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(D) The power consumed by a resistor (bulb) is given by the formula $P = \frac{V^2}{R}$,where $R$ is the resistance of the bulb.Initially,the bulb ope

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

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(D) The power consumed by a resistor (bulb) is given by the formula $P = \frac{V^2}{R}$,where $R$ is the resistance of the bulb.Initially,the bulb operates at voltage $V_0$ and power $P_0$. Therefore,the resistance $R$ is given by:$R = \frac{V_0^2}{P_0} \quad \dots(1)$When the bulb is connected to a new voltage $V$,it consumes a new power $P$. Since the resistance $R$ of the bulb remains constant,we have:$P = \frac{V^2}{R}$Substituting the value of $R$ from equation $(1)$ into this expression:$P = \frac{V^2}{(V_0^2 / P_0)}$$P = \left( \frac{V^2}{V_0^2} \right) P_0$$P = \left( \frac{V}{V_0} \right)^2 P_0$

$A$ light bulb operates at a voltage $V_0$ and consumes power $P_0$. If the voltage is changed to $V$,what is the new power $P$ consumed by the bulb?

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