Physics, asked by saurabh5350, 7 months ago

when two cells of emfs E1 and E2 are connected in series so as assist each other, their balancing length on a potentiometer is found be 2.7. when the cells are connected in series so as to oppose each other, the balancing length is found to be 0.3m . Compare the emfs of the two cells.

Answers

Answered by knjroopa

Explanation:

Given when two cells of emfs E1 and E2 are connected in series so as assist each other, their balancing length on a potentiometer is found be 2.7. 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 emfs of the two cells.

The two cells of Emf’s E1 and E2 are connected in series to assist each other and the balancing length is 2.7 m
So we have I1 = 2.7 m
When cells are opposing each other the balancing length is 0.3 m
Therefore E1 + E2 = K1 and E1 – E2 = K2
So E1 + E2 / E1 – E2 = K1 / K2
So E1 / E2 = I1 + I2 / I1 – I2
= 2.7 + 0.3 / 2.7 – 0.3
= 3 / 2.4
= 1.25
Therefore the comparison of the emf’s of two cells will be 1.25

Reference link will be

https://brainly.in/question/19244518

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when two cells of emfs E1 and E2 are connected in series so as assist each other, their balancing length on a potentiometer is found be 2.7. when the cells are connected in series so as to oppose each other, the balancing length is found to be 0.3m . Compare the emfs of the two cells.​

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when two cells of emfs E1 and E2 are connected in series so as assist each other, their balancing length on a potentiometer is found be 2.7. when the cells are connected in series so as to oppose each other, the balancing length is found to be 0.3m . Compare the emfs of the two cells.