An electric bulb rated for `500 W` at `100 V` is used in a circuit having a `200 V` supply. The reistance `R` that must be put in series with bulb, so that the bulb delivers `500 `W is ……….`Omega`.

Explanation

Resistance of the given bulb
`R_b=V^2/P=(100)^2/500=20 Omega`
To get 100V out of 200V across the bulb,
`R=R_b=20 Omega`