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 $R

Vedclass · Accepted Answer

The galvanometer resistance is $150 \ \Omega$.

Vedclass · Answer

(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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