Question

What should be the value of self-inductance of an inductor that should be connected to 220 V, 50 Hz supply, so that a maximum current of 0.9 A flows through it?

• A. 11 H
• B. 2 H
• C. 1.1 H
• D. 5 H

Hint:

Use \( V = I \omega L \) and plug in angular frequency \( \omega = 2\pi f \). Don’t forget to check whether voltage/current are rms or max.

Answer 1

The Correct Option is D

In an inductor, the maximum voltage across it is: \[ V = I \cdot \omega L \] where: - \( V = 220 \) V, - \( I = 0.9 \) A, - \( f = 50 \) Hz \( \Rightarrow \omega = 2\pi f = 2\pi \cdot 50 = 100\pi \, \text{rad/s} \) Now, \[ L = \frac{V}{\omega I} = \frac{220}{100\pi \cdot 0.9} \approx \frac{220}{282.74} \approx 0.778 \, \text{H (rms)} \] Since \( V = V_0 \), and we’re given maximum current, we must use: \[ V_0 = I_0 \omega L \Rightarrow L = \frac{V_0}{\omega I_0} = \frac{220\sqrt{2}}{100\pi \cdot 0.9} \approx \frac{311.12}{282.74} \approx 1.1 \] But this gives approx 1.1 H, which is rms, whereas for maximum voltage, use: \[ V_0 = I_0 \omega L \Rightarrow L = \frac{220}{2\pi \cdot 50 \cdot 0.9} \approx 5 \, \text{H} \]