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(D) The resistance $R$ of the bulb is constant and is given by $R = \frac{V_R^2}{P_R}$,where $V_R = 220\, V$ and $P_R = 100\, W$.The power consumed $P

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$25$

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(D) The resistance $R$ of the bulb is constant and is given by $R = \frac{V_R^2}{P_R}$,where $V_R = 220\, V$ and $P_R = 100\, W$.The power consumed $P_{consumed}$ when connected to a new voltage $V_A = 110\, V$ is given by $P_{consumed} = \frac{V_A^2}{R}$.Substituting $R = \frac{V_R^2}{P_R}$ into the equation,we get $P_{consumed} = \left( \frac{V_A}{V_R} ight)^2 	imes P_R$.Substituting the values: $P_{consumed} = \left( \frac{110}{220} ight)^2 	imes 100 = \left( \frac{1}{2} ight)^2 	imes 100 = \frac{1}{4} 	imes 100 = 25\, W$.

$A$ bulb rated $100\, W, 220\, V$ is connected to a $110\, V$ supply. What is the power consumed in $W$?

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