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`.

Options

A. `10Omega`
B. `20Omega`
C. `30Omega`
D. `40Omega`

Answer

B

Explanation

`R_(B)=(100^(2))/(500)=20 Omega`
Since power consumed in bulb`=500W`
`p.d.` across bulb=`100V` p.d across `R = 100V`
`100 = R/(R+R_(B)) xx 200` ltbr. `R = R_(B) = 20 Omega`.