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(A) For two batteries with $e.m.f.s$ $E_1, E_2$ and internal resistances $r_1, r_2$ connected in parallel,the equivalent $e.m.f.$ $E_{eq}$ is given by

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$(i)$ and $(ii)$ both are correct.

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(A) For two batteries with $e.m.f.s$ $E_1, E_2$ and internal resistances $r_1, r_2$ connected in parallel,the equivalent $e.m.f.$ $E_{eq}$ is given by $E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$.This value $E_{eq}$ always lies between $E_1$ and $E_2$,meaning it is less than the larger $e.m.f.$ and greater than the smaller $e.m.f.$ Thus,statement $(i)$ is correct.The equivalent internal resistance $r_{eq}$ for parallel connection is given by $\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}$,which implies $r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$.Since $r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$,it is mathematically certain that $r_{eq} < r_1$ and $r_{eq} < r_2$. Thus,statement $(ii)$ is also correct.

Two batteries with different $e.m.f.$ are connected in parallel. Consider the following statements: $(i)$ The equivalent $e.m.f.$ is less than the individual $e.m.f.s$. $(ii)$ The equivalent internal resistance is less than the individual internal resistances.

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