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: Set up with angular frequency:
Write the angular frequency first: $\omega = 2\pi f = 100\pi\ \text{rad/s} \approx 314\ \text{rad/s}$.

Step 2: Use Ohm's law for a.c.:
For a pure inductor, $V_{rms} = I_{rms}\, \omega L$. So $I_{rms} = \frac{V_{rms}}{\omega L}$.

Step 3: Put in the numbers:
$\omega L = 314 \times 0.25 = 78.5\ \Omega$. Then \[ I_{rms} = \frac{220}{78.5} = 2.80\ \text{A} \]

Step 4: Match with the choices:
The value 2.8 A appears as the third option. The other values differ by a large margin, so no rounding can turn them into 2.8 A.

Final Answer:
The current in the circuit is close to 2.8 A. \[ \boxed{2.8\ \text{A}} \]