A filament bulb `(500 W, 100 V)` is to be used in a `230 V` main supply. When a resistance `R` is connected in series, it works perfectly and the bulb consumers `500 W`. The value of `R` is

Options

A. `230Omega`
B. `46Omega`
C. `26Omega`
D. `13Omega`

Answer

C

Explanation

Resistance of bulb, `R_(B)=(V^(2))/(P)=((100)^(2))/(500)=20Omega`
Power of the bulb in the circuit, `P=VI`
`I=(P)/(V_(B))=(500)/(100)=5A`
`V_(R)=IRimplies(230-100)=5xxR`
`therefore R=26Omega`