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Emf of a cell is 1.5V and its internal resistance is 1ohm.For what current drawn from the cell will the terminal potential difference be half of its emf?

Terminal potential of the battery is given by:
V = E - Ir
where E is the emf of the battery
I is the current
r is the internal resistance of the battery.

V = E2r = 1 ohmE2 = E -I×1I = E -E/2E = 1.5 vI = 1.5/2 = 0.75 A

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we know that emf- ir= V... where r is internal resistance of cell... there fore v= 1.5/2=0.75....

therefore 1.5-i=.75.. i=.75 A..... hope it helps u.. hav a nyc day..

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