150 \). If \( V_s = 115 \, \text{V} \) (rms), what is the primary voltage?

\( \frac{V_s}{V_p} = \frac{N_s}{N_p} \).

\( V_p = V_s \times \frac{N_p}{N_s} = 115 \times \frac{300}{150} = 230 \, \text{V} \).

210 V
220 V
240 V
230 V
4

A \( 16 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the capacitive reactance?

\( X_C = \frac{1}{\omega C} \), \( \omega = 2\pi \times 60 = 376.8 \, \text{rad/s} \).

\( C = 16 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{376.8 \times 16 \times 10^{-6}} \approx 165.8 \, \Omega \).

160 Ω
162 Ω
165.8 Ω
170 Ω
3

What happens to the current in a purely capacitive AC circuit when the frequency of the source increases?

In a purely capacitive circuit, the capacitive reactance (\( X_C = \frac{1}{\omega C} \)) decreases as the frequency (\( f \), where \( \omega = 2\pi f \)) increases. Since current is inversely proportional to reactance (\( I = \frac{V}{X_C} \)), the current increases.

It decreases
It increases
It remains constant
It becomes zero
2

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0