Question

A bulb is rated at 100V, 100W. It can be treated as a resistor. Find out the inductance of an inductor (called choke coil) that should be connected in series with the bulb at its rated power with the help of an ac source of 200V and 50Hz.

Answer 1

Correct Answer - `(sqrt(3))/(pi)H`
From the rating of the bulb, the resistance of the blub is
`R=(V^2)/(P) = 100 Omega`

For the blub to be operated at its rated value the rms current through it should be 1A.
Also, `I_(rms)=(V_(rms))/(Z)` ltbr gt`:. 1=(200)/(sqrt(100^(2)+(2 pi 50 L)^(2)) implies L= (sqrt(3))/(pi) H`.