Calculate Internal Resistance of a Cell

To find the internal resistance of the cell, let:

• $E$ be the electromotive force (emf) of the cell,
• $r$ be the internal resistance of the cell.

The current $I$ in a circuit containing a cell of emf $E$, internal resistance $r$, and an external resistance $R$ is given by Ohm's law: $I = \frac{E}{R + r}$

Case 1: When $R_1 = 5\ \Omega$, the current $I_1 = 0.25\text{ A}$. $0.25 = \frac{E}{5 + r}$ $E = 0.25 \times (5 + r) \quad \text{--- (1)}$

Case 2: When $R_2 = 2\ \Omega$, the current $I_2 = 0.5\text{ A}$. $0.5 = \frac{E}{2 + r}$ $E = 0.5 \times (2 + r) \quad \text{--- (2)}$

Equating equations (1) and (2) since the emf $E$ remains constant: $0.25(5 + r) = 0.5(2 + r)$

Dividing both sides by $0.25$: $5 + r = 2(2 + r)$ $5 + r = 4 + 2r$ $2r - r = 5 - 4$ $r = 1\ \Omega$

Thus, the internal resistance of the cell is $1\ \Omega$.