Alternating Currents Chapter-Wise Test 21

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the significance of the power factor being zero in an AC circuit?

A power factor of zero (\( \cos \phi = 0 \)) means the phase difference between voltage and current is 90°, as in purely inductive or capacitive circuits. This indicates no average power is dissipated, as energy is only stored and released, not consumed.

Maximum power is dissipated
The circuit is purely resistive
No average power is dissipated
The current is zero
3

A series LCR circuit has \( R = 20 \, \Omega \), \( X_L = 40 \, \Omega \), \( X_C = 60 \, \Omega \). What is the phase angle?

\( \tan \phi = \frac{X_C - X_L}{R} = \frac{60 - 40}{20} = 1 \).

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

Since \( X_C > X_L \), the circuit is capacitive, and current leads voltage.

30°
40°
45°
60°
3

A transformer has \( N_p = 600 \), \( N_s = 300 \). If \( V_p = 240 \, \text{V} \) (rms), what is the secondary voltage?

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

\( V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{300}{600} = 120 \, \text{V} \).

100 V
120 V
140 V
160 V
2

A series LCR circuit has \( L = 2.5 \, \text{H} \), \( C = 40 \, \mu\text{F} \). What is the resonant angular frequency?

\( \omega_0 = \frac{1}{\sqrt{L C}} \).

\( L = 2.5 \, \text{H} \), \( C = 40 \times 10^{-6} \, \text{F} \).

\( \omega_0 = \frac{1}{\sqrt{2.5 \times 40 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-4}}} = 100 \, \text{rad/s} \).

100 rad/s
110 rad/s
120 rad/s
130 rad/s
1

In an AC circuit with a series LCR combination, why does the power dissipated depend only on the resistive component?

Power dissipation in an AC circuit (\( P = I^2 R \cos \phi \)) occurs only through resistance, as inductors and capacitors store and release energy without converting it to heat. The reactive components affect the current and phase, but only \( R \) dissipates power.

Because inductance is infinite
Because capacitance is zero
Because frequency is constant
Because only resistance converts energy to heat
4

A \( 13 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the capacitive reactance?

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

\( C = 13 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{314 \times 13 \times 10^{-6}} \approx 245.1 \, \Omega \).

240 Ω
242 Ω
245.1 Ω
250 Ω
3

A \( 60 \, \Omega \) resistor and \( 15 \, \mu\text{F} \) capacitor are in series with a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the impedance?

\( X_C = \frac{1}{\omega C} = \frac{1}{314 \times 15 \times 10^{-6}} \approx 212.3 \, \Omega \).

\( Z = \sqrt{R^2 + X_C^2} = \sqrt{60^2 + 212.3^2} = \sqrt{3600 + 45071.29} \approx 220.8 \, \Omega \).

215 Ω
218 Ω
222 Ω
220.8 Ω
4

A series LCR circuit has \( R = 200 \, \Omega \), \( C = 15 \, \mu\text{F} \), and is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the impedance?

\( X_C = \frac{1}{\omega C} \), \( \omega = 2\pi \times 50 = 314 \, \text{rad/s} \), \( C = 15 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{314 \times 15 \times 10^{-6}} \approx 212.3 \, \Omega \).

No inductor, so \( X_L = 0 \).

Impedance: \( Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{200^2 + 212.3^2} \approx 291.5 \, \Omega \).

280 Ω
291.5 Ω
300 Ω
310 Ω
2

A \( 200 \, \text{V} \) (rms) source supplies a \( 100 \, \Omega \) resistor. What is the peak current?

RMS current: \( I = \frac{V}{R} = \frac{200}{100} = 2 \, \text{A} \).

Peak current: \( i_m = \sqrt{2} I = 1.414 \times 2 \approx 2.828 \, \text{A} \).

2.5 A
2.7 A
3 A
2.828 A
4

A series LCR circuit with \( R = 100 \, \Omega \), \( L = 1 \, \text{H} \), \( C = 1 \, \mu\text{F} \) is at resonance. What is the impedance?

At resonance, \( X_L = X_C \), so \( Z = R \).

Given: \( R = 100 \, \Omega \).

Impedance \( Z = 100 \, \Omega \).

80 Ω
90 Ω
100 Ω
110 Ω
3

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