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
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
- A
$P = {\left( {\frac{{{V_0}}}{V}} \right)^2}{P_0}$
- B
$P = {\left( {\frac{V}{{{V_0}}}} \right)^2}{P_0}$
- C
$P = \left( {\frac{V}{{{V_0}}}} \right)\,{P_0}$
- D
$P = \left( {\frac{{{V_0}}}{V}} \right)\,{P_0}$
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