Two cells of unequal emfs $ε_{1}$ & and $ε_{2}$ , and internal resistances $r_{1}$ and $r_{2}$ are joined as shown. $V_{A}$ and $V_{B}$ are the potentials at A and B respectively.

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One cell will supply energy to the other

The potential difference across both the cells will be equal

The potential difference across one cell will be greater than its emf.

$V A - V B = \frac{(ε_{1} r_{2} + ε_{2} r_{1})}{r_{1} + r_{2}}$

answer is A, B, C, D.

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Detailed Solution

$V_{A} - V_{B} = E_{e q} = \frac{r_{1}}{\frac{1}{r_{1}}} \frac{r_{2}}{\frac{1}{r_{2}}} = \frac{ε_{1} r_{2} + ε_{2} r_{1}}{r_{1} + r_{2}}$
If $ε_{1} > ε_{2}$ source $ε_{1}$ act as a source and $ε_{2}$ act as a load.
And $\begin{array}{l} V_{1} = ε_{1} - i r_{1} \\ V_{2} = ε_{2} + i r_{2} \end{array}$
$V_{1} = V_{2} a s i = \frac{ε_{1} - ε_{2}}{r_{1} + r_{2}}$ for $ε_{1} > ε_{2}$