To solve this problem, we need to analyze the circuit and apply Kirchhoff's laws.$\\$The circuit consists of three resistors: $r_1, r_2,$ and $r_3. \\$Resistor $r_1$ is in series with a parallel combination of $r_2$ and $r_3. \\$Using Kirchhoff's current law at the junction where the current splits:$\\ i_1 = i_2 + i_3 \\$Using Ohm's law, the voltage across $r_2$ and $r_3$ is the same:$\\ i_2 \cdot r_2 = i_3 \cdot r_3 \\$From this, we can express $i_2$ in terms of $i_3: \\ i_2 = \frac{r_3}{r_2} \cdot i_3 \\$Substitute $i_2$ in the current equation:$\\ i_1 = \frac{r_3}{r_2} \cdot i_3 + i_3 \\$Factor out $i_3: \\ i_1 = i_3 \left( \frac{r_3}{r_2} + 1 \right) \\$Simplify the expression:$\\ i_1 = i_3 \left( \frac{r_3 + r_2}{r_2} \right) \\$Rearrange to find the ratio $\frac{i_3}{i_1}: \\ \frac{i_3}{i_1} = \frac{r_2}{r_2 + r_3} \\$Therefore, the correct option is Option 2: $\frac{r_2}{r_2 + r_3}.$