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(A) First, calculate the equivalent resistance of the circuit. The resistors $R_2$ and $R_3$ are in parallel, so their equivalent resistance $R_p$ is

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(A) First, calculate the equivalent resistance of the circuit. The resistors $R_2$ and $R_3$ are in parallel, so their equivalent resistance $R_p$ is given by:$R_p = \frac{R_2 	imes R_3}{R_2 + R_3} = \frac{4 	imes 4}{4 + 4} = \frac{16}{8} = 2 \,\Omega$Now, the total equivalent resistance $R_{eq}$ of the circuit is the sum of $R_1$, $R_p$, and $R_4$ in series:$R_{eq} = R_1 + R_p + R_4 = 2 \,\Omega + 2 \,\Omega + 1 \,\Omega = 5 \,\Omega$The total current $i$ flowing through the circuit is:$i = \frac{V}{R_{eq}} = \frac{10 \,V}{5 \,\Omega} = 2 \,A$This total current $i = 2 \,A$ flows through $R_1$, then splits into two parallel branches $R_2$ and $R_3$. Since $R_2 = R_3 = 4 \,\Omega$, the current divides equally between them:$i_{R_3} = i 	imes \left( \frac{R_2}{R_2 + R_3} ight) = 2 \,A 	imes \left( \frac{4}{4 + 4} ight) = 2 	imes \frac{4}{8} = 1 \,A$

In the given circuit, the current in resistance $R_3$ is: (in $\,A$)

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