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

Options

A. `P = ((V_0)/(V))P_(0)`
B. `P = ((V)/(V_0))P_(0)`
C. `P = ((V)/(V_0))^(2)P_(0)`
D. `P = ((V_0)/(V))^(2)P_(0)`

Answer

C

Explanation

`R_(B) = (V_0^2)/(P_0)`
`P = (V^2)/(R_B) = ((V)/(V_0))^(2)P_(0)`.