JEE Physics 2023 Question with Solution

Given: emf of each cell $\varepsilon = 1.5 \, \text{V}$, external resistance $R = 10 \, \Omega$, and voltmeter reading across $R$ is $V_R = 1.5 \, \text{V}$.

Find: the internal resistance $r$ of each cell.

The two cells are connected in series, so total emf is

$$2\varepsilon = 2 \times 1.5 = 3 \, \text{V}$$

and total internal resistance is $2r$.

The voltmeter reads the potential difference across the external resistance, so

$$V_R = IR$$

Hence,

$$I = \frac{V_R}{R} = \frac{1.5}{10} = 0.15 \, \text{A}$$

Now use the circuit current relation

$$I = \frac{2\varepsilon}{R + 2r}$$

Substituting the known values,

$$0.15 = \frac{3}{10 + 2r}$$

So,

$$10 + 2r = \frac{3}{0.15} = 20$$

Therefore,

$$2r = 20 - 10 = 10$$

and

$$r = \frac{10}{2} = 5 \, \Omega$$

Therefore, the internal resistance of each cell is $5 \, \Omega$.