Question

Find the necessary inductance. if 110 V, 10 W rating bulb is to be used with 220 V A.C source having frequency 50 Hz.

• A. L=8.90 H
• B. L=6.75 H
• C. L=7.25 H
• D. L=6.5 H

Answer 1

The correct option is B L=6.75 H

Given, $V=110v$ ,$P=10W$ ,$V_0=220w$ ,$f=50Hz$

Current through series inductor,

Current through bulb=$\frac{P}{V}=\frac{10W}{110V}=0.09A$

Voltage across inductor,$V_{ind}=\sqrt{V_0^2-V^2}$=$\sqrt{{220}^2-{110}^2}=191V$

Reactance of inductor,$R=\frac{V_{ind}}{I}=\frac{191}{0.09}=2122.22$

Also,$R=2\pi fL$ or $L$=$\frac{R}{2\pi f}$=$\frac{2122.22}{2\pi 50}$=$6.75H$