A non-zero current passes through the galvano | vedclass

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(B, D) Let the battery voltage be $V = 10 \ V$. When the key $K$ is open, the circuit is a series-parallel combination. The equivalent resistance is $

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The galvanometer resistance is $150 \ \Omega$.

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(B, D) Let the battery voltage be $V = 10 \ V$. When the key $K$ is open, the circuit is a series-parallel combination. The equivalent resistance is $R_{eq} = (200 + 300) \parallel (100 + G) = \frac{500(100+G)}{600+G}$. The current through the galvanometer branch is $I_{G, open} = \frac{V}{100+G}$.When the key $K$ is closed, the circuit becomes a Wheatstone bridge. The condition that the current through the galvanometer does not change means the bridge is balanced, i.e.,$\frac{200}{100} = \frac{300}{G}$.Solving for $G$: $G = \frac{300 	imes 100}{200} = 150 \ \Omega$. Thus, option $D$ is correct.With $G = 150 \ \Omega$, the current through the galvanometer when the key is closed is $I_{G, closed} = \frac{V}{R_{eq}'}$. Since the bridge is balanced, the potential at the two ends of the galvanometer is the same, and the current is determined by the potential divider: $I_G = \frac{10}{100+150} = \frac{10}{250} = 0.04 \ A = 40 \ mA$. Thus, option $B$ is correct.Since the bridge is balanced, the ratio of resistances in the arms is equal, so the currents in the upper and lower branches are independent of the galvanometer connection, but the current through $200 \ \Omega$ is not necessarily equal to the current through $300 \ \Omega$ unless the resistances are equal. Here $200 \ \Omega 
eq 300 \ \Omega$, so option $C$ is incorrect.

$A$ non-zero current passes through the galvanometer $G$ shown in the circuit when the key $K$ is closed and its value does not change when the key is opened. Then, which of the following statement$(s)$ is/are true?

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