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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. - Physics

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प्रश्न

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.

संख्यात्मक

उत्तर

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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अध्याय 9: Current Electricity - Exercises [पृष्ठ २२९]

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बालभारती Physics [English] 12 Standard HSC Maharashtra State Board
अध्याय 9 Current Electricity
Exercises | Q 15 | पृष्ठ २२९

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