Three electric bulbs of 200,W, 200,W and 400, | vedclass

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(C) Let the two bulbs of $200\,W$ be $A$ and $B$,and the bulb of $400\,W$ be $C$.Bulbs $A$ and $B$ are connected in parallel. The effective power $P^{

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$200$

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(C) Let the two bulbs of $200\,W$ be $A$ and $B$,and the bulb of $400\,W$ be $C$.Bulbs $A$ and $B$ are connected in parallel. The effective power $P^{\prime}$ of this parallel combination is given by the sum of their individual powers:$P^{\prime} = P_{A} + P_{B} = 200\,W + 200\,W = 400\,W$.Now,this combination $P^{\prime}$ is in series with bulb $C$ $(400\,W)$. The resultant power $P_{R}$ of two components in series is given by the formula $\frac{1}{P_{R}} = \frac{1}{P^{\prime}} + \frac{1}{P_{C}}$,which simplifies to $P_{R} = \frac{P^{\prime} 	imes P_{C}}{P^{\prime} + P_{C}}$.Substituting the values:$P_{R} = \frac{400\,W 	imes 400\,W}{400\,W + 400\,W} = \frac{160000\,W^2}{800\,W} = 200\,W$.

Three electric bulbs of $200\,W, 200\,W$ and $400\,W$ are shown in the figure. The resultant power of the combination is ................ $W$.

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