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`.

Answer

`30 Omega`

Explanation

resistance of bulb `R_(0) = V^(2) //l^(2)`
`= 100 xx 100//500= 20 Omega`
Current in bulb `l = V//R_(0) = 100//20 = 5A`
For the same power dissipated in bulb the current in the circuit must be `5A` when both the safe resistance of the circuit of the circuit `(R)` is given by
`R' = (V)/(I) = (250)/(3) = 50 Omega`
resistance required to be put in series will be
`R = R' - 20 = 50 - 20= 30 Omega`