In the circuit shown, if a conducting wire is | vedclass

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(A) Let the potential of the left junction be $V_L = 0$ and the right junction be $V_R = V$.The potential at point $A$ is determined by the voltage di

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Flow from $A$ to $B$

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(A) Let the potential of the left junction be $V_L = 0$ and the right junction be $V_R = V$.The potential at point $A$ is determined by the voltage divider of the top two resistors ($4 \, \Omega$ and $4 \, \Omega$):$V_A = V_L + \frac{4}{4+4} 	imes (V_R - V_L) = 0 + \frac{4}{8} 	imes V = \frac{V}{2}$.The potential at point $B$ is determined by the voltage divider of the bottom two resistors ($1 \, \Omega$ and $3 \, \Omega$):$V_B = V_L + \frac{1}{1+3} 	imes (V_R - V_L) = 0 + \frac{1}{4} 	imes V = \frac{V}{4}$.Comparing the potentials, $V_A = \frac{V}{2}$ and $V_B = \frac{V}{4}$.Since $V_A > V_B$, the current will flow from the higher potential to the lower potential, i.e., from $A$ to $B$.

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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