JEE Physics 2025 Question with Solution

Given: Two nonideal batteries are connected in parallel with emfs $E_1$ and $E_2$, and internal resistances $r_1$ and $r_2$.

Find: Which of the two statements is correct.

For two nonideal batteries in parallel, the equivalent emf is a weighted average:

$$E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2}$$

So $E_{eq}$ lies between $E_1$ and $E_2$. It is not necessarily smaller than either of the two emfs. Therefore, Statement-I is false.

The equivalent internal resistance is:

$$r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$$

Since resistances in parallel combine to give a value smaller than either individual resistance, $r_{eq} < r_1$ and $r_{eq} < r_2$. Therefore, Statement-II is true.

Hence, the correct conclusion is: Statement-I is false but Statement-II is true.

The correct option is D.

Given:

• Statement-I concerns the equivalent emf of two nonideal batteries in parallel.
• Statement-II concerns the equivalent internal resistance of two nonideal batteries in parallel.

Find: Determine the truth value of each statement.

Step 1: Use the result for parallel combination of nonideal batteries.

$$E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2}$$ $$r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$$

Step 2: Examine Statement-II. Because $r_{eq}$ is the parallel combination of $r_1$ and $r_2$, it is always less than each individual resistance. So Statement-II is true.

Step 3: Examine Statement-I. The expression for $E_{eq}$ shows that it is a weighted average of $E_1$ and $E_2$.

• If $E_1 = E_2$, then $E_{eq} = E_1 = E_2$.
• If $E_1 \neq E_2$, then $E_{eq}$ lies between the two emfs.

Therefore, it is incorrect to say that the equivalent emf is smaller than either of the two emfs. So Statement-I is false.

Conclusion: Statement-I is false but Statement-II is true, so the correct option is D.