In the network shown,the cell of emf E has in | vedclass

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(B) In the given circuit diagram,the galvanometer shows zero deflection,which means the Wheatstone bridge is balanced.For a balanced Wheatstone bridge

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The galvanometer will show zero deflection.

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(B) In the given circuit diagram,the galvanometer shows zero deflection,which means the Wheatstone bridge is balanced.For a balanced Wheatstone bridge,the condition is $\frac{R_1}{R_3} = \frac{R_2}{R_4}$.This implies that the potentials at points $A$ and $B$ are equal,i.e.,$V_A = V_B$.The potential difference between points $A$ and $B$ is zero,so no current flows through the galvanometer.When the cell is replaced by a new cell of emf $2E$ and internal resistance $3r$,the total current in the main circuit changes,but the ratio of potentials at $A$ and $B$ remains the same because the bridge balance condition $\frac{R_1}{R_3} = \frac{R_2}{R_4}$ depends only on the resistors $R_1, R_2, R_3,$ and $R_4$,not on the source emf or internal resistance.Therefore,the condition $V_A = V_B$ still holds,and the galvanometer will continue to show zero deflection.

In the network shown,the cell of emf $E$ has internal resistance $r$ and the galvanometer shows zero deflection. If the cell is replaced by a new cell of emf $2E$ and internal resistance $3r$,keeping everything else identical,then:

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