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(D) The resistance $R$ of the bulb is constant and is given by $R = \frac{V^2}{P}$.For the given bulb,$R = \frac{220^2}{1000} = \frac{48400}{1000} = 4

Vedclass · Accepted Answer

$250$

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(D) The resistance $R$ of the bulb is constant and is given by $R = \frac{V^2}{P}$.For the given bulb,$R = \frac{220^2}{1000} = \frac{48400}{1000} = 48.4\, \Omega$.When connected to a $110\, V$ supply,the power consumed $P'$ is given by $P' = \frac{V'^2}{R}$.$P' = \frac{110^2}{48.4} = \frac{12100}{48.4} = 250\, W$.Alternatively,using the ratio formula: $\frac{P_1}{P_2} = \left( \frac{V_1}{V_2} \right)^2$.$\frac{1000}{P_2} = \left( \frac{220}{110} \right)^2 = 2^2 = 4$.$P_2 = \frac{1000}{4} = 250\, W$.

$A$ $220\, V$,$1000\, W$ bulb is connected across a $110\, V$ mains supply. The power consumed will be ............ $W$.

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