The terminal voltages of a cell are found to be 29 V and 28 V when it delivers currents of 1 A and 2 A, respectively. The emf and the internal resistance of the cell are, respectively

30 V; 1 Ω

The correct answer is Option (2) → 30 V; 1 Ω

Given:

Terminal voltages: $V_1 = 29\ \text{V}$ at $I_1 = 1\ \text{A}$, $V_2 = 28\ \text{V}$ at $I_2 = 2\ \text{A}$

Terminal voltage: $V = \mathcal{E} - I r$, where $\mathcal{E}$ = emf, $r$ = internal resistance.

Set up equations:

$29 = \mathcal{E} - 1 \cdot r$

$28 = \mathcal{E} - 2 \cdot r$

Subtract second equation from first:

$29 - 28 = (\mathcal{E} - r) - (\mathcal{E} - 2r)$

$1 = r$

∴ Internal resistance, $r = 1\ \Omega$

Substitute $r$ in first equation:

$29 = \mathcal{E} - 1 \Rightarrow \mathcal{E} = 30\ \text{V}$

∴ Emf = 30 V, Internal resistance = 1 Ω