An electric bulb rated for 500 watt at 100 volt is used in a circuit having a 200 volt supply. The resistance R that must be put in series with the bulb, so that the bulb delivers 500 watt is ……….. Ohm.

Explanation

Resistance of the bulb

`R_(0)=(V_(0)^(2))/(P_(0))=(100xx100)/(500)=20Omega`
To deliver 500 W long the given bulb 100 V P.d. must drop at the bulb. As the given source has emf 200 V, remaining 100 V P.D. must drop at the resistance R. As the potential differences are equal across `R and R_(0)` and they are in series, `R=R_(0)`
`therefore R=20Omega`