An electric bulb is rated `220 V and 100 W`. When it is operated on `110 V`, the power consumed will be :

Options

A. 100 w
B. 75 W
C. 50 W
D. 25W

Answer

D

Explanation

As we know that the energy consumed by bulb
`rarr P=(v^(2))/(R )`
`rarr R=(V^(2))/(P)`
`R=(220)^(2)/(100)=484 omega`
The resistance of the bulb remains constant if the supply voltage is reduced to 110 v if the bulb is is given by the expression for power
`P=(v^(2))/(R )=(110)^(2)/(484)=12100/484=25 W`