A room heater is rated `500 W, 220V`. (a) Find the resistance of its coil. (b) If the supply voltage drops to `200 V`, what will be the power consumed? ( c) If an electric bulb rated `100 W, 220 V` is connected in series with this heater, what will be the power consumed by the heater and by the bulb when the supply is at `220 V`?

Explanation

(a) The power consumed by a coil of resistance `R` when connected across a supply `V` is
`P = V^2/R`
The resistance of the heater coil is, therefore,
`R= (220 V)^2 /(500 W) = 96.8 Omega`. (b) If the supply voltage drops to `200 V`, the power consumed will be
`P = V^2/R = (200 V)^2/ (96.8 Omega) = 413 W`.
(c) The resistance of the `100 W, 220 V` bulb is
`R = (220 V)^2/(100 W) = 484 Omega.
If this is connected in series with the heater of `96.8 Omega`, the current `i` will be
`i = (220V)/(484 Omega + 96.8 Omega) = 0.379 A`.
Thus, the power consumed by the heater
`= i^2 xx 96.8 Omega = 0.144 xx 96.8 W = 13.9 W`
and that by the bulb
`= i^2 xx 484 Omega = 69.7 W`.