Alternating Currents Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Why does the voltage across an inductor in a series LCR circuit potentially exceed the source voltage at resonance?

At resonance, current is maximized (\( I = \frac{V}{R} \)), and the voltage across the inductor (\( V_L = I X_L \)) can exceed the source voltage if \( X_L > R \). This is possible because \( V_L \) and \( V_C \) are out of phase with the source voltage, and their vector sum with \( V_R \) equals the source voltage.

Because resistance is zero
Because frequency is low
Because capacitance is infinite
Because current is maximized
4

In a purely inductive AC circuit, the current lags the voltage by what phase angle?

For a pure inductor, \( i = i_m \sin (\omega t - \frac{\pi}{2}) \), while \( v = v_m \sin \omega t \).

Phase difference: \( \phi = -\frac{\pi}{2} \), meaning current lags voltage by \( \frac{\pi}{2} \) or \( 90^\circ \).

0°
45°
90°
180°
3

Why does the voltage across the inductor in a series LCR circuit equal the voltage across the capacitor at resonance, despite their phase difference?

At resonance, \( X_L = X_C \), and since \( V_L = I X_L \) and \( V_C = I X_C \) with the same current \( I \), their magnitudes are equal. However, they are 180° out of phase, so their vector sum is zero, not their magnitudes.

Because resistance is zero
Because reactances are equal
Because frequency is high
Because current is zero
2

In a series LCR circuit, why does resonance not occur if either the inductor or capacitor is absent?

Resonance in a series LCR circuit requires both an inductor and capacitor to create a condition where \( X_L = X_C \), canceling the reactive components. Without either, there’s no opposing reactance to balance, preventing resonance.

Because resistance becomes zero
Because frequency cannot change
Because voltage increases
Because there’s no balancing of reactive components
4

A series LCR circuit has \( R = 15 \, \Omega \), \( X_L = 25 \, \Omega \), \( X_C = 40 \, \Omega \). What is the phase angle?

\( \tan \phi = \frac{X_C - X_L}{R} = \frac{40 - 25}{15} = 1 \).

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

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

40°
42°
45°
50°
3

A series LCR circuit with \( R = 90 \, \Omega \), \( X_L = 120 \, \Omega \), \( X_C = 60 \, \Omega \) has a \( 270 \, \text{V} \) (rms) source. What is the power dissipated?

\( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{90^2 + (120 - 60)^2} = \sqrt{8100 + 3600} = \sqrt{11700} \approx 108.17 \, \Omega \).

RMS current: \( I = \frac{V}{Z} = \frac{270}{108.17} \approx 2.496 \, \text{A} \).

Power: \( P = I^2 R = (2.496)^2 \times 90 \approx 560.6 \, \text{W} \).

540 W
550 W
570 W
560.6 W
4

A \( 100 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the inductive reactance?

\( X_L = \omega L \), \( \omega = 2\pi f \).

\( f = 50 \, \text{Hz} \), \( L = 100 \times 10^{-3} \, \text{H} \).

\( \omega = 2 \times 3.14 \times 50 = 314 \, \text{rad/s} \).

\( X_L = 314 \times 0.1 = 31.4 \, \Omega \).

31.4 Ω
32 Ω
33 Ω
35 Ω
1

A series LCR circuit has \( R = 10 \, \Omega \), \( X_L = 15 \, \Omega \), \( X_C = 5 \, \Omega \). What is the power factor?

\( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{10^2 + (15 - 5)^2} = \sqrt{100 + 100} = 14.14 \, \Omega \).

Power factor: \( \cos \phi = \frac{R}{Z} = \frac{10}{14.14} \approx 0.707 \).

0.5
0.6
0.8
0.707
4

A \( 35 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the rms current?

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

\( L = 35 \times 10^{-3} \, \text{H} \).

\( X_L = 376.8 \times 0.035 = 13.19 \, \Omega \).

RMS current: \( I = \frac{V}{X_L} = \frac{110}{13.19} \approx 8.34 \, \text{A} \).

8 A
8.34 A
8.5 A
9 A
2

A \( 198.1 \, \text{V} \) (peak) AC source is connected to a \( 70 \, \Omega \) resistor. What is the average power consumed?

RMS voltage: \( V = \frac{v_m}{\sqrt{2}} = \frac{198.1}{1.414} \approx 140 \, \text{V} \).

RMS current: \( I = \frac{V}{R} = \frac{140}{70} = 2 \, \text{A} \).

Average power: \( P = I^2 R = 2^2 \times 70 = 280 \, \text{W} \).

280 W
300 W
320 W
340 W
1

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