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  4. Alternating Current Chapter-Wise Test 2

Alternating Current Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 165 \, \text{V} \) (rms) AC source supplies a \( 55 \, \Omega \) resistor. What is the average power consumed?

RMS current: \( I = \frac{V}{R} = \frac{165}{55} = 3 \, \text{A} \).

Average power: \( P = I^2 R = 3^2 \times 55 = 495 \, \text{W} \).

480 W
495 W
510 W
525 W
2

A \( 268.7 \, \text{V} \) (peak) AC source is connected to a \( 95 \, \Omega \) resistor. What is the average power consumed?

RMS voltage: \( V = \frac{v_m}{\sqrt{2}} = \frac{268.7}{1.414} \approx 190 \, \text{V} \).

RMS current: \( I = \frac{V}{R} = \frac{190}{95} = 2 \, \text{A} \).

Average power: \( P = I^2 R = 2^2 \times 95 = 380 \, \text{W} \).

380 W
400 W
420 W
440 W
1

A \( 200 \, \text{V} \) (rms) AC source is connected to a series LCR circuit with \( R = 20 \, \Omega \) at resonance. What is the power dissipated?

At resonance, \( Z = R = 20 \, \Omega \).

RMS current: \( I = \frac{V}{R} = \frac{200}{20} = 10 \, \text{A} \).

Power: \( P = I^2 R = 10^2 \times 20 = 2000 \, \text{W} \).

1500 W
1800 W
2000 W
2200 W
3

A \( 17 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the rms current?

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

\( C = 17 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{376.8 \times 17 \times 10^{-6}} \approx 156 \, \Omega \).

RMS current: \( I = \frac{V}{X_C} = \frac{110}{156} \approx 0.705 \, \text{A} \).

0.705 A
0.8 A
0.9 A
1 A
1

In an AC circuit containing only a resistor, what happens to the power dissipated if the frequency of the source is doubled?

In a purely resistive AC circuit, power dissipated is \( P = I^2 R \), where \( I = \frac{V}{R} \), and \( R \) is constant. Since resistance does not depend on frequency, and assuming the rms voltage remains constant, the power dissipated remains unchanged when frequency doubles.

It remains constant
It doubles
It halves
It becomes zero
1

A series LCR circuit has \( L = 1.5 \, \text{H} \), \( C = 35 \, \mu\text{F} \). What is the resonant frequency in Hz?

Resonant angular frequency: \( \omega_0 = \frac{1}{\sqrt{L C}} \).

\( L = 1.5 \, \text{H} \), \( C = 35 \times 10^{-6} \, \text{F} \).

\( \omega_0 = \frac{1}{\sqrt{1.5 \times 35 \times 10^{-6}}} \approx 138.3 \, \text{rad/s} \).

\( f_0 = \frac{\omega_0}{2\pi} = \frac{138.3}{6.28} \approx 22 \, \text{Hz} \).

20 Hz
22 Hz
25 Hz
28 Hz
2

A \( 60 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC supply. What is the peak current in the circuit?

Capacitive reactance: \( X_C = \frac{1}{\omega C} \), where \( \omega = 2\pi f \).

\( f = 60 \, \text{Hz} \), \( C = 60 \times 10^{-6} \, \text{F} \).

\( \omega = 2 \times 3.14 \times 60 = 376.8 \, \text{rad/s} \).

\( X_C = \frac{1}{376.8 \times 60 \times 10^{-6}} = 44.24 \, \Omega \).

RMS current: \( I = \frac{V}{X_C} = \frac{110}{44.24} = 2.487 \, \text{A} \).

Peak current: \( i_m = \sqrt{2} I = 1.414 \times 2.487 \approx 3.52 \, \text{A} \).

3 A
3.52 A
4 A
4.5 A
2

A \( 70 \, \Omega \) resistor and \( 14 \, \mu\text{F} \) capacitor are in series with a \( 210 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the impedance?

\( X_C = \frac{1}{\omega C} = \frac{1}{314 \times 14 \times 10^{-6}} \approx 227.5 \, \Omega \).

\( Z = \sqrt{R^2 + X_C^2} = \sqrt{70^2 + 227.5^2} = \sqrt{4900 + 51756.25} \approx 238.2 \, \Omega \).

235 Ω
236 Ω
240 Ω
238.2 Ω
4

A \( 23 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the rms current?

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

\( C = 23 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{376.8 \times 23 \times 10^{-6}} \approx 115.4 \, \Omega \).

RMS current: \( I = \frac{V}{X_C} = \frac{110}{115.4} \approx 0.953 \, \text{A} \).

0.9 A
0.92 A
1 A
0.953 A
4

A \( 180 \, \text{V} \) (rms) AC source supplies a \( 90 \, \Omega \) resistor. What is the average power consumed?

RMS current: \( I = \frac{V}{R} = \frac{180}{90} = 2 \, \text{A} \).

Average power: \( P = I^2 R = 2^2 \times 90 = 360 \, \text{W} \).

300 W
360 W
400 W
450 W
2

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