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(C) Case $1$: When the key $K$ is open.The circuit consists of two parallel branches. The left branch has resistors $R$ and $2R$ in series,and the rig

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$\frac{8}{9}$

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(C) Case $1$: When the key $K$ is open.The circuit consists of two parallel branches. The left branch has resistors $R$ and $2R$ in series,and the right branch has resistors $2R$ and $R$ in series.Equivalent resistance of the left branch: $R_L = R + 2R = 3R$.Equivalent resistance of the right branch: $R_R = 2R + R = 3R$.These two branches are in parallel,so the total equivalent resistance is $R_{eq1} = \frac{3R 	imes 3R}{3R + 3R} = \frac{9R^2}{6R} = \frac{3R}{2}$.The current measured by the ammeter is $I_1 = \frac{V}{R_{eq1}} = \frac{V}{3R/2} = \frac{2V}{3R}$.Case $2$: When the key $K$ is closed.The key $K$ connects the middle points of the two branches. This effectively creates two parallel combinations in series.The left side is a parallel combination of $R$ and $2R$: $R_{p1} = \frac{R 	imes 2R}{R + 2R} = \frac{2R}{3}$.The right side is a parallel combination of $2R$ and $R$: $R_{p2} = \frac{2R 	imes R}{2R + R} = \frac{2R}{3}$.These two combinations are in series,so the total equivalent resistance is $R_{eq2} = R_{p1} + R_{p2} = \frac{2R}{3} + \frac{2R}{3} = \frac{4R}{3}$.The current measured by the ammeter is $I_2 = \frac{V}{R_{eq2}} = \frac{V}{4R/3} = \frac{3V}{4R}$.Ratio of currents: $\frac{I_1}{I_2} = \frac{2V/3R}{3V/4R} = \frac{2}{3} 	imes \frac{4}{3} = \frac{8}{9}$.

Find the ratio of currents as measured by the ammeter in two cases: when the key $K$ is open and when the key $K$ is closed.

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