Alternating Currents Chapter-Wise Test 20

Correct answer Carries: 4.

Wrong Answer Carries: -1.

In an LCR series circuit, if the frequency is below the resonant frequency, which component dominates the circuit's behavior?

Below resonance, \( \omega \) is low, making \( X_C = \frac{1}{\omega C} \) larger than \( X_L = \omega L \). Thus, the capacitive reactance dominates, and the circuit behaves as predominantly capacitive, with current leading voltage.

Resistance
Inductance
Capacitance
None of the components
3

A series LCR circuit has \( L = 6 \, \text{H} \), \( C = 10 \, \mu\text{F} \). What is the resonant angular frequency?

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

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

\( \omega_0 = \frac{1}{\sqrt{6 \times 10 \times 10^{-6}}} = \frac{1}{\sqrt{6 \times 10^{-5}}} \approx 129.1 \, \text{rad/s} \).

129.1 rad/s
140 rad/s
150 rad/s
160 rad/s
1

Why does an ideal transformer maintain constant power across its primary and secondary coils?

In an ideal transformer, there are no losses (e.g., resistance, flux leakage). Energy conservation dictates that input power (\( V_p I_p \)) equals output power (\( V_s I_s \)), so the power remains constant despite changes in voltage and current.

Because resistance is zero
Because frequency increases
Because voltage is constant
Because of energy conservation
3

A \( 15 \, \Omega \) resistor is connected to a \( 75 \, \text{V} \) (rms) AC source. What is the rms current?

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

Given: \( V = 75 \, \text{V} \), \( R = 15 \, \Omega \).

\( I = \frac{75}{15} = 5 \, \text{A} \).

5 A
5.5 A
6 A
6.5 A
1

A \( 169.7 \, \text{V} \) (peak) AC source is connected to a \( 60 \, \Omega \) resistor. What is the average power consumed?

RMS voltage: \( V = \frac{v_m}{\sqrt{2}} = \frac{169.7}{1.414} \approx 120 \, \text{V} \).

RMS current: \( I = \frac{V}{R} = \frac{120}{60} = 2 \, \text{A} \).

Average power: \( P = I^2 R = 2^2 \times 60 = 240 \, \text{W} \).

240 W
260 W
280 W
300 W
1

A series LCR circuit with \( R = 50 \, \Omega \), \( X_L = 70 \, \Omega \), \( X_C = 30 \, \Omega \) has a \( 200 \, \text{V} \) (rms) source. What is the power dissipated?

\( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{50^2 + (70 - 30)^2} = \sqrt{2500 + 1600} = \sqrt{4100} \approx 64 \, \Omega \).

RMS current: \( I = \frac{V}{Z} = \frac{200}{64} \approx 3.125 \, \text{A} \).

Power: \( P = I^2 R = (3.125)^2 \times 50 \approx 488.28 \, \text{W} \).

450 W
470 W
500 W
488.28 W
4

A \( 240 \, \text{V} \) (rms) source supplies a \( 120 \, \Omega \) resistor. What is the peak current?

RMS current: \( I = \frac{V}{R} = \frac{240}{120} = 2 \, \text{A} \).

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 has \( R = 80 \, \Omega \), \( X_L = 60 \, \Omega \), \( X_C = 40 \, \Omega \). What is the impedance?

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

\( Z = \sqrt{80^2 + (60 - 40)^2} = \sqrt{6400 + 400} = \sqrt{6800} \approx 82.46 \, \Omega \).

80 Ω
82.46 Ω
85 Ω
90 Ω
2

A \( 50 \, \Omega \) resistor and \( 20 \, \mu\text{F} \) capacitor are in series with a \( 110 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the voltage across the resistor?

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

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

RMS current: \( I = \frac{V}{Z} = \frac{110}{166.6} \approx 0.66 \, \text{A} \).

Voltage across resistor: \( V_R = I R = 0.66 \times 50 \approx 33 \, \text{V} \).

30 V
32 V
35 V
33 V
4

A series LCR circuit with \( R = 50 \, \Omega \) is at resonance with a \( 250 \, \text{V} \) (rms) source. What is the rms current?

At resonance, \( Z = R = 50 \, \Omega \).

RMS current: \( I = \frac{V}{R} = \frac{250}{50} = 5 \, \text{A} \).

4.5 A
4.8 A
5 A
5.2 A
3

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