Two cells of emf e1 and e2 are joined in series and the balancing length of the potentiometer. wire is 625 cm. if the terminals of e1 are reversed the balancing length obtained is 125 cm. given e2>e1 the ratio e1:e2 will be?

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Answer 1 · 2018-03-22T17:21:34

#e_2/e_1 = 3/2#

The net emf of the two cells in the two cases #e_2+e_1# and #e_2-e_1#, respectively. Since the emf is proportional to the balance length

#(e_2+e_1)/(e_2-e_1)=625/125 = 5/1#

Thus

#e_2/e_1 ={(e_2+e_1)+(e_2-e_1)}/{(e_2+e_1)-(e_2-e_1)} = (5+1)/(5-1) = 3/2#