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(B) Let the potential at point $C$ be $V_C$ and at point $D$ be $V_D$. Let $V_C - V_D = V$.The circuit consists of two parallel branches: $CAD$ and $C

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flow from $B$ to $A$

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(B) Let the potential at point $C$ be $V_C$ and at point $D$ be $V_D$. Let $V_C - V_D = V$.The circuit consists of two parallel branches: $CAD$ and $CBD$.In branch $CAD$,the total resistance is $4 \, \Omega + 4 \, \Omega = 8 \, \Omega$. The current is $I = \frac{V}{8} \, A$.The potential at $A$ relative to $C$ is $V_C - V_A = I 	imes 4 \, \Omega = \frac{V}{8} 	imes 4 = \frac{V}{2}$. Thus,$V_A = V_C - \frac{V}{2}$.In branch $CBD$,the total resistance is $1 \, \Omega + 3 \, \Omega = 4 \, \Omega$. The current is $I' = \frac{V}{4} \, A$.The potential at $B$ relative to $C$ is $V_C - V_B = I' 	imes 1 \, \Omega = \frac{V}{4} 	imes 1 = \frac{V}{4}$. Thus,$V_B = V_C - \frac{V}{4}$.Comparing the potentials,$V_B = V_C - 0.25V$ and $V_A = V_C - 0.5V$. Since $V_B > V_A$,the current will flow from $B$ to $A$ when a wire is connected between them.

In the circuit shown,if a conducting wire is connected between points $A$ and $B,$ the current in this wire will

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