JEE Physics 2023 Question with Solution

Given: The circuit has three resistors of $10 \, \Omega$ each, with currents $I_1$, $I_2$, and $I_3$ as marked.

Find: The value of $\left| \frac{I_1 + I_3}{I_2} \right|$.

From the extracted solution, the currents are taken as

and

Now substitute these values into the required expression:

Therefore, the value of the given expression is $2$.

Given: The annotated figure indicates the left side is at $10 \, \text{V}$, the middle junction is at $20 \, \text{V}$, and the right side is at $0 \, \text{V}$.

Find: $\left| \frac{I_1 + I_3}{I_2} \right|$.

Each of the two inner resistors of $10 \, \Omega$ is connected between the $20 \, \text{V}$ node and the $10 \, \text{V}$ node, so the potential difference across each is $10 \, \text{V}$. Hence,

The bottom $10 \, \Omega$ resistor is connected between $10 \, \text{V}$ and $0 \, \text{V}$, so

Now evaluate the ratio:

The source HTML also contains an inconsistent intermediate value $3$, but its final conclusion and the second approach both give $2$, which is the accepted answer.

Therefore, the correct numerical value is $2$.