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(B) Let the potential at node $F$ be $0\,V$. Then the potential at node $A$ is $30\,V$. Let the potential at node $C$ be $V_C$.Applying Kirchhoff's Cu

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$1$

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(B) Let the potential at node $F$ be $0\,V$. Then the potential at node $A$ is $30\,V$. Let the potential at node $C$ be $V_C$.Applying Kirchhoff's Current Law $(KCL)$ at node $C$:$\frac{V_C - 30}{4 + 2} + \frac{V_C - 0}{3} + \frac{V_C - 15}{1 + 2} = 0$$\frac{V_C - 30}{6} + \frac{V_C}{3} + \frac{V_C - 15}{3} = 0$Multiplying the entire equation by $6$:$(V_C - 30) + 2V_C + 2(V_C - 15) = 0$$V_C - 30 + 2V_C + 2V_C - 30 = 0$$5V_C = 60 \implies V_C = 12\,V$.The current through the $15\,V$ battery branch (from $C$ to $D$ to $E$) is given by $I = \frac{V_C - 15}{1 + 2} = \frac{12 - 15}{3} = \frac{-3}{3} = -1\,A$.The magnitude of the current is $1\,A$.

The figure shows a network of five resistances and two batteries. The current through the $15\,V$ battery is .............. $A$.

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