JEE Physics 2023 Question with Solution

Given: An $8 \, \text{V}$ source is connected to the resistor network. We need the current through $R_2$.

Find: The current flowing through $R_2$.

From the extracted solution, the equivalent resistance of the circuit is taken as

$$R_{\text{eq}} = 4 \, \Omega$$

Using Ohm's law, the total current is

$$i = \frac{V}{R_{\text{eq}}}$$

Substituting the values,

$$i = \frac{8}{4} = 2 \, \text{A}$$

Now apply current division as shown in the solution:

$$i_1 = i \cdot \frac{R_2}{R_1 + R_2}$$

where the branch resistances used are $6 \, \Omega$ and $3 \, \Omega$. Therefore,

$$i_1 = 2 \cdot \frac{3}{3 + 6} = 2 \cdot \frac{3}{9} = \frac{6}{9} = \frac{2}{3} \, \text{A}$$

The current through $R_2$ is then given as

$$i_2 = \frac{i_1}{2}$$

So,

$$i_2 = \frac{\frac{2}{3}}{2} = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3} \, \text{A}$$

Therefore, the current through $R_2$ is $\frac{1}{3} \, \text{A}$. The correct option is A.

Given: the solution identifies the correct option as A and lists the current values $i = 2 \, \text{A}$, $i_1 = \frac{2}{3} \, \text{A}$, and $i_2 = \frac{1}{3} \, \text{A}$.

Find: Current through $R_2$.

1. Equivalent resistance used in the solution:

$$R_{\text{eq}} = 4 \, \Omega$$

2. Total current through the circuit:

$$i = \frac{V}{R_{\text{eq}}} = \frac{8}{4} = 2 \, \text{A}$$

3. Current division step shown in the solution:

$$i_1 = i \cdot \frac{3}{3 + 6} = 2 \cdot \frac{3}{9} = \frac{2}{3} \, \text{A}$$

4. Final halving step used for the branch containing $R_2$:

$$i_2 = \frac{i_1}{2} = \frac{1}{3} \, \text{A}$$

Hence, the current flowing through $R_2$ is $\frac{1}{3} \, \text{A}$.