1.

An electric bulb is designed to draw `P_(0)` power at `V_(0)` voltage. If the voltage is `V`, it drawas power. ThenA. `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» Correct Answer - C
`R_(B) = (V_0^2)/(P_0)`
`P = (V^2)/(R_B) = ((V)/(V_0))^(2)P_(0)`.


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