A `100 W` bulb designed to operate on `100 V` is to be connected across a `500 V` source . Find the resistance to be put in series so that bulb consumes `100 W` only.

Options

A. `100 Omega`
B. `500 Omega`
C. `400 Omega`
D. `300 Omega`

Answer

C

Explanation

Resistance of bulb is `R_(b) = (V^(2))/( P) = ((100)^(2))/(100) = 100 Omega`
If bulb consumes `100 W` , then potential difference across the bulb should be `100 V`. So `I = 100 // 100 = 1 A`. And potential difference across `R is 500 - 100 = 400 V`. So ltbr `R = ( 400)/(I) = ( 400)/(1) = 400 Omega`