1. Home
    2. >
  2. Neet Physics Class 11 Mock Test Papers
    3. >
  3. Physics Chapter Wise Test Papers
    4. >
  4. Alternating Current Chapter-Wise Test 3

Alternating Current Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A series LCR circuit has \( R = 60 \, \Omega \), \( X_L = 50 \, \Omega \), \( X_C = 30 \, \Omega \). What is the impedance?

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

\( Z = \sqrt{60^2 + (50 - 30)^2} = \sqrt{3600 + 400} = \sqrt{4000} \approx 63.25 \, \Omega \).

60 Ω
62 Ω
63.25 Ω
65 Ω
2

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

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

\( L = 60 \times 10^{-3} \, \text{H} \).

\( X_L = 376.8 \times 0.06 = 22.61 \, \Omega \).

RMS current: \( I = \frac{V}{X_L} = \frac{110}{22.61} \approx 4.87 \, \text{A} \).

Peak current: \( i_m = \sqrt{2} I = 1.414 \times 4.87 \approx 6.88 \, \text{A} \).

6.5 A
6.7 A
6.88 A
7 A
3

A series LCR circuit has \( L = 4 \, \text{H} \), \( C = 25 \, \mu\text{F} \). What is the resonant angular frequency?

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

\( L = 4 \, \text{H} \), \( C = 25 \times 10^{-6} \, \text{F} \).

\( \omega_0 = \frac{1}{\sqrt{4 \times 25 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-4}}} = 100 \, \text{rad/s} \).

100 rad/s
120 rad/s
150 rad/s
200 rad/s
1

A \( 26 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the rms current?

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

\( C = 26 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{314 \times 26 \times 10^{-6}} \approx 122.4 \, \Omega \).

RMS current: \( I = \frac{V}{X_C} = \frac{230}{122.4} \approx 1.879 \, \text{A} \).

1.7 A
1.8 A
1.9 A
1.879 A
4

In an AC circuit with a resistor and capacitor in series, what happens to the total voltage across the components compared to the source voltage?

In an RC series circuit, the voltages across the resistor (\( V_R \)) and capacitor (\( V_C \)) are 90° out of phase. The total source voltage is the vector sum, \( V = \sqrt{V_R^2 + V_C^2} \), which equals the applied voltage, not the algebraic sum, due to the phase difference.

It is less than the source voltage
It equals the source voltage when added vectorially
It is greater than the source voltage
It is zero
2

What happens to the current in an AC circuit containing only an inductor when the source frequency approaches zero?

In a purely inductive circuit, \( X_L = \omega L \), and \( I = \frac{V}{X_L} \). As frequency (\( f \)) approaches zero, \( \omega = 2\pi f \) also approaches zero, making \( X_L \) very small. Thus, the current increases significantly, approaching a maximum limited only by resistance (which is zero in an ideal inductor).

It becomes zero
It increases significantly
It remains constant
It oscillates rapidly
1

A \( 24 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the rms current?

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

\( C = 24 \times 10^{-6} \, \text{F} \).

\( X_C = \frac{1}{314 \times 24 \times 10^{-6}} \approx 132.6 \, \Omega \).

RMS current: \( I = \frac{V}{X_C} = \frac{230}{132.6} \approx 1.734 \, \text{A} \).

1.5 A
1.6 A
1.8 A
1.734 A
4

In an ideal AC circuit with only a capacitor, what is the relationship between the rates of change of voltage and current?

In a capacitor, \( I = C \frac{dV}{dt} \), meaning the current is directly proportional to the rate of change of voltage. Conversely, the rate of change of current relates to the second derivative of voltage, but the primary relationship is that current depends on \( \frac{dV}{dt} \).

Current is proportional to the rate of change of voltage
Voltage is proportional to the rate of change of current
They are equal
They are out of phase by 180°
1

A \( 220 \, \text{V} \) (rms), \( 50 \, \text{Hz} \) AC source is connected to a \( 44 \, \text{mH} \) inductor. Calculate the inductive reactance.

Inductive reactance: \( X_L = \omega L \), where \( \omega = 2\pi f \).

Given: \( f = 50 \, \text{Hz} \), \( L = 44 \, \text{mH} = 0.044 \, \text{H} \).

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

\( X_L = 314 \times 0.044 = 13.816 \, \Omega \approx 13.82 \, \Omega \).

12.5 Ω
13.82 Ω
15 Ω
16.5 Ω
2

In an LCR circuit with \( R = 3 \, \Omega \), \( X_L = 8 \, \Omega \), \( X_C = 4 \, \Omega \), what is the power factor?

Impedance: \( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{3^2 + (8 - 4)^2} = \sqrt{9 + 16} = 5 \, \Omega \).

Power factor: \( \cos \phi = \frac{R}{Z} = \frac{3}{5} = 0.6 \).

0.5
0.6
0.7
0.8
2

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0
error: Content is protected !!