Question

A pure inductor of 0.25 H is connected to a source of 220 V. If the frequency of the source is 50 Hz, then the rms current in the circuit will be.

• A. 1.6 A
• B. 4.0 A
• C. 2.8 A
• D. 28 A

Hint:

Use \(I_{rms} = V_{rms}/(2\pi f L)\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A pure inductor opposes an a.c. current through its reactance. The rms current equals the rms voltage divided by this reactance.

Step 2: Key Formula or Approach:
Inductive reactance is \(X_L = 2\pi f L\). Then \(I_{rms} = \frac{V_{rms}}{X_L}\).

Step 3: Find the reactance:
\[ X_L = 2\pi \times 50 \times 0.25 = 25\pi \approx 78.5\ \Omega \]

Step 4: Find the current:
\[ I_{rms} = \frac{220}{78.5} \approx 2.80\ \text{A} \]

Step 5: Check the options:
1.6 A would need a reactance of about 137 ohm. 4.0 A would need 55 ohm. 28 A would need 7.85 ohm, which is off by a factor of 10 and comes from a decimal slip. Only 2.8 A fits.

Final Answer:
The rms current is about 2.8 A, which is option 3. \[ \boxed{I_{rms} \approx 2.8\ \text{A}} \]