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. `100Omega`
B. `50Omega`
C. `20Omega`
D. `10Omega`

Answer

C

Explanation

In order that the bulb draws 500 W, there should be 100 V applied across it. The remaining 100 V should be dropped in the sereis resistance R. Now, the current in the bulbe is given by
P=V.I
`therefore` `I=(P)/(V)`= `(500)/(100)`=5A
=5A The same current flows through R.
Hence, R =`(V)/(1)`=`(100)/(5)`=`20Omega`