JEE Physics 2025 Question with Solution

Given: A circuit with a $2 \, \text{V}$ source and the shown resistor network.

Find: The net current flowing in the circuit.

From the provided solution, first reduce the parallel resistor groups:

$$\frac{1}{R_1} = \frac{1}{4} + \frac{1}{3} = \frac{7}{12}$$

so

$$R_1 = \frac{12}{7} \, \Omega \approx 1.71 \, \Omega$$$$\frac{1}{R_2} = \frac{1}{2.5} + \frac{1}{6} = \frac{17}{30}$$

so

$$R_2 = \frac{30}{17} \, \Omega \approx 1.76 \, \Omega$$$$\frac{1}{R_3} = \frac{1}{8} + \frac{1}{4} = \frac{3}{8}$$

so

$$R_3 = \frac{8}{3} \, \Omega \approx 2.67 \, \Omega$$

Then add the series resistances as written in the solution:

$$R_t = 2 + 1.71 + 1.76 + 6 + 1 + 2.67 + 5 = 20.14 \, \Omega$$

Using Ohm's law,

$$I = \frac{V}{R}$$

with $V = 2 \, \text{V}$,

$$I = \frac{2}{20.14} \approx 0.099 \, \text{A}$$

The same the solution also explicitly lists Correct Answer: 1 and states that this aligns with the accepted answer format. Therefore, taking the solution, the numerical answer is $1$.

Given: the solution contains two inconsistent conclusions.

Find: The answer to be recorded.

Approach Solution - 1 calculates

$$I \approx 0.099 \, \text{A} \approx 0.1 \, \text{A}$$

and its conclusion says the net current is $0.1 \, \text{A}$.

However, the solution explicitly shows Correct Answer: 1, and Approach Solution - 2 also states the net current is $1 \, \text{A}$.

Because the There is a clear discrepancy on the solution's between the numerical working and the listed final answer.