Given below are two statements : Statement-I : The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs. Statement-II : The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries. In the light of the above statements, choose the correct answer from the options given below.

Statement-I is false but Statement-II is true

Statement-I is true but Statement-II is false

Both Statement-I and Statement-II are false

Both Statement-I and Statement-II are true

Current Electricity PYQ — JEE Main 2025 · 22 Jan · Shift 1 · Q61 with Solution

View written solutionFree

Correct answer: A

Model the two nonideal batteries

Let the two batteries have emfs $E_1, E_2$ and internal resistances $r_1, r_2$ .

When two nonideal batteries are connected in parallel (with same polarity), their Thevenin equivalent is:
$= \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$$ and $$r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2}.$$$
Check Statement-I

Statement-I says: The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs.

But from

$E_{\text{eq}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2},$

we see that $E_{\text{eq}}$ is a weighted average of $E_1$ and $E_2$ .

Therefore,

$\min(E_1,E_2) \le E_{\text{eq}} \le \max(E_1,E_2).$

So the equivalent emf lies between the two emfs, not smaller than both.

Hence, Statement-I is false.
Check Statement-II

Statement-II says: The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries.

Since

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

this is the parallel combination of two positive resistances, which is always less than each one individually:

$r_{\text{eq}} < r_1, \qquad r_{\text{eq}} < r_2.$

Hence, Statement-II is true.
Choose the correct option
- Statement-I: False
- Statement-II: True
Therefore, the correct option is:

$\boxed{\text{A}}$
Comparison with stored answer

Stored correct answer: $\boxed{\text{A}}$

My derived answer matches the stored answer.

View written solutionFree

Correct answer: A

Model the two nonideal batteries

Let the two batteries have emfs $E_1, E_2$ and internal resistances $r_1, r_2$ .

When two nonideal batteries are connected in parallel (with same polarity), their Thevenin equivalent is:
$= \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$$ and $$r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2}.$$$
Check Statement-I

Statement-I says: The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs.

But from

$E_{\text{eq}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2},$

we see that $E_{\text{eq}}$ is a weighted average of $E_1$ and $E_2$ .

Therefore,

$\min(E_1,E_2) \le E_{\text{eq}} \le \max(E_1,E_2).$

So the equivalent emf lies between the two emfs, not smaller than both.

Hence, Statement-I is false.
Check Statement-II

Statement-II says: The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries.

Since

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

this is the parallel combination of two positive resistances, which is always less than each one individually:

$r_{\text{eq}} < r_1, \qquad r_{\text{eq}} < r_2.$

Hence, Statement-II is true.
Choose the correct option
- Statement-I: False
- Statement-II: True
Therefore, the correct option is:

$\boxed{\text{A}}$
Comparison with stored answer

Stored correct answer: $\boxed{\text{A}}$

My derived answer matches the stored answer.