In the following Wheatstone bridge,P/Q = R/S. | vedclass

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(D) The condition for a balanced Wheatstone bridge is $P/Q = R/S$. When the bridge is balanced,the potential difference between the two points connect

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In either side

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(D) The condition for a balanced Wheatstone bridge is $P/Q = R/S$. When the bridge is balanced,the potential difference between the two points connected by the galvanometer is zero. Closing the key $K$ connects the two points,but since the potentials are already equal,no additional current flows through the galvanometer branch. Therefore,the galvanometer will maintain its existing deflection (if any) or show no change in its state. Given the options,the deflection remains as it was,which could be in either direction depending on the initial state of the circuit.

In the following Wheatstone bridge,$P/Q = R/S$. If key $K$ is closed,then the galvanometer will show deflection:

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