Question: 35. Three resistors having resistances \mathrm{r}_{1}, \mathrm{r}_{2} and \mathrm{r}_{3} are connected as shown in the given circuit. The ratio \frac{i_{3}}{i_{1}} of currents in terms of resistances used in the circuit is :
(1) \frac{r_{1}}{r_{2}+r_{3}}
(2) \frac{r_{2}}{r_{2}+r_{3}}
(3) \frac{r_{1}}{r_{1}+r_{2}}
(4) \frac{r_{2}}{r_{1}+r_{3}}
Answer: Option (2)
Explanation:
The resistor r_{1} is in series with a parallel combination of resistors r_{2} and r_{3}.
The current i_{1} entering the junction splits into currents i_{2} and
i_{3} through resistors r_{2} and r_{3} respectively.
Since r_{2} and r_{3} are in parallel,
the potential difference across them is the same.
Using Ohm’s law, i_{2}=\frac{V}{r_{2}} and i_{3}=\frac{V}{r_{3}}.
The total current entering the parallel combination is
i_{1}=i_{2}+i_{3}=\frac{V}{r_{2}}+\frac{V}{r_{3}}.
So, i_{1}=V\left(\frac{r_{2}+r_{3}}{r_{2}r_{3}}\right).
The ratio of currents is \frac{i_{3}}{i_{1}}=\frac{\frac{V}{r_{3}}}{V\left(\frac{r_{2}+r_{3}}{r_{2}r_{3}}\right)}.
Simplifying, we get \frac{i_{3}}{i_{1}}=\frac{r_{2}}{r_{2}+r_{3}}.
Hence, the correct answer is Option (2).