JEE Physics 2024 Question with Solution

Given: Cell $E_1 = 8 \, \text{V}$, $r_1 = 2 \, \Omega$; cell $E_2 = 2 \, \text{V}$, $r_2 = 4 \, \Omega$. The cells are connected in opposition.

Find: The terminal potential difference of cell $E_2$.

Since the cells are connected in opposition, the net emf is

$$E_{\text{net}} = E_1 - E_2 = 8 - 2 = 6 \, \text{V}$$

The total internal resistance is

$$R_{\text{total}} = r_1 + r_2 = 2 + 4 = 6 \, \Omega$$

Hence the current in the circuit is

$$I = \frac{E_{\text{net}}}{R_{\text{total}}} = \frac{6}{6} = 1 \, \text{A}$$

Because $E_1 > E_2$, cell $E_2$ is being charged. Therefore its terminal potential difference is

$$V_{T2} = E_2 + I r_2$$

Substituting the values,

$$V_{T2} = 2 + (1)(4) = 6 \, \text{V}$$

Therefore, the terminal potential difference of cell $E_2$ is $6 \, \text{V}$.

The solution gives $6 \, \text{V}$, but this value is not present in the listed options. The answer key key marks option A. Since the source options and solution are inconsistent, the most defensible recorded answer is A with this discrepancy noted.