An electric bulb of 100 W - 300V is connected with on A.C. supply of 500 V and 150 /π Hz. The required inductance to save the electric bulb is
Answer & explanation
Correct answer: option 2
$\text{Resistance of the bulb is } R = \frac{V^2}{P} = \frac{300^2}{100} = 900\Omega$
$ I = \frac{P}{V} = \frac{1}{3} A$
$\text{Inductive Resistance }X_L = 2\pi fL = 2\pi \frac{150}{\pi}L =300L$
$ I = \frac{1}{3} = \frac{V}{\sqrt{R^2 + X_L^2}} = \frac{500}{\sqrt{900^2+ {300L}^2}}$
$ 900^2 + 300^2L^2 = 1500^2 $
$ L^2 = \frac{600\times 2400}{300\times 300} = 16$
$ \Rightarrow L = 4H$