An electric bulb rated for `500 W` at `100 V` is used in a circuit having a `200 V` supply. The reistance `R` that must be put in series with bulb, so that the bulb delivers `500 `W is ……….`Omega`.

Options

A. `10Omega`
B. `15Omega`
C. `20Omega`
D. `25Omega`

Answer

C

Explanation

Resistance of a bulb = `("(Rated voltage)"^(2))/(("Rated power"))`
`therefore R_(B)=((100V)^(2))/((500W))=20Omega`
When a resistance R to be put in series with the bulb, it will deliver 500 W,only if voltage across it remains 100 V.
`therefore` The voltage across resistance R is
`V_(R)=V-V_(B)=200V-100V=100V`
In series, current flowing through a resistance R and `R_(B)` remains the same.
`therefore (V_(R))/(R)=(V_(B))/(R_(B))`
`R=(V_(R)R_(B))/(V_(B))=(100Vxx20Omega)/(100V)=20Omega`