Question

A pure inductor of inductance \( 0.1 \, \text{H} \) is connected to an AC source (of rms voltage 220 V and angular frequency 300 Hz). The rms current is

• A. \( \frac{3}{22} \, \text{A} \)
• B. \( \frac{22}{3} \, \text{A} \)
• C. \( \frac{11}{150} \, \text{A} \)
• D. \( \frac{150}{11} \, \text{A} \)
• E. \( \frac{22}{6\pi} \, \text{A} \)

Hint:

For an AC circuit with a pure inductor, the rms current depends on the voltage and the inductive reactance \( X_L \), which is related to the frequency and inductance of the circuit.

Answer 1

Answer: (E)

Step 1: Using the formula for rms current.
The formula for the rms current \( I_{\text{rms}} \) in an AC circuit with a pure inductor is given by:
\[ I_{\text{rms}} = \frac{V_{\text{rms}}}{X_L} \] Where:
- \( V_{\text{rms}} = 220 \, \text{V} \) (rms voltage)
- \( X_L = \omega L = 2\pi f L \) is the inductive reactance, where:
- \( f = 300 \, \text{Hz} \) (frequency)
- \( L = 0.1 \, \text{H} \) (inductance)
- \( \omega = 2\pi f \) is the angular frequency.

Step 2: Calculate the inductive reactance \( X_L \).
First, calculate the angular frequency:
\[ \omega = 2\pi f = 2\pi \times 300 = 600\pi \, \text{rad/s} \] Now, calculate the inductive reactance \( X_L \):
\[ X_L = \omega L = 600\pi \times 0.1 = 60\pi \, \Omega \]

Step 3: Calculate the rms current.
Now we can calculate the rms current:
\[ I_{\text{rms}} = \frac{V_{\text{rms}}}{X_L} = \frac{220}{60\pi} \]

Step 4: Final calculation.
Simplifying the above expression:
\[ I_{\text{rms}} = \frac{220}{60\pi} = \frac{22}{6\pi} \, \text{A} \]

Final Answer: (E) \( \frac{22}{6\pi} \, \text{A} \)