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Question
When two cells of emf's E1 and E2 are connected in series so as to assist each other, their balancing length on a potentiometer wire is found to be 2.7 m. When the cells are connected in series so as to oppose each other, the balancing length is found to be 0.3 m. Compare the emf's of the two cells.
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Solution
Data: l1 = 2.7 m (cells assisting),
l2 = 0.3 m (cells opposing)
`"E"_1 + "E"_2 = "Kl"_1` and `"E"_1 - "E"_2 = "Kl"_2`
∴ `("E"_1 + "E"_2)/("E"_1 - "E"_2) = "Kl"_1/"Kl"_2`
∴`"E"_1/"E"_2 = ("l"_1 + "l"_2)/("l"_1 - "l"_2) = (2.7 + 0.3)/(2.7 - 0.3) = 3/2.4 = 30/24` = 1.25
The ratio of the emf's of the two cells is 1.25.
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