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Two cells of unequal emfs epsi(1)" and "...

Two cells of unequal emfs `epsi_(1)" and "epsi_(2)`, and internal resistances `r_(1)" and "r_(2)` are joined as shown. `V_(A)" and "V_(B)` are the potentials as A and B respectively.

A

One cell will supply energy to the other

B

The potential difference across both the cells will be equal

C

The potential difference across one cell will be greater then its emf

D

`D_(A)-V_(B)=((epsi_(1)r_(2)+epsi_(2)r_(1)))/(r_(1)+r_(2))`

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A, B, C, D
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RESONANCE-CURRENT ELECTRICITY-Exercise-2 (Part-3)
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