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

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(C) In the circuit,the left loop consists of the battery $V_{A}$,resistor $R_{1}$,and resistor $R$ in series.Since the galvanometer $(G)$ shows no def

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$2$

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(C) In the circuit,the left loop consists of the battery $V_{A}$,resistor $R_{1}$,and resistor $R$ in series.Since the galvanometer $(G)$ shows no deflection,no current flows through the branch containing the galvanometer and battery $V_{B}$.Therefore,the current $I$ flowing through the left loop is determined by the series combination of $R_{1}$ and $R$:$I = \frac{V_{A}}{R_{1} + R}$Substituting the given values:$I = \frac{12}{500 + 100} = \frac{12}{600} = 0.02 \; A$Now,the potential difference across the resistor $R$ is the voltage at the junction point relative to the common negative terminal:$V = I 	imes R$$V = 0.02 	imes 100 = 2 \; V$Since there is no current through the galvanometer,the potential difference across it must be zero. Thus,the potential of battery $V_{B}$ must be equal to the potential across resistor $R$.$V_{B} = V = 2 \; V$.

In the circuit shown,the cells $A$ and $B$ have negligible resistance. For $V_{A} = 12 \; V$,$R_{1} = 500 \; \Omega$ and $R = 100 \; \Omega$,the galvanometer $(G)$ shows no deflection. The value of $V_{B}$ is .... $V$.

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