Two nonideal batteries are wired in parallel. Evaluate the...

For two batteries connected in parallel with emfs $E_1$ and $E_2$ and internal resistances $r_1$ and $r_2$, the equivalent emf is given by $E_{eq} = \dfrac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$.

This expression shows that $E_{eq}$ is a weighted average of $E_1$ and $E_2$. Thus, $E_{eq}$ always lies between $E_1$ and $E_2$ and cannot be smaller than both individual emfs. Hence, statement A is incorrect.

The equivalent internal resistance is given by the parallel combination of the individual internal resistances, $r_{eq} = \dfrac{r_1 r_2}{r_1 + r_2}$.

In a parallel combination of resistors, the equivalent resistance is always strictly less than the smallest individual resistance. Hence, statement B is correct.