Three resistors having resistances r1, r2 and r3 are connected as shown in the given circuit. The ratio i₃/i₁ of currents in terms of resistances used in the circuit is :

The correct option is r₂/(r₂ + r₃).

Step-by-step Explanation:

1. Identify the current division:
From the given circuit diagram, we can see that the total current entering terminal A is $i_1$, which passes through resistor $r_1$. At the junction, this current splits into two parallel branches containing resistors $r_2$ and $r_3$ carrying currents $i_2$ and $i_3$ respectively. According to Kirchhoff's Current Law (KCL):

$i_1=i_2+i_3$

2. Apply the parallel voltage relation:
Since the resistors $r_2$ and $r_3$ are connected in parallel, the potential difference (voltage drop) across both resistors is equal. By Ohm's law:

$i_2{r}_2=i_3{r}_3$

3. Express $i_2$ in terms of $i_3$:

$i_2=i_3\frac{r_3}{r_2}$

4. Substitute $i_2$ back into the current equation:

$i_1=i_3\frac{r_3}{r_2}+i_3$

Factoring out $i_3$ gives:

$i_1=i_3\left(\frac{r_3}{r_2}+1\right)$

Simplifying the expression in the parentheses:

$i_1=i_3\left(\frac{r_2+r_3}{r_2}\right)$

5. Find the ratio $\frac{i_3}{i_1}$:
Rearranging the equation to solve for the ratio of $i_3$ to $i_1$:

$\frac{i_3}{i_1}=\frac{r_2}{r_2+r_3}$