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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. |
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Answer» (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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