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
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(b) $P \propto {V^2}$ $ \Rightarrow $ $\frac{P}{{{P_0}}} = {\left( {\frac{V}{{{V_0}}}} \right)^2} \Rightarrow P = {\left( {\frac{V}{{{V_0}}}} \right)^2}{P_0}$
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