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 otherB. The potential difference across both the cells will be equalC. The potential difference across one cell will be greater then its emfD. `D_(A)-V_(B)=((epsi_(1)r_(2)+epsi_(2)r_(1)))/(r_(1)+r_(2))`

Two cells of unequal emfs `epsi_(1)" and "epsi_(2)`, and internal resi

1.	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 otherB. The potential difference across both the cells will be equalC. The potential difference across one cell will be greater then its emfD. `D_(A)-V_(B)=((epsi_(1)r_(2)+epsi_(2)r_(1)))/(r_(1)+r_(2))`
Answer» Correct Answer - A::B::C::D

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