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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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