Two bulbs,one of 200 ,W and the other of 100 | vedclass

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(D) The resistance $R$ of a bulb is related to its rated power $P_{	ext{rated}}$ and rated voltage $V_{	ext{rated}}$ by the formula $R = \frac{V_{

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the power dissipation in the $100 \,W$ bulb is more than that in the $200 \,W$ bulb

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(D) The resistance $R$ of a bulb is related to its rated power $P_{	ext{rated}}$ and rated voltage $V_{	ext{rated}}$ by the formula $R = \frac{V_{	ext{rated}}^2}{P_{	ext{rated}}}$.Since the rated voltage is the same for both bulbs,$R \propto \frac{1}{P_{	ext{rated}}}$.Thus,the $100 \,W$ bulb has a higher resistance than the $200 \,W$ bulb.When connected in series,the same current $I$ flows through both bulbs.The power dissipated in each bulb is given by $P = I^2 R$.Since $I$ is constant,$P \propto R$.Because the $100 \,W$ bulb has a higher resistance,it will dissipate more power than the $200 \,W$ bulb.Therefore,option $(d)$ is correct.

Two bulbs,one of $200 \,W$ and the other of $100 \,W$ are connected in series with a $100 \,V$ battery which has no internal resistance. Then,

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