Alternating Currents Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

In an AC circuit with a resistor and inductor in series, what happens to the power factor if the inductance is increased?

In an RL series circuit, power factor \( \cos \phi = \frac{R}{Z} \), where \( Z = \sqrt{R^2 + X_L^2} \) and \( X_L = \omega L \). Increasing inductance increases \( X_L \), which increases \( Z \), reducing \( \cos \phi \) (closer to 0), as the circuit becomes more inductive.

It increases
It remains constant
It becomes 1
It decreases
4

A \( 8 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the capacitive reactance?

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

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

\( X_C = \frac{1}{376.8 \times 8 \times 10^{-6}} \approx 331.6 \, \Omega \).

320 Ω
325 Ω
331.6 Ω
340 Ω
3

What happens to the current in a purely inductive AC circuit when the frequency of the source decreases?

In a purely inductive circuit, \( X_L = \omega L \), and current \( I = \frac{V}{X_L} \). Decreasing frequency reduces \( \omega \), lowering \( X_L \), which increases the current since \( I \) is inversely proportional to \( X_L \).

It decreases
It remains constant
It increases
It becomes zero
3

A sinusoidal voltage of peak value \( 283 \, \text{V} \) is applied to a \( 5 \, \Omega \) resistor. What is the rms current?

RMS voltage: \( V = \frac{v_m}{\sqrt{2}} = \frac{283}{\sqrt{2}} \approx 200 \, \text{V} \).

RMS current: \( I = \frac{V}{R} = \frac{200}{5} = 40 \, \text{A} \).

35 A
40 A
45 A
50 A
2

What is the key requirement for an AC circuit to exhibit resonance?

Resonance in an AC circuit requires both inductance and capacitance to be present (e.g., in an LCR circuit). This allows the inductive and capacitive reactances to balance at a specific frequency, a condition not possible with only one reactive component.

Presence of both inductance and capacitance
High resistance
Constant voltage source
Zero frequency
1

A \( 15 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the capacitive reactance?

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

\( C = 15 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{314 \times 15 \times 10^{-6}} \approx 212 \, \Omega \).

200 Ω
212 Ω
220 Ω
230 Ω
2

A \( 55 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the peak current?

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

\( L = 55 \times 10^{-3} \, \text{H} \).

\( X_L = 376.8 \times 0.055 = 20.72 \, \Omega \).

RMS current: \( I = \frac{V}{X_L} = \frac{110}{20.72} \approx 5.31 \, \text{A} \).

Peak current: \( i_m = \sqrt{2} I = 1.414 \times 5.31 \approx 7.51 \, \text{A} \).

7 A
7.2 A
7.51 A
7.8 A
3

What is the role of a capacitor in improving the power factor of an AC circuit with an inductive load?

An inductive load causes the current to lag the voltage, reducing the power factor. Adding a capacitor in parallel introduces a leading current that counteracts the lagging current, bringing the net phase difference closer to zero, thus improving the power factor.

It increases resistance
It reduces frequency
It increases inductance
It counteracts the lagging current
4

A \( 90 \, \text{mH} \) inductor is connected to a \( 220 \, \text{V} \), \( 60 \, \text{Hz} \) AC source. What is the rms current?

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

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

\( X_L = 376.8 \times 0.09 = 33.91 \, \Omega \).

RMS current: \( I = \frac{V}{X_L} = \frac{220}{33.91} \approx 6.49 \, \text{A} \).

6.2 A
6.49 A
6.6 A
6.8 A
2

A series LCR circuit has \( L = 4 \, \text{H} \), \( C = 10 \, \mu\text{F} \). What is the resonant angular frequency?

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

\( L = 4 \, \text{H} \), \( C = 10 \times 10^{-6} \, \text{F} \).

\( \omega_0 = \frac{1}{\sqrt{4 \times 10 \times 10^{-6}}} = \frac{1}{\sqrt{4 \times 10^{-5}}} \approx 158.1 \, \text{rad/s} \).

158.1 rad/s
160 rad/s
165 rad/s
170 rad/s
1

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