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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