A servo voltage stabiliser restricts the voltage output to `220 V +- 1%`. If an electric bulb rated at `220 V, 100 W` is connected to it, what will be the minimum and maximum power consumed by it?

Explanation

Let stabilizer give voltage `V` , then the power produced by the bulb is
`P = ((V^(2))/(220))100 or (dP)/(P) = 2(dV)/(V)`
If `( dV)/( V) = +- 1% , then (dP)/(P) = +- 2%`
Hence, the maximum power produced is
`P_(max) = 100% + 2% = 102 W`
`Minimum power produced is `P_(min) = 100% - 2% = 98 W`