Alternating Currents Chapter-Wise Test 10

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A transformer has \( N_p = 450 \), \( N_s = 900 \). If \( V_p = 120 \, \text{V} \) (rms), what is the secondary voltage?

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

\( V_s = V_p \times \frac{N_s}{N_p} = 120 \times \frac{900}{450} = 240 \, \text{V} \).

220 V
240 V
260 V
280 V
2

A \( 65 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the rms current?

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

\( L = 65 \times 10^{-3} \, \text{H} \).

\( X_L = 376.8 \times 0.065 = 24.49 \, \Omega \).

RMS current: \( I = \frac{V}{X_L} = \frac{110}{24.49} \approx 4.49 \, \text{A} \).

4.2 A
4.49 A
4.6 A
4.8 A
2

A \( 21 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the rms current?

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

\( C = 21 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{314 \times 21 \times 10^{-6}} \approx 151.6 \, \Omega \).

RMS current: \( I = \frac{V}{X_C} = \frac{230}{151.6} \approx 1.517 \, \text{A} \).

1.517 A
1.6 A
1.7 A
1.8 A
1

A \( 100 \, \text{W} \) bulb is connected to a \( 110 \, \text{V} \) (rms) AC supply. What is the resistance of the bulb?

Average power: \( P = \frac{V^2}{R} \).

\( R = \frac{V^2}{P} = \frac{(110)^2}{100} = \frac{12100}{100} = 121 \, \Omega \).

100 Ω
121 Ω
130 Ω
150 Ω
2

A series LCR circuit has \( L = 2 \, \text{H} \), \( C = 50 \, \mu\text{F} \). What is the resonant angular frequency?

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

\( L = 2 \, \text{H} \), \( C = 50 \times 10^{-6} \, \text{F} \).

\( \omega_0 = \frac{1}{\sqrt{2 \times 50 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-4}}} = 100 \, \text{rad/s} \).

100 rad/s
120 rad/s
150 rad/s
200 rad/s
1

A series LCR circuit has \( R = 130 \, \Omega \), \( X_L = 110 \, \Omega \), \( X_C = 90 \, \Omega \). What is the impedance?

\( Z = \sqrt{R^2 + (X_L - X_C)^2} \).

\( Z = \sqrt{130^2 + (110 - 90)^2} = \sqrt{16900 + 400} = \sqrt{17300} \approx 131.5 \, \Omega \).

130 Ω
131.5 Ω
135 Ω
140 Ω
2

A \( 80 \, \Omega \) resistor is connected to a \( 160 \, \text{V} \) (rms) AC source. What is the rms current?

RMS current: \( I = \frac{V}{R} \).

Given: \( V = 160 \, \text{V} \), \( R = 80 \, \Omega \).

\( I = \frac{160}{80} = 2 \, \text{A} \).

2 A
2.5 A
3 A
3.5 A
1

In a series LCR circuit at resonance, the power dissipated is \( 4800 \, \text{W} \) with \( R = 3 \, \Omega \). What is the rms current?

At resonance, \( Z = R \), and power \( P = I^2 R \).

\( 4800 = I^2 \times 3 \).

\( I^2 = \frac{4800}{3} = 1600 \).

\( I = \sqrt{1600} = 40 \, \text{A} \).

35 A
40 A
45 A
50 A
2

What is the primary reason AC voltage is preferred over DC voltage for long-distance power transmission?

AC voltage is preferred for long-distance power transmission because it can be easily stepped up or down using transformers. This allows for high-voltage transmission to reduce energy losses due to resistance, followed by stepping down to safer levels for distribution.

It can be transformed to different voltages easily
It has a constant magnitude
It does not require conductors
It generates less heat
1

Why does an AC circuit with only an inductor exhibit a "wattless" current?

In a purely inductive AC circuit, the current lags the voltage by 90°. The power factor (\( \cos 90^\circ = 0 \)) is zero, meaning no average power is dissipated; the current is "wattless" because it only stores and releases energy without consuming it.

Because the inductor has high resistance
Because the phase difference is 90°
Because the current is constant
Because the voltage is zero
2

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