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(A) Let the resistance of each identical bulb be $R$. Case $1$: When switch $S$ is closed,bulbs $B$ and $C$ are in parallel,and this combination is in

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$\frac{9 P}{4}$

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(A) Let the resistance of each identical bulb be $R$. Case $1$: When switch $S$ is closed,bulbs $B$ and $C$ are in parallel,and this combination is in series with bulb $A$. The equivalent resistance is $R_{eq} = R + \frac{R \cdot R}{R + R} = R + \frac{R}{2} = \frac{3R}{2}$. The total current from the battery is $I = \frac{E}{R_{eq}} = \frac{2E}{3R}$. The voltage across the parallel combination of $B$ and $C$ is $V_{BC} = I \cdot \frac{R}{2} = \left(\frac{2E}{3R}\right) \cdot \frac{R}{2} = \frac{E}{3}$. The power consumed by bulb $B$ is $P = \frac{V_{BC}^2}{R} = \frac{(E/3)^2}{R} = \frac{E^2}{9R}$. Case $2$: When switch $S$ is opened,bulb $C$ is disconnected. Bulbs $A$ and $B$ are in series. The equivalent resistance is $R'_{eq} = R + R = 2R$. The current in the circuit is $I' = \frac{E}{2R}$. The power consumed by bulb $B$ is $P' = (I')^2 R = \left(\frac{E}{2R}\right)^2 R = \frac{E^2}{4R^2} \cdot R = \frac{E^2}{4R}$. Comparing $P$ and $P'$,we have $\frac{P'}{P} = \frac{E^2/4R}{E^2/9R} = \frac{9}{4}$. Therefore,$P' = \frac{9P}{4}$.

Three identical bulbs are connected as shown in the figure. When switch $S$ is closed,the power consumed in bulb $B$ is $P$. What will be the power consumed by the same bulb when switch $S$ is opened?

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