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(D) First,simplify the circuit. The resistors $R_1$ $(2\,\Omega)$ and $R_2$ $(4\,\Omega)$ are in series,so their equivalent resistance is $R_{12} = 2

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$\frac{1}{3}\,A$

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(D) First,simplify the circuit. The resistors $R_1$ $(2\,\Omega)$ and $R_2$ $(4\,\Omega)$ are in series,so their equivalent resistance is $R_{12} = 2 + 4 = 6\,\Omega$.This $R_{12}$ is in parallel with $R_5$ $(6\,\Omega)$,so the equivalent resistance of this branch is $R_{AC} = \frac{6 	imes 6}{6 + 6} = 3\,\Omega$.Now,$R_{AC}$ $(3\,\Omega)$ is in series with $R_4$ $(3\,\Omega)$,giving $R_{DC} = 3 + 3 = 6\,\Omega$.This $R_{DC}$ is in parallel with $R_7$ $(3\,\Omega)$,so $R_{AD} = \frac{6 	imes 3}{6 + 3} = 2\,\Omega$.Finally,$R_{AD}$ $(2\,\Omega)$ is in series with $R_3$ $(2\,\Omega)$,so the total equivalent resistance is $R_{eq} = 2 + 2 = 4\,\Omega$.The total current from the battery is $i = \frac{V}{R_{eq}} = \frac{8}{4} = 2\,A$.Using the current divider rule at node $D$,the current $i_1$ flowing through the branch $DC$ is $i_1 = i 	imes \frac{R_7}{R_7 + R_{DC}} = 2 	imes \frac{3}{3 + 6} = 2 	imes \frac{3}{9} = \frac{2}{3}\,A$.At node $C$,the current $i_1$ splits between $R_5$ and the series combination of $R_1$ and $R_2$. The current $i_2$ flowing through $R_2$ is $i_2 = i_1 	imes \frac{R_5}{R_5 + (R_1 + R_2)} = \frac{2}{3} 	imes \frac{6}{6 + (2 + 4)} = \frac{2}{3} 	imes \frac{6}{12} = \frac{2}{3} 	imes \frac{1}{2} = \frac{1}{3}\,A$.

The current flowing through $R_2$ is:

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