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(C) Let the potential at node $F$ be $V_F = 0 \ V$. Then the potential at node $C$ is $V_C = 6 \ V$.Applying Kirchhoff's Current Law $(KCL)$ at node $

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$7$

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(C) Let the potential at node $F$ be $V_F = 0 \ V$. Then the potential at node $C$ is $V_C = 6 \ V$.Applying Kirchhoff's Current Law $(KCL)$ at node $C$:Let $I_1$ be the current from the left loop through the $4 \ \Omega$ resistor and $I_2$ be the current from the right loop through the $1 \ \Omega$ resistor,both flowing towards $C$.For the left loop $(ABCF)$: $30 - 4I_1 - 6 - 2I_1 = 0 \implies 6I_1 = 24 \implies I_1 = 4 \ A$.For the right loop $(CDEF)$: $6 - 1I_2 - 15 + 2I_2 = 0 \implies I_2 = 9 \ A$ (Wait,let's re-evaluate).Correct $KCL$ at node $C$: Current flowing into $C$ from left is $I_1 = (30 - 6) / (4 + 2) = 24 / 6 = 4 \ A$.Current flowing into $C$ from right: The loop $CDEF$ has $6 \ V$ and $15 \ V$ batteries. The net $EMF$ is $15 - 6 = 9 \ V$ and total resistance is $1 + 2 = 3 \ \Omega$. So,$I_2 = 9 / 3 = 3 \ A$.Total current through branch $CF$ is $I = I_1 + I_2 = 4 + 3 = 7 \ A$.

The figure shows a network of four resistances and three batteries. The current through the branch $CF$ is ............... $A$.

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