In the circuit of the adjoining figure,the cu | vedclass

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(D) Let the potential at the top junction be $V_T$ and the potential at the junction between the two batteries be $V_M$. Let the potential at $A$ be $

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$0$

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(D) Let the potential at the top junction be $V_T$ and the potential at the junction between the two batteries be $V_M$. Let the potential at $A$ be $V_A = 0 \ V$. Then the potential at $C$ is $V_C = 0 \ V$ because the circuit is symmetric.By symmetry,the potential at the top junction $V_T$ is such that no current flows through the $10 \ \Omega$ resistor,or simply by observing the symmetry of the circuit,the potential at $A$ and $C$ are equal.Since the potential at $A$ and $C$ are equal $(V_A = V_C)$,the potential difference across the $12 \ \Omega$ resistor connected between $E$ and $F$ (which are at the same potential as $A$ and $C$ respectively) is $V_A - V_C = 0 \ V$.Therefore,the current $i$ through the $12 \ \Omega$ resistor is $i = \frac{V_A - V_C}{12} = \frac{0}{12} = 0 \ A$.

In the circuit of the adjoining figure,the current through the $12 \ \Omega$ resistor will be ................ $A$.

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