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# An electric bulb is rated 220 volt and 100 watt, power consumed by it when operated on 110 volt isa) 50Wb) 85Wc) 90Wd) 25W  Verified
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Hint: The power dissipated across the bulb is given when operated at a particular potential difference. But when the bulb is operated at a different potential difference the power dissipated across it will change. The property of the bulb that does not change is its resistance. Hence from the given data lets calculate the resistance from the power expression and further use this resistance to calculate the power dissipated across it.

$\text{P=VI}....\text{(1)}$ where V is the potential generated across the bulb and I is the current flowing through the bulb. By Ohms law we can write the current through a resistor as, $I=\dfrac{V}{R}$ where V is the potential generated across the bulb and r is the resistance of the bulb. After substituting for current in equation 1 we get,
$\text{P=V}\dfrac{\text{V}}{\text{R}}=\dfrac{{{\text{V}}^{2}}}{\text{R}}\text{J/sec }....\text{(2)}$
\begin{align} & \text{P}=\dfrac{{{\text{V}}^{2}}}{\text{R}} \\ & 100=\dfrac{{{220}^{2}}}{\text{R}} \\ & R=\dfrac{{{220}^{2}}}{100}=\dfrac{48400}{100}=484\Omega \\ \end{align}
\begin{align} & \text{P}=\dfrac{{{\text{V}}^{2}}}{\text{R}} \\ & P=\dfrac{{{110}^{2}}}{484}=\dfrac{12100}{484}=25\text{W} \\ \end{align}