Question

Potential differences across the terminals of a cell were measured (in volt) against different currents (in ampere) flowing through the cell. A graph was drawn which was a straight line ABC as shown in figure

Determine from graph (i) emf of the cell (ii) maximum current obtained from the cell and (iii) internal resistance of the cell.

Answer 1

(i) EMF of the cell is equal to maximum potential difference across the two electrodes of cell corresponding to zero current. Thus emf of the cell, `epsilon = 1.4 V`.
(ii) Max. current is drawn from the cell when the terminal pot. Diff. is zero. Therefore
`I_(max) = 0.28 A`
(iii) Internal resistance,
`r=epsilon/I_(max) = (1.4 V)/(0.28A)= 5 Omega`