In the circuit diagram shown in the figure, the resistance $R_1 = 10Ω, R_2 = 15Ω$ and $E= 20 V$. When the switch S is closed, the ratio of $i_1$ to $i_3$ will be

5 : 3

The correct answer is Option (4) → 5 : 3

The circuit has two resistors $R_1=10\,\Omega$ and $R_2=15\,\Omega$ connected in parallel between A and B when switch S is closed.

Total current from battery: $i_1$

This current divides into two parallel branches: $i_2$ through $R_2$ and $i_3$ through $R_1$.

So, $i_1 = i_2 + i_3$

Since both resistors are in parallel, the voltage across each is the same: $V=20\,$V.

Current in $R_1$: $i_3 = \frac{V}{R_1}$

Current in $R_2$: $i_2 = \frac{V}{R_2}$

Therefore,

$\displaystyle \frac{i_1}{i_3} = \frac{i_2 + i_3}{i_3} = 1 + \frac{i_2}{i_3}$

$= 1 + \frac{V/R_2}{V/R_1} = 1 + \frac{R_1}{R_2}$

$= \frac{R_1 + R_2}{R_2}$

Substituting values: $\frac{10+15}{15}=\frac{25}{15}=\frac{5}{3}$

Hence, $i_1 : i_3 = 5:3$