An electric bulb is designed to draw power `P_(0)` at voltage `V_(0)`. If the voltage is V, it draws a power P. Then

Options

A. `P = ((V_(0))/(V))^(2) P_(0)`
B. `P = ((V)/(V_(0)))^(2) P_(0)`
C. `P = ((V)/(V_(0))) P_(0)`
D. `P = ((V_(0))/(V))P_(0)`

Answer

B

Explanation

`P_(0) = (V_(0)^(2))/(R )`
`P = (V^(2))/(R )`, putting the value of R we get
`(V^(2))/(V_(0)^(2)//P_(0)) = ((V)/(V_(0)))^(2) P_(0)`