Alternating Currents Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A transformer has \( N_p = 400 \), \( N_s = 800 \). If \( V_p = 120 \, \text{V} \) (rms), what is the secondary voltage?

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

\( V_s = V_p \times \frac{N_s}{N_p} = 120 \times \frac{800}{400} = 240 \, \text{V} \).

200 V
240 V
260 V
300 V
2

A \( 100 \, \Omega \) resistor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC supply. What is the average power consumed?

Average power: \( P = I^2 R \), where \( I = \frac{V}{R} \).

\( I = \frac{220}{100} = 2.2 \, \text{A} \).

\( P = (2.2)^2 \times 100 = 4.84 \times 100 = 484 \, \text{W} \).

450 W
484 W
500 W
520 W
2

In an AC circuit with only a capacitor, why does the current lead the voltage?

In a capacitor, current flows to charge or discharge the plates before the voltage across it changes. This causes the current to reach its maximum 90° ahead of the voltage, as the voltage builds up after the current starts flowing.

Because of high resistance
Because of inductance
Because current charges the capacitor before voltage builds
Because frequency is low
3

A series LCR circuit with \( R = 40 \, \Omega \), \( X_L = 50 \, \Omega \), \( X_C = 30 \, \Omega \) has a \( 120 \, \text{V} \) (rms) source. What is the power dissipated?

\( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (50 - 30)^2} = \sqrt{1600 + 400} = \sqrt{2000} \approx 44.72 \, \Omega \).

RMS current: \( I = \frac{V}{Z} = \frac{120}{44.72} \approx 2.68 \, \text{A} \).

Power: \( P = I^2 R = (2.68)^2 \times 40 \approx 287.3 \, \text{W} \).

250 W
270 W
300 W
287.3 W
4

A \( 60 \, \Omega \) resistor and \( 40 \, \mu\text{F} \) capacitor are in series with a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the impedance?

\( X_C = \frac{1}{\omega C} = \frac{1}{314 \times 40 \times 10^{-6}} \approx 79.6 \, \Omega \).

\( Z = \sqrt{R^2 + X_C^2} = \sqrt{60^2 + 79.6^2} = \sqrt{3600 + 6336.16} \approx 99.8 \, \Omega \).

95 Ω
97 Ω
100 Ω
99.8 Ω
4

A \( 14 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the rms current?

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

\( C = 14 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{314 \times 14 \times 10^{-6}} \approx 227.5 \, \Omega \).

RMS current: \( I = \frac{V}{X_C} = \frac{230}{227.5} \approx 1.01 \, \text{A} \).

1.01 A
1.2 A
1.3 A
1.5 A
1

A \( 100 \, \Omega \) resistor and \( 50 \, \mu\text{F} \) capacitor are in series with a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the impedance?

\( X_C = \frac{1}{\omega C} = \frac{1}{314 \times 50 \times 10^{-6}} \approx 63.7 \, \Omega \).

\( Z = \sqrt{R^2 + X_C^2} = \sqrt{100^2 + 63.7^2} = \sqrt{10000 + 4057.69} \approx 118.7 \, \Omega \).

110 Ω
115 Ω
120 Ω
118.7 Ω
4

A \( 9 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the capacitive reactance?

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

\( C = 9 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{314 \times 9 \times 10^{-6}} \approx 353.7 \, \Omega \).

340 Ω
350 Ω
353.7 Ω
360 Ω
3

A transformer has \( N_p = 400 \), \( N_s = 800 \). If \( V_p = 120 \, \text{V} \) (rms), what is the secondary voltage?

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

\( V_s = V_p \times \frac{N_s}{N_p} = 120 \times \frac{800}{400} = 240 \, \text{V} \).

220 V
240 V
260 V
280 V
2

In a series LCR circuit, what happens to the total voltage across the inductor and capacitor at resonance?

At resonance, \( X_L = X_C \), and the voltages across the inductor (\( V_L = I X_L \)) and capacitor (\( V_C = I X_C \)) are equal in magnitude but 180° out of phase. Thus, they cancel each other, making the net voltage across the LC combination zero.

It equals the source voltage
It is maximum
It is zero
It doubles
3

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