In the circuit shown,the cells A and B have n | vedclass

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(B) When the galvanometer shows no deflection,it means no current flows through the branch containing the galvanometer and cell $B$.This implies that

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$4$

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(B) When the galvanometer shows no deflection,it means no current flows through the branch containing the galvanometer and cell $B$.This implies that the potential difference across the resistor $R$ must be equal to the electromotive force $V_B$ of the cell $B$.The circuit formed by $V_A$,$R_1$,and $R$ acts as a potential divider.The voltage drop across $R$ is given by the formula:$V_R = \left( \frac{R}{R + R_1} ight) V_A$Substituting the given values:$V_R = \left( \frac{200}{200 + 600} ight) 	imes 16 \ V$$V_R = \left( \frac{200}{800} ight) 	imes 16 \ V$$V_R = \frac{1}{4} 	imes 16 \ V = 4 \ V$Since the galvanometer shows no deflection,$V_B = V_R = 4 \ V$.

In the circuit shown,the cells $A$ and $B$ have negligible internal resistances. For $V_A = 16 \ V$,$R_1 = 600 \ \Omega$,and $R = 200 \ \Omega$,the galvanometer shows no deflection. The value of $V_B$ is $...V$.

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