Alternating Currents Chapter-Wise Test 16

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A series LCR circuit has \( L = 3 \, \text{H} \), \( C = 12 \, \mu\text{F} \). What is the resonant angular frequency?

\( \omega_0 = \frac{1}{\sqrt{L C}} \).

\( L = 3 \, \text{H} \), \( C = 12 \times 10^{-6} \, \text{F} \).

\( \omega_0 = \frac{1}{\sqrt{3 \times 12 \times 10^{-6}}} = \frac{1}{\sqrt{36 \times 10^{-6}}} = \frac{10^3}{6} \approx 166.67 \, \text{rad/s} \).

166.67 rad/s
180 rad/s
200 rad/s
220 rad/s
1

A series LCR circuit has \( R = 20 \, \Omega \), \( X_L = 30 \, \Omega \), \( X_C = 50 \, \Omega \). What is the power factor?

\( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{20^2 + (30 - 50)^2} = \sqrt{400 + 400} = \sqrt{800} \approx 28.28 \, \Omega \).

Power factor: \( \cos \phi = \frac{R}{Z} = \frac{20}{28.28} \approx 0.707 \).

0.6
0.65
0.707
0.75
3

In an AC circuit with a series combination of resistor and inductor, what happens to the impedance if the frequency decreases?

In an RL series circuit, impedance \( Z = \sqrt{R^2 + X_L^2} \), where \( X_L = \omega L \). Decreasing frequency reduces \( \omega \), lowering \( X_L \), which decreases the impedance since \( R \) remains constant.

It increases
It decreases
It remains constant
It becomes infinite
2

A \( 60 \, \Omega \) resistor is connected to a \( 180 \, \text{V} \) (rms) AC source. What is the rms current?

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

Given: \( V = 180 \, \text{V} \), \( R = 60 \, \Omega \).

\( I = \frac{180}{60} = 3 \, \text{A} \).

3 A
3.5 A
4 A
4.5 A
1

In an AC circuit with a series combination of resistor and capacitor, what determines whether the circuit is predominantly capacitive?

In an RC series circuit, the circuit is predominantly capacitive if the capacitive reactance (\( X_C = \frac{1}{\omega C} \)) is greater than the resistance (\( R \)). This makes the impedance \( Z = \sqrt{R^2 + X_C^2} \) dominated by \( X_C \), and the current leads the voltage significantly.

The voltage amplitude
The frequency alone
The relative magnitude of capacitive reactance to resistance
The inductance of the circuit
3

A \( 200 \, \text{V} \) (rms) AC source supplies a \( 100 \, \Omega \) resistor. What is the peak voltage?

Peak voltage: \( v_m = \sqrt{2} V \).

\( V = 200 \, \text{V} \).

\( v_m = 1.414 \times 200 = 282.8 \, \text{V} \approx 283 \, \text{V} \).

283 V
300 V
320 V
350 V
1

A series LCR circuit with \( R = 60 \, \Omega \), \( X_L = 80 \, \Omega \), \( X_C = 20 \, \Omega \) has a \( 180 \, \text{V} \) (rms) source. What is the power dissipated?

\( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{60^2 + (80 - 20)^2} = \sqrt{3600 + 3600} = \sqrt{7200} \approx 84.85 \, \Omega \).

RMS current: \( I = \frac{V}{Z} = \frac{180}{84.85} \approx 2.12 \, \text{A} \).

Power: \( P = I^2 R = (2.12)^2 \times 60 \approx 269.66 \, \text{W} \).

250 W
260 W
280 W
269.66 W
4

A \( 130 \, \text{V} \) (rms) AC source supplies a \( 65 \, \Omega \) resistor. What is the average power consumed?

RMS current: \( I = \frac{V}{R} = \frac{130}{65} = 2 \, \text{A} \).

Average power: \( P = I^2 R = 2^2 \times 65 = 260 \, \text{W} \).

240 W
260 W
280 W
300 W
2

What is the primary energy loss mechanism in a real transformer that reduces its efficiency?

In a real transformer, energy losses occur due to resistance in the windings (\( I^2 R \) losses), flux leakage, eddy currents, and hysteresis. Among these, resistance in the windings (copper loss) is a primary mechanism, converting electrical energy into heat.

Increase in frequency
Capacitance of the coils
Voltage drop
Resistance in the windings
4

A transformer has \( N_p = 700 \), \( N_s = 350 \). If \( V_p = 280 \, \text{V} \) (rms), what is the secondary voltage?

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

\( V_s = V_p \times \frac{N_s}{N_p} = 280 \times \frac{350}{700} = 140 \, \text{V} \).

130 V
140 V
150 V
160 V
2

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