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(D) Given that point $A$ is earthed,its potential $V_A = 0\,V$.Let the potentials at points $B$,$C$,and $D$ be $V_B$,$V_C$,and $V_D$ respectively.From

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$1\,A$ from $B$ to $C$

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(D) Given that point $A$ is earthed,its potential $V_A = 0\,V$.Let the potentials at points $B$,$C$,and $D$ be $V_B$,$V_C$,and $V_D$ respectively.From the circuit,the potential at $B$ is determined by the $2\,V$ battery connected to $A$: $V_B - V_A = 2\,V \implies V_B = 2\,V$.The potential at $D$ is determined by the $8\,V$ battery connected to $B$: $V_B - V_D = 8\,V \implies 2 - V_D = 8 \implies V_D = -6\,V$.Now,consider the branch $CD$. The potential difference across the $4\,V$ battery is $V_C - V_D = 4\,V \implies V_C - (-6) = 4 \implies V_C = -2\,V$.The current $I$ through the $4\,\Omega$ resistor (between $B$ and $C$) is given by $I = \frac{V_B - V_C}{R} = \frac{2 - (-2)}{4} = \frac{4}{4} = 1\,A$.Since $V_B > V_C$ $(2\,V > -2\,V)$,the current flows from $B$ to $C$.

The figure shows a network of resistances in which the point $A$ is earthed. The current through the $4\,\Omega$ resistor is

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