Alternating Currents Chapter-Wise Test 8

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the physical significance of the term "wattless current" in an AC circuit?

"Wattless current" refers to the component of current in a purely reactive (inductive or capacitive) AC circuit that is 90° out of phase with the voltage. It does not contribute to average power dissipation, as power is zero when \( \cos 90° = 0 \), hence "wattless."

It is the current with maximum power
It is the current in a resistive circuit
It is the current that heats the circuit
It is the current that does not contribute to average power
4

A \( 90 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the inductive reactance?

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

\( f = 50 \, \text{Hz} \), \( L = 90 \times 10^{-3} \, \text{H} \).

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

\( X_L = 314 \times 0.09 = 28.26 \, \Omega \).

25 Ω
28.26 Ω
30 Ω
32 Ω
1

A series LCR circuit has \( L = 8 \, \text{H} \), \( C = 5 \, \mu\text{F} \). What is the resonant angular frequency?

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

\( L = 8 \, \text{H} \), \( C = 5 \times 10^{-6} \, \text{F} \).

\( \omega_0 = \frac{1}{\sqrt{8 \times 5 \times 10^{-6}}} = \frac{1}{\sqrt{40 \times 10^{-6}}} \approx 158.1 \, \text{rad/s} \).

158.1 rad/s
160 rad/s
170 rad/s
180 rad/s
1

A series LCR circuit has \( R = 40 \, \Omega \), \( X_L = 65 \, \Omega \), \( X_C = 25 \, \Omega \). What is the power factor?

\( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (65 - 25)^2} = \sqrt{1600 + 1600} = \sqrt{3200} \approx 56.57 \, \Omega \).

Power factor: \( \cos \phi = \frac{R}{Z} = \frac{40}{56.57} \approx 0.707 \).

0.6
0.65
0.707
0.75
3

A \( 40 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the capacitive reactance?

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

\( C = 40 \times 10^{-6} \, \text{F} \).

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

75 Ω
78 Ω
79.6 Ω
82 Ω
3

A \( 16 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the peak current?

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

\( C = 16 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{314 \times 16 \times 10^{-6}} \approx 199 \, \Omega \).

RMS current: \( I = \frac{V}{X_C} = \frac{230}{199} \approx 1.156 \, \text{A} \).

Peak current: \( i_m = \sqrt{2} I = 1.414 \times 1.156 \approx 1.63 \, \text{A} \).

1.5 A
1.63 A
1.7 A
1.8 A
2

A \( 95 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is the inductive reactance?

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

\( f = 50 \, \text{Hz} \), \( L = 95 \times 10^{-3} \, \text{H} \).

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

\( X_L = 314 \times 0.095 = 29.83 \, \Omega \).

29.83 Ω
30 Ω
32 Ω
35 Ω
1

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

\( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{30^2 + (50 - 20)^2} = \sqrt{900 + 900} = \sqrt{1800} \approx 42.43 \, \Omega \).

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

0.6
0.65
0.707
0.75
3

A series LCR circuit has \( R = 30 \, \Omega \), \( X_L = 40 \, \Omega \), \( X_C = 60 \, \Omega \). What is the phase angle?

\( \tan \phi = \frac{X_C - X_L}{R} = \frac{60 - 40}{30} = \frac{20}{30} = 0.667 \).

\( \phi = \tan^{-1}(0.667) \approx 33.7^\circ \).

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

30°
32°
33.7°
35°
3

A series LCR circuit has \( R = 90 \, \Omega \), \( X_L = 70 \, \Omega \), \( X_C = 50 \, \Omega \). What is the impedance?

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

\( Z = \sqrt{90^2 + (70 - 50)^2} = \sqrt{8100 + 400} = \sqrt{8500} \approx 92.2 \, \Omega \).

90 Ω
92.2 Ω
95 Ω
100 Ω
2

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