Alternating Current Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 45 \, \text{mH} \) inductor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the peak current?

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

\( L = 45 \times 10^{-3} \, \text{H} \).

\( X_L = 314 \times 0.045 = 14.13 \, \Omega \).

RMS current: \( I = \frac{V}{X_L} = \frac{230}{14.13} \approx 16.28 \, \text{A} \).

Peak current: \( i_m = \sqrt{2} I = 1.414 \times 16.28 \approx 23.02 \, \text{A} \).

22 A
22.5 A
23.02 A
23.5 A
3

A \( 10 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the rms current?

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

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

\( X_C = \frac{1}{314 \times 10 \times 10^{-6}} \approx 318.5 \, \Omega \).

RMS current: \( I = \frac{V}{X_C} = \frac{230}{318.5} \approx 0.722 \, \text{A} \).

0.6 A
0.7 A
0.8 A
0.722 A
4

A \( 35 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the peak current?

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

\( C = 35 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{314 \times 35 \times 10^{-6}} \approx 91 \, \Omega \).

RMS current: \( I = \frac{V}{X_C} = \frac{230}{91} \approx 2.527 \, \text{A} \).

Peak current: \( i_m = \sqrt{2} I = 1.414 \times 2.527 \approx 3.57 \, \text{A} \).

3.2 A
3.57 A
3.8 A
4 A
2

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

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

Given: \( V = 240 \, \text{V} \), \( R = 120 \, \Omega \).

\( I = \frac{240}{120} = 2 \, \text{A} \).

2 A
2.5 A
3 A
3.5 A
1

A series LCR circuit has \( R = 25 \, \Omega \), \( X_L = 45 \, \Omega \), \( X_C = 20 \, \Omega \). What is the power factor?

\( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{25^2 + (45 - 20)^2} = \sqrt{625 + 625} = \sqrt{1250} \approx 35.36 \, \Omega \).

Power factor: \( \cos \phi = \frac{R}{Z} = \frac{25}{35.36} \approx 0.707 \).

0.6
0.65
0.707
0.75
3

A \( 40 \, \text{mH} \) inductor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the rms current?

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

\( L = 40 \times 10^{-3} \, \text{H} \).

\( X_L = 314 \times 0.04 = 12.56 \, \Omega \).

RMS current: \( I = \frac{V}{X_L} = \frac{230}{12.56} \approx 18.31 \, \text{A} \).

16 A
18.31 A
20 A
22 A
2

A series LCR circuit has \( R = 70 \, \Omega \), \( X_L = 40 \, \Omega \), \( X_C = 20 \, \Omega \). What is the impedance?

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

\( Z = \sqrt{70^2 + (40 - 20)^2} = \sqrt{4900 + 400} = \sqrt{5300} \approx 72.8 \, \Omega \).

70 Ω
72.8 Ω
75 Ω
80 Ω
2

What is the primary purpose of using laminated cores in transformers?

Laminated cores in transformers reduce eddy current losses. By dividing the core into thin, insulated layers, the induced currents are confined to smaller paths, minimizing energy loss as heat and improving efficiency.

To increase resistance
To enhance capacitance
To increase voltage
To reduce eddy current losses
4

What is the effect on the impedance of a series LCR circuit when the frequency is increased beyond the resonant frequency?

Beyond the resonant frequency, the inductive reactance (\( X_L = \omega L \)) increases with frequency, while the capacitive reactance (\( X_C = \frac{1}{\omega C} \)) decreases. Since \( X_L \) grows faster than \( X_C \) decreases, the net reactance (\( X_L - X_C \)) becomes positive, increasing the impedance (\( Z = \sqrt{R^2 + (X_L - X_C)^2} \)).

It decreases
It increases
It remains constant
It becomes zero
2

Why does a step-down transformer increase the current in the secondary coil compared to the primary coil?

In a step-down transformer (\( V_s < V_p \), \( N_s < N_p \)), power is conserved (\( V_p I_p = V_s I_s \)). Since \( V_s \) is lower, \( I_s \) must be higher than \( I_p \) (\( I_s = I_p \times \frac{N_p}{N_s} \), where \( \frac{N_p}{N_s} > 1 \)) to maintain the same power output.

Because resistance increases
Because voltage increases
Because frequency decreases
Because power is conserved
4

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