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?

Answer

A, B

Explanation

Out put voltage `
`= 220 V +- 1%`
`1% of 220 V = 2.2 V`
` The resistance of bulb `
`R = (V^2)/(P) = ((220)^2/(100)) `
`= (48400)/(100) = 484`
(a) For minimum power consumed V'
`=((220)) - 1% = ((220-2.2)V = 217.8`
` I' = (V')/(r ) = (217.8)/(484) =0.45 A`
power consumed
`=ixx V' `
`= 0.45xx 217.8' = 98.0 W`
` (b) For maximum power consumed V'' `
`=220 + 1% =220 + 2.2`
`= 222.2 V`
` `i''= (V'')/(r ) = (222.2)/(484) = 0.459 A`
` Power consumed `
`i''xxV''`
`= 0.459xx222.2 = 102W`