Question

An electric bulb is designed to consume 100 W at 200 V. It is connected to a 300 V and $\omega =100r/s$ A.C. source. For what value of inductance (in henry) of the coil bulb, will we get correct voltage?

Answer 1

Since $P=Vi=\frac{V^2}{R}\Rightarrow i=\frac{100}{200}=\frac{1}{2}\text{current for proper voltage}$

& Resistance of bulb
$R=\frac{V^2}{P}=\frac{{200}^2}{100}=400$

For A.C $\Rightarrow V=iZ\Rightarrow V=i\sqrt{{X_L}^2+R^2}$

$300=\frac{1}{2}\sqrt{{\omega }^2{L}^2+R^2}\Rightarrow 600=\sqrt{{400}^2+{100}^2{L}^2}$

${600}^2={400}^2+{100}^2{L}^2\Rightarrow 36=16+L^2$

L=4.47