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`.

Explanation

From the rating of the bulb, the resistance of the bulb is `R=(V_("rms"))/(P)=100Omega`
For the bulb to be operated as its rated value of the rms current through it should be 1A.
`"Also" I_(rms)=(V_(rms))/(Z)`
`therefore 1=(200)/(sqrt(100^(2)+(2pi50L)^(2))), L=(sqrt3)/(pi)H`