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

Options

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

Answer

A

Explanation

`26 Omega`
Resistance of bulb, `R_(B) = (V^(2))/(P) = ((100)^(2))/(500)`
`therefore R_(B) = 20 Omega`
Current is same in series, therefor , R `prop` V
`therefore (R)/(R_(B)) = (V_(R))/(V_(B))`
`therefore R = R_(B) xx (V_(R))/(V_(B))`
`= 20 xx (130)/(100)`
`therefore R = 26 Omega`