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(A) Given that point $A$ is earthed,the potential at $A$ is $V_A = 0\,V$.Let the potential at point $D$ be $V_D$ and at point $B$ be $V_B$.From the ci

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$2\,A$ from $D$ to $A$

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(A) Given that point $A$ is earthed,the potential at $A$ is $V_A = 0\,V$.Let the potential at point $D$ be $V_D$ and at point $B$ be $V_B$.From the circuit,the battery of $8\,V$ is connected between $B$ and $D$ such that $V_B - V_D = 8\,V$,so $V_B = V_D + 8$.Applying Kirchhoff's Current Law $(KCL)$ at node $D$:$\frac{V_D - V_B}{8} + \frac{V_D - V_C}{4} + \frac{V_D - V_A}{3} = 0$ (Note: The $8\,V$ battery is in the branch $BD$).Actually,let's use nodal analysis at $D$ and $B$ directly.At node $D$: $\frac{V_D - V_B}{8} + \frac{V_D - V_C}{4} + \frac{V_D - 0}{3} = 0$. Since $V_C - V_D = 4\,V$,then $V_D - V_C = -4\,V$.Substituting: $\frac{V_D - (V_D + 8)}{8} + \frac{-4}{4} + \frac{V_D}{3} = 0$$-1 - 1 + \frac{V_D}{3} = 0$$\frac{V_D}{3} = 2 \implies V_D = 6\,V$.The current through the $3\,\Omega$ resistor is $I = \frac{V_D - V_A}{3} = \frac{6 - 0}{3} = 2\,A$.Since $V_D = 6\,V$ and $V_A = 0\,V$,the current flows from $D$ to $A$.

The figure shows a network of resistances in which the point $A$ is earthed. The current through the $3\,\Omega$ resistor is

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