Alternating Currents Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A series LCR circuit with \( R = 80 \, \Omega \), \( X_L = 100 \, \Omega \), \( X_C = 40 \, \Omega \) has a \( 240 \, \text{V} \) (rms) source. What is the power dissipated?

\( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{80^2 + (100 - 40)^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \, \Omega \).

RMS current: \( I = \frac{V}{Z} = \frac{240}{100} = 2.4 \, \text{A} \).

Power: \( P = I^2 R = (2.4)^2 \times 80 = 5.76 \times 80 = 460.8 \, \text{W} \).

440 W
450 W
470 W
460.8 W
4

A \( 45 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the peak current?

\( X_C = \frac{1}{\omega C} \), \( \omega = 2\pi \times 60 = 376.8 \, \text{rad/s} \).

\( C = 45 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{376.8 \times 45 \times 10^{-6}} \approx 59 \, \Omega \).

RMS current: \( I = \frac{V}{X_C} = \frac{110}{59} \approx 1.864 \, \text{A} \).

Peak current: \( i_m = \sqrt{2} I = 1.414 \times 1.864 \approx 2.64 \, \text{A} \).

2.5 A
2.64 A
2.8 A
3 A
2

What happens to the average power dissipated in an AC circuit with only a capacitor over one complete cycle?

In a purely capacitive AC circuit, the current leads the voltage by 90°. The instantaneous power alternates between positive and negative, and over a complete cycle, these values cancel out, resulting in zero average power dissipation. The capacitor stores and releases energy without dissipating it as heat.

It is maximum
It is zero
It depends on the frequency
It is half of the peak power
1

A \( 15 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the rms current?

\( X_C = \frac{1}{\omega C} \), \( \omega = 2\pi \times 60 = 376.8 \, \text{rad/s} \).

\( C = 15 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{376.8 \times 15 \times 10^{-6}} \approx 177 \, \Omega \).

RMS current: \( I = \frac{V}{X_C} = \frac{110}{177} \approx 0.621 \, \text{A} \).

0.5 A
0.6 A
0.7 A
0.621 A
4

In an AC circuit with only a resistor, how does the current behave relative to the applied voltage?

In a purely resistive AC circuit, the current and voltage oscillate in phase, meaning they reach their peak, zero, and minimum values simultaneously. This occurs because a resistor does not introduce any phase shift between voltage and current.

It is in phase with the voltage
It lags the voltage by 90°
It leads the voltage by 90°
It is 180° out of phase with the voltage
1

A transformer has \( N_p = 300 \), \( N_s = 150 \). If \( V_s = 120 \, \text{V} \) (rms), what is the primary voltage?

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

\( V_p = V_s \times \frac{N_p}{N_s} = 120 \times \frac{300}{150} = 240 \, \text{V} \).

200 V
220 V
260 V
240 V
4

What is the significance of the rms value in specifying AC quantities?

The rms (root mean square) value of an AC quantity (e.g., voltage or current) is the equivalent DC value that produces the same average power in a resistive load. It accounts for the time-varying nature of AC, making it a standard measure for power calculations.

It indicates the peak voltage
It measures the frequency
It represents the equivalent DC value for power
It shows the phase difference
3

A \( 200 \, \text{V} \) (rms) source supplies a \( 80 \, \Omega \) resistor. What is the peak current?

RMS current: \( I = \frac{V}{R} = \frac{200}{80} = 2.5 \, \text{A} \).

Peak current: \( i_m = \sqrt{2} I = 1.414 \times 2.5 \approx 3.535 \, \text{A} \).

3 A
3.2 A
3.8 A
3.535 A
4

A \( 70 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the inductive reactance?

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

\( f = 50 \, \text{Hz} \), \( L = 70 \times 10^{-3} \, \text{H} \).

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

\( X_L = 314 \times 0.07 = 21.98 \, \Omega \).

21.98 Ω
25 Ω
28 Ω
30 Ω
1

A \( 226.3 \, \text{V} \) (peak) AC source is connected to a \( 80 \, \Omega \) resistor. What is the average power consumed?

RMS voltage: \( V = \frac{v_m}{\sqrt{2}} = \frac{226.3}{1.414} \approx 160 \, \text{V} \).

RMS current: \( I = \frac{V}{R} = \frac{160}{80} = 2 \, \text{A} \).

Average power: \( P = I^2 R = 2^2 \times 80 = 320 \, \text{W} \).

320 W
340 W
360 W
380 W
1

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