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(A) battery is getting charged when the current enters through its positive terminal. Let the potential at point $D$ be $0\,V$. Since the $12\,V$ batt

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$8\,V$ battery

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(A) battery is getting charged when the current enters through its positive terminal. Let the potential at point $D$ be $0\,V$. Since the $12\,V$ battery is connected between $A$ and $C$,let $V_A = 12\,V$ and $V_C = 0\,V$. The circuit is a Wheatstone bridge. The resistances are $R_{AB} = 4\,\Omega$,$R_{BC} = 4\,\Omega$,$R_{AD} = 4\,\Omega$,and $R_{DC} = 4\,\Omega$. Since $R_{AB}/R_{AD} = R_{BC}/R_{DC} = 4/4 = 1$,the bridge is balanced. The potential at $B$ is the same as the potential at $D$ if there were no battery. With the $8\,V$ battery connected between $B$ and $D$ (positive terminal at $B$),the potential $V_B = 8\,V$ and $V_D = 0\,V$. For the $12\,V$ battery,current flows from $A$ to $C$ through the external circuit,meaning it is discharging. For the $8\,V$ battery,the potential at $B$ is $8\,V$ and the potential at $D$ is $0\,V$. The current from the $12\,V$ battery flows through the branches $AB$ and $AD$. Calculating the potentials: $V_B = V_A - I_{AB}R_{AB} = 12 - I_{AB}(4)$. In a balanced bridge,the current flows from $A$ to $B$ and $A$ to $D$. The potential at $B$ due to the $12\,V$ source is $V_B = 12 	imes (4/(4+4)) = 6\,V$. Since the actual potential at $B$ is $8\,V$ (due to the battery),current flows from $B$ to $A$ and $B$ to $C$. Thus,current enters the $8\,V$ battery through its positive terminal,so the $8\,V$ battery is getting charged.

The given figure shows a network of resistances and two batteries. Which of the two batteries is getting charged?

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