Alternating Currents Chapter-Wise Test 9

Correct answer Carries: 4.

Wrong Answer Carries: -1.

In an AC circuit with only an inductor, what is the phase relationship between the current and the voltage?

In a purely inductive AC circuit, the current lags the voltage by 90°. This is because the inductor opposes changes in current, causing the current to reach its peak after the voltage.

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

A transformer has \( N_p = 500 \), \( N_s = 250 \). If \( V_s = 120 \, \text{V} \) (rms), what is the primary voltage?

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

\( V_p = V_s \times \frac{N_p}{N_s} = 120 \times \frac{500}{250} = 240 \, \text{V} \).

220 V
230 V
250 V
240 V
4

A series LCR circuit with \( R = 60 \, \Omega \) is at resonance with a \( 240 \, \text{V} \) (rms) source. What is the rms current?

At resonance, \( Z = R = 60 \, \Omega \).

RMS current: \( I = \frac{V}{R} = \frac{240}{60} = 4 \, \text{A} \).

3.5 A
3.8 A
4 A
4.5 A
3

A transformer has \( N_p = 800 \), \( N_s = 400 \). If \( V_s = 110 \, \text{V} \) (rms), what is the primary voltage?

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

\( V_p = V_s \times \frac{N_p}{N_s} = 110 \times \frac{800}{400} = 220 \, \text{V} \).

200 V
210 V
230 V
220 V
4

A \( 70 \, \Omega \) resistor and \( 30 \, \mu\text{F} \) capacitor are in series with a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the impedance?

\( X_C = \frac{1}{\omega C} = \frac{1}{314 \times 30 \times 10^{-6}} \approx 106.1 \, \Omega \).

\( Z = \sqrt{R^2 + X_C^2} = \sqrt{70^2 + 106.1^2} = \sqrt{4900 + 11257.21} \approx 127.3 \, \Omega \).

120 Ω
125 Ω
130 Ω
127.3 Ω
4

A \( 230 \, \text{V} \) (rms) source supplies a \( 115 \, \Omega \) resistor. What is the peak current?

RMS current: \( I = \frac{V}{R} = \frac{230}{115} = 2 \, \text{A} \).

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 \( 22 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the rms current?

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

\( C = 22 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{314 \times 22 \times 10^{-6}} \approx 144.7 \, \Omega \).

RMS current: \( I = \frac{V}{X_C} = \frac{220}{144.7} \approx 1.52 \, \text{A} \).

1.52 A
1.6 A
1.8 A
2 A
1

A series LCR circuit has \( R = 35 \, \Omega \), \( X_L = 60 \, \Omega \), \( X_C = 25 \, \Omega \). What is the power factor?

\( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{35^2 + (60 - 25)^2} = \sqrt{1225 + 1225} = \sqrt{2450} \approx 49.5 \, \Omega \).

Power factor: \( \cos \phi = \frac{R}{Z} = \frac{35}{49.5} \approx 0.707 \).

0.6
0.65
0.707
0.75
3

In an LCR series circuit, what happens to the circuit’s behavior when the frequency is significantly above the resonant frequency?

Above resonance, \( X_L = \omega L \) becomes much larger than \( X_C = \frac{1}{\omega C} \), making the net reactance positive. The circuit behaves as predominantly inductive, with the current lagging the voltage.

It becomes purely resistive
It becomes predominantly capacitive
It becomes predominantly inductive
It stops conducting
3

In an AC circuit with only a capacitor, what happens to the impedance if the capacitance is decreased?

In a purely capacitive circuit, impedance is \( X_C = \frac{1}{\omega C} \). Decreasing capacitance (\( C \)) increases \( X_C \) because they are inversely proportional, thus increasing the impedance.

It decreases
It increases
It remains constant
It becomes zero
2

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