Alternating Currents Chapter-Wise Test 19

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 190 \, \text{V} \) (rms) source supplies a \( 95 \, \Omega \) resistor. What is the peak current?

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

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 transformer has \( N_p = 250 \), \( N_s = 500 \). If \( V_s = 440 \, \text{V} \) (rms), what is the primary voltage?

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

\( V_p = V_s \times \frac{N_p}{N_s} = 440 \times \frac{250}{500} = 220 \, \text{V} \).

200 V
210 V
230 V
220 V
4

A \( 95 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the peak current?

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

\( L = 95 \times 10^{-3} \, \text{H} \).

\( X_L = 314 \times 0.095 = 29.83 \, \Omega \).

RMS current: \( I = \frac{V}{X_L} = \frac{220}{29.83} \approx 7.375 \, \text{A} \).

Peak current: \( i_m = \sqrt{2} I = 1.414 \times 7.375 \approx 10.43 \, \text{A} \).

10 A
10.2 A
10.43 A
10.6 A
3

A \( 175 \, \text{V} \) (rms) AC source supplies a \( 35 \, \Omega \) resistor. What is the average power consumed?

RMS current: \( I = \frac{V}{R} = \frac{175}{35} = 5 \, \text{A} \).

Average power: \( P = I^2 R = 5^2 \times 35 = 875 \, \text{W} \).

850 W
875 W
900 W
925 W
2

A series LCR circuit with \( R = 110 \, \Omega \), \( X_L = 140 \, \Omega \), \( X_C = 80 \, \Omega \) has a \( 330 \, \text{V} \) (rms) source. What is the power dissipated?

\( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{110^2 + (140 - 80)^2} = \sqrt{12100 + 3600} = \sqrt{15700} \approx 125.3 \, \Omega \).

RMS current: \( I = \frac{V}{Z} = \frac{330}{125.3} \approx 2.634 \, \text{A} \).

Power: \( P = I^2 R = (2.634)^2 \times 110 \approx 763.2 \, \text{W} \).

740 W
750 W
770 W
763.2 W
4

A series LCR circuit has \( R = 15 \, \Omega \), \( X_L = 30 \, \Omega \), \( X_C = 45 \, \Omega \). What is the phase angle?

\( \tan \phi = \frac{X_C - X_L}{R} = \frac{45 - 30}{15} = 1 \).

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

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

40°
42°
45°
50°
3

A series LCR circuit has \( R = 120 \, \Omega \), \( X_L = 100 \, \Omega \), \( X_C = 80 \, \Omega \). What is the impedance?

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

\( Z = \sqrt{120^2 + (100 - 80)^2} = \sqrt{14400 + 400} = \sqrt{14800} \approx 121.66 \, \Omega \).

120 Ω
121.66 Ω
125 Ω
130 Ω
2

In an AC circuit with a pure inductor, why is the average power dissipated zero over a complete cycle?

In a pure inductor, the current lags the voltage by 90°. The instantaneous power oscillates between positive (energy stored) and negative (energy returned), averaging to zero over a cycle because the inductor does not dissipate energy as heat but stores and releases it.

Because the inductor has high resistance
Because energy is stored and returned without dissipation
Because the current is constant
Because the voltage is zero
2

A transformer has \( N_p = 600 \), \( N_s = 1200 \). If \( V_s = 480 \, \text{V} \) (rms), what is the primary voltage?

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

\( V_p = V_s \times \frac{N_p}{N_s} = 480 \times \frac{600}{1200} = 240 \, \text{V} \).

220 V
230 V
250 V
240 V
4

What is the primary factor determining the resonant frequency in a series LCR circuit?

The resonant frequency in a series LCR circuit is determined by the inductance (\( L \)) and capacitance (\( C \)), given by \( f_0 = \frac{1}{2\pi \sqrt{L C}} \). Resistance affects damping but not the resonant frequency itself.

Resistance of the circuit
Values of inductance and capacitance
Voltage of the source
Current in the circuit
2

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