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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.

Text Solution

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(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`
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