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(D) To find the current through the $2 \Omega$ resistor, we first simplify the circuit.$1$. The rightmost part of the circuit consists of two $1 \Omeg

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$0.3$

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(D) To find the current through the $2 \Omega$ resistor, we first simplify the circuit.$1$. The rightmost part of the circuit consists of two $1 \Omega$ resistors in series, which are in parallel with another $1 \Omega$ resistor. The equivalent resistance of this part is $R_1 = \frac{(1+1) 	imes 1}{(1+1) + 1} = \frac{2}{3} \Omega$.$2$. This $R_1$ is in series with the $1 \Omega$ resistor, giving $R_2 = 1 + \frac{2}{3} = \frac{5}{3} \Omega$.$3$. This $R_2$ is in parallel with the $3 \Omega$ resistor. The equivalent resistance is $R_3 = \frac{3 	imes (5/3)}{3 + (5/3)} = \frac{5}{14/3} = \frac{15}{14} \Omega$.$4$. Finally, this $R_3$ is in series with the $1 \Omega$ and $2 \Omega$ resistors. The total equivalent resistance is $R_{eq} = 1 + 2 + \frac{15}{14} = 3 + \frac{15}{14} = \frac{42+15}{14} = \frac{57}{14} \Omega$.$5$. The total current from the battery is $I = \frac{V}{R_{eq}} = \frac{1.2}{57/14} = \frac{1.2 	imes 14}{57} \approx 0.2947 	ext{ A}$.$6$. Since the $2 \Omega$ resistor is in series with the battery, the total current flows through it. Rounding to the nearest option, the current is approximately $0.3 	ext{ A}$.

In the given circuit, the current through the $2 \Omega$ resistor is: (in $\text{ A}$)

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