Two bulbs of 100, W and 200, W working at 220 | vedclass

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(A) The power rating of a bulb is given by $P = \frac{V^2}{R}$,where $V$ is the rated voltage and $R$ is the resistance of the bulb.For the first bulb

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$65$

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(A) The power rating of a bulb is given by $P = \frac{V^2}{R}$,where $V$ is the rated voltage and $R$ is the resistance of the bulb.For the first bulb,$R_1 = \frac{V^2}{P_1} = \frac{220^2}{100}$.For the second bulb,$R_2 = \frac{V^2}{P_2} = \frac{220^2}{200}$.When connected in series,the total resistance is $R_{eq} = R_1 + R_2 = \frac{V^2}{P_1} + \frac{V^2}{P_2} = V^2 \left( \frac{P_1 + P_2}{P_1 P_2} ight)$.The total power consumed in the series combination is $P_{total} = \frac{V^2}{R_{eq}} = \frac{V^2}{V^2 \left( \frac{P_1 + P_2}{P_1 P_2} ight)} = \frac{P_1 P_2}{P_1 + P_2}$.Substituting the given values: $P_{total} = \frac{100 	imes 200}{100 + 200} = \frac{20000}{300} = \frac{200}{3} \approx 66.67\, W$.Rounding to the nearest given option,the total power consumed is approximately $65\, W$.

Two bulbs of $100\, W$ and $200\, W$ working at $220\, V$ are joined in series with a $220\, V$ supply. The total power consumed will be approximately ........... $W$.

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