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(B) Case $1$: Switch $S$ is open.The circuit consists of two parallel branches. The upper branch has resistors $R$ and $2R$ in series,and the lower br

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$9$

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(B) Case $1$: Switch $S$ is open.The circuit consists of two parallel branches. The upper branch has resistors $R$ and $2R$ in series,and the lower branch has resistors $2R$ and $R$ in series.Equivalent resistance of upper branch = $R + 2R = 3R$.Equivalent resistance of lower branch = $2R + R = 3R$.Since these two branches are in parallel,$R_{eq, open} = \frac{3R 	imes 3R}{3R + 3R} = \frac{9R^2}{6R} = \frac{3R}{2}$.Case $2$: Switch $S$ is closed.The circuit can be viewed as two parallel combinations connected in series. The left side has $R$ and $2R$ in parallel,and the right side has $2R$ and $R$ in parallel.Equivalent resistance of left part = $\frac{R 	imes 2R}{R + 2R} = \frac{2R}{3}$.Equivalent resistance of right part = $\frac{2R 	imes R}{2R + R} = \frac{2R}{3}$.Since these two parts are in series,$R_{eq, closed} = \frac{2R}{3} + \frac{2R}{3} = \frac{4R}{3}$.Ratio = $\frac{R_{eq, open}}{R_{eq, closed}} = \frac{3R/2}{4R/3} = \frac{3R}{2} 	imes \frac{3}{4R} = \frac{9}{8}$.Given the ratio is $x : 8$,therefore $x = 9$.

The ratio of the equivalent resistance of the network (shown in figure) between the points $a$ and $b$ when the switch is open and when the switch is closed is $x : 8$. The value of $x$ is ..... .

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