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(D) Let the potential at the left junction be $V_C$ and the right junction be $V_D$. Let $V_C = V$ and $V_D = 0$.The potential at point $A$ is determi

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From $B$ to $A$

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(D) Let the potential at the left junction be $V_C$ and the right junction be $V_D$. Let $V_C = V$ and $V_D = 0$.The potential at point $A$ is determined by the voltage divider rule in the upper branch: $V_A = V 	imes \frac{4}{4+4} = \frac{V}{2}$.The potential at point $B$ is determined by the voltage divider rule in the lower branch: $V_B = V 	imes \frac{3}{1+3} = \frac{3V}{4}$.Comparing the potentials,$V_B = 0.75V$ and $V_A = 0.5V$. Since $V_B > V_A$,the potential difference is $V_B - V_A = 0.25V > 0$.When a conducting wire is connected between $A$ and $B$,current flows from a higher potential to a lower potential. Therefore,the current flows from $B$ to $A$.

As shown in the circuit,if a conducting wire is connected between points $A$ and $B$,then the electric current in this wire will flow:

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