Alternating Currents Chapter-Wise Test 15

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A series LCR circuit has \( R = 25 \, \Omega \), \( X_L = 50 \, \Omega \), \( X_C = 70 \, \Omega \). What is the phase angle?

\( \tan \phi = \frac{X_C - X_L}{R} = \frac{70 - 50}{25} = \frac{20}{25} = 0.8 \).

\( \phi = \tan^{-1}(0.8) \approx 38.66^\circ \).

Since \( X_C > X_L \), current leads voltage.

35°
37°
38.66°
40°
3

A \( 110 \, \text{mH} \) inductor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the inductive reactance?

\( X_L = \omega L \), \( \omega = 2\pi f \).

\( f = 50 \, \text{Hz} \), \( L = 110 \times 10^{-3} \, \text{H} \).

\( \omega = 2 \times 3.14 \times 50 = 314 \, \text{rad/s} \).

\( X_L = 314 \times 0.11 = 34.54 \, \Omega \).

34.54 Ω
35 Ω
36 Ω
38 Ω
1

What is the significance of the phase difference being 90° in a purely reactive AC circuit?

A 90° phase difference (in purely inductive or capacitive circuits) means the power factor (\( \cos 90^\circ = 0 \)) is zero, indicating no average power is dissipated. Energy oscillates between the source and the reactive element without being converted to heat.

Maximum power is dissipated
Resistance is maximum
No average power is dissipated
Current is constant
3

Why does the impedance of an AC circuit with only an inductor increase with frequency?

The impedance in a purely inductive circuit is the inductive reactance (\( X_L = \omega L \)), where \( \omega = 2\pi f \). As frequency (\( f \)) increases, \( \omega \) increases linearly, causing \( X_L \) (and thus impedance) to increase proportionally.

Because resistance increases
Because inductive reactance increases with frequency
Because capacitance decreases
Because voltage decreases
2

A \( 200 \, \Omega \) resistor and \( 15 \, \mu\text{F} \) capacitor are in series with a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the voltage across the capacitor?

\( X_C = \frac{1}{\omega C} = \frac{1}{314 \times 15 \times 10^{-6}} \approx 212.3 \, \Omega \).

\( Z = \sqrt{200^2 + 212.3^2} \approx 291.5 \, \Omega \).

RMS current: \( I = \frac{V}{Z} = \frac{220}{291.5} \approx 0.755 \, \text{A} \).

Voltage across capacitor: \( V_C = I X_C = 0.755 \times 212.3 \approx 160.3 \, \text{V} \).

150 V
160.3 V
170 V
180 V
2

Why does a transformer not work with direct current (DC)?

A transformer relies on a changing magnetic flux to induce voltage in the secondary coil via mutual induction. DC provides a constant current, producing a steady magnetic field with no flux change, so no voltage is induced.

Because DC has no resistance
Because DC has high frequency
Because DC generates heat
Because DC does not produce a changing magnetic flux
4

A \( 20 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the peak current?

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

\( C = 20 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{376.8 \times 20 \times 10^{-6}} \approx 132.7 \, \Omega \).

RMS current: \( I = \frac{V}{X_C} = \frac{110}{132.7} \approx 0.829 \, \text{A} \).

Peak current: \( i_m = \sqrt{2} I = 1.414 \times 0.829 \approx 1.17 \, \text{A} \).

1 A
1.17 A
1.3 A
1.5 A
2

A \( 75 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the rms current?

\( X_L = \omega L \), \( \omega = 2\pi \times 60 = 376.8 \, \text{rad/s} \).

\( L = 75 \times 10^{-3} \, \text{H} \).

\( X_L = 376.8 \times 0.075 = 28.26 \, \Omega \).

RMS current: \( I = \frac{V}{X_L} = \frac{110}{28.26} \approx 3.89 \, \text{A} \).

3.5 A
3.89 A
4 A
4.2 A
2

An AC voltage \( v = 141.4 \sin (100\pi t) \) V is applied across a \( 50 \, \Omega \) resistor. What is the rms current through the resistor?

Given: \( v = v_m \sin \omega t \), where \( v_m = 141.4 \, \text{V} \), \( R = 50 \, \Omega \).

RMS voltage: \( V = \frac{v_m}{\sqrt{2}} = \frac{141.4}{\sqrt{2}} = 100 \, \text{V} \).

RMS current: \( I = \frac{V}{R} = \frac{100}{50} = 2 \, \text{A} \).

1.5 A
2 A
2.5 A
3 A
2

A \( 20 \, \Omega \) resistor is connected to a \( 100 \, \text{V} \) (rms) AC source. What is the rms current?

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

Given: \( V = 100 \, \text{V} \), \( R = 20 \, \Omega \).

\( I = \frac{100}{20} = 5 \, \text{A} \).

5 A
5.5 A
6 A
6.5 A
1

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