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(D) Let the potential at point $A$ be $V_A = 0 \, V$. Then the potential at point $C$ is $V_C = 0 \, V$ because the circuit is symmetric and the two b

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$0$

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(D) Let the potential at point $A$ be $V_A = 0 \, V$. Then the potential at point $C$ is $V_C = 0 \, V$ because the circuit is symmetric and the two batteries of $5 \, V$ are connected in opposition relative to the central node.Since the potential at point $A$ is $0 \, V$ and the potential at point $C$ is $0 \, V$,the potential difference across the $12 \, \Omega$ resistor connected between $E$ and $F$ (which are connected to $A$ and $C$ respectively) is $V_E - V_F = 0 \, V - 0 \, V = 0 \, V$.According to Ohm's law,$I = V/R = 0 \, V / 12 \, \Omega = 0 \, A$.Thus,the current passing through the $12 \, \Omega$ resistor is $0 \, A$.

In the given circuit,the current passing through the $12\, \Omega$ resistor is............ $A$.

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