Alternating Currents Chapter-Wise Test 17

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 11 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the capacitive reactance?

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

\( C = 11 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{376.8 \times 11 \times 10^{-6}} \approx 241.2 \, \Omega \).

235 Ω
238 Ω
241.2 Ω
245 Ω
3

A transformer has \( N_p = 200 \), \( N_s = 400 \). If \( V_p = 110 \, \text{V} \) (rms), what is the secondary voltage?

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

\( V_s = V_p \times \frac{N_s}{N_p} = 110 \times \frac{400}{200} = 220 \, \text{V} \).

200 V
220 V
240 V
260 V
2

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

\( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{30^2 + (45 - 15)^2} = \sqrt{900 + 900} = \sqrt{1800} \approx 42.43 \, \Omega \).

Power factor: \( \cos \phi = \frac{R}{Z} = \frac{30}{42.43} \approx 0.707 \).

0.6
0.65
0.707
0.75
3

In an AC circuit with a resistor and inductor in series, what determines the magnitude of the phase difference between voltage and current?

In an RL series circuit, the phase angle \( \phi = \tan^{-1} \left( \frac{X_L}{R} \right) \). The magnitude of this angle depends on the ratio of inductive reactance (\( X_L = \omega L \)) to resistance (\( R \)), as it reflects the relative contributions of inductance and resistance.

The frequency alone
The voltage amplitude
The ratio of inductive reactance to resistance
The capacitance of the circuit
3

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
240 V
260 V
220 V
4

A \( 25 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the peak current?

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

\( C = 25 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{314 \times 25 \times 10^{-6}} \approx 127.4 \, \Omega \).

RMS current: \( I = \frac{V}{X_C} = \frac{230}{127.4} \approx 1.805 \, \text{A} \).

Peak current: \( i_m = \sqrt{2} I = 1.414 \times 1.805 \approx 2.55 \, \text{A} \).

2.4 A
2.55 A
2.7 A
2.9 A
2

A \( 6 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the capacitive reactance?

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

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

\( X_C = \frac{1}{314 \times 6 \times 10^{-6}} \approx 530.5 \, \Omega \).

520 Ω
525 Ω
530.5 Ω
540 Ω
3

What is the effect on an ideal transformer’s secondary voltage if the number of turns in the secondary coil is halved?

In an ideal transformer, \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \). If the number of secondary turns (\( N_s \)) is halved, the secondary voltage (\( V_s \)) becomes half its original value, assuming primary voltage (\( V_p \)) remains constant.

It doubles
It halves
It remains unchanged
It becomes zero
2

A transformer has \( N_p = 300 \), \( N_s = 150 \). If \( V_p = 240 \, \text{V} \) (rms), what is the secondary voltage?

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

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

100 V
120 V
140 V
160 V
2

A \( 40 \, \Omega \) resistor is connected to a \( 120 \, \text{V} \) (rms) AC source. What is the rms current?

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

Given: \( V = 120 \, \text{V} \), \( R = 40 \, \Omega \).

\( I = \frac{120}{40} = 3 \, \text{A} \).

3 A
3.5 A
4 A
4.5 A
1

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