JEE Physics 2026 Question with Solution

Given: Two identical cells, each of emf $E$ and internal resistance $r$, are connected to an external resistance of $6\,\Omega$. The current is the same for series and parallel combinations.

Find: The value of internal resistance $r$.

For cells in series:

$$E_s = 2E$$ $$r_s = 2r$$

So the current is

$$I_s = \frac{2E}{6+2r}$$

For cells in parallel:

$$E_p = E$$ $$r_p = \frac{r}{2}$$

So the current is

$$I_p = \frac{E}{6+\frac{r}{2}}$$

Since the currents are equal,

$$\frac{2E}{6+2r} = \frac{E}{6+\frac{r}{2}}$$

Cancelling $E$,

$$\frac{2}{6+2r} = \frac{1}{6+\frac{r}{2}}$$

Cross-multiplying,

$$2\left(6+\frac{r}{2}\right) = 6+2r$$ $$12+r = 6+2r$$ $$r = 3$$

Therefore, the value of the internal resistance is $3\,\Omega$, so the correct option is B.

Given: The same external resistor $6\,\Omega$ is connected to two identical cells arranged once in series and once in parallel.

Find: Internal resistance $r$ such that both currents are equal.

Use the current formula $I = \frac{\text{total emf}}{\text{external resistance} + \text{internal resistance}}$.

1. In series, total emf becomes $2E$ and total internal resistance becomes $2r$.

$$I_s = \frac{2E}{6+2r}$$

2. In parallel, equivalent emf remains $E$ and equivalent internal resistance becomes $\frac{r}{2}$.

$$I_p = \frac{E}{6+\frac{r}{2}}$$

3. Since both currents are same,

$$\frac{2E}{6+2r} = \frac{E}{6+\frac{r}{2}}$$

4. Divide both sides by $E$:

$$\frac{2}{6+2r} = \frac{1}{6+\frac{r}{2}}$$

5. Cross-multiply carefully:

$$2\left(6+\frac{r}{2}\right) = 6+2r$$ $$12+r = 6+2r$$ $$r = 3$$

Hence, the required internal resistance is $3\,\Omega$.