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(A) Let the potential at junction $A$ be $V_A = 3 	ext{ V}$ and at junction $C$ be $V_C = 0 	ext{ V}$.Let the potential at junction $B$ be $V_B$.App

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$0.25$

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(A) Let the potential at junction $A$ be $V_A = 3 	ext{ V}$ and at junction $C$ be $V_C = 0 	ext{ V}$.Let the potential at junction $B$ be $V_B$.Applying Kirchhoff's Current Law $(KCL)$ at junction $B$:$\frac{V_B - V_A}{6} + \frac{V_B - V_C}{12} + \frac{V_B - V_C}{12} = 0$$\frac{V_B - 3}{6} + \frac{V_B}{12} + \frac{V_B}{12} = 0$Multiplying by $12$:$2(V_B - 3) + V_B + V_B = 0$$2V_B - 6 + 2V_B = 0$$4V_B = 6 \implies V_B = 1.5 	ext{ V}$.The current through the $6 \Omega$ resistor between $A$ and $B$ is $I = \frac{V_A - V_B}{6} = \frac{3 - 1.5}{6} = \frac{1.5}{6} = 0.25 	ext{ A}$.

For the circuit shown in the figure,the current through the $6 \Omega$ resistor connected between the junctions $A$ and $B$ is (in $A$)

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