JEE Physics 2023 Question with Solution

Given: A battery of $24 \, \text{V}$ with internal resistance $3\Omega$ is connected to the resistor network.

Find: The values of $I_4$ and $I_5$.

From the extracted solution, the equivalent resistance is taken as

$$R_{\text{eq}} = 3 + 1 + 2 + 4 + 2 = 12 \, \Omega$$

So the current through the battery is

$$I = \frac{24}{12} = 2 \, \text{A}$$

Using current division for $R_4$,

$$I_4 = \frac{R_5}{R_4 + R_5} \times 2 = \frac{5}{20 + 5} \times 2 = \frac{2}{5} \, \text{A}$$

Then

$$I_5 = 2 - I_4 = 2 - \frac{2}{5} = \frac{8}{5} \, \text{A}$$

Therefore, the currents are $I_4 = \frac{2}{5} \, \text{A}$ and $I_5 = \frac{8}{5} \, \text{A}$. The solution concludes this pair, so the correct option is A. The solution's label saying option C is inconsistent with the worked values.

Given: Total current entering the branch containing $R_4$ and $R_5$ is $2 \, \text{A}$.

Find: Branch currents through $R_4$ and $R_5$.

When two resistors are in parallel, current divides inversely in proportion to their resistances. Hence the current through $R_4$ is written as

$$I_4 = \frac{R_5}{R_4 + R_5} \times I$$

Substituting the values used in the solution,

$$I_4 = \frac{5}{20 + 5} \times 2 = \frac{2}{5} \, \text{A}$$

The remaining current goes through $R_5$:

$$I_5 = I - I_4 = 2 - \frac{2}{5} = \frac{8}{5} \, \text{A}$$

Thus the worked solution supports option A by value, even though the page heading marks C.