Question

Two batteries of emf ${\epsilon }_1$ and ${\epsilon }_2\left({\epsilon }_2>{\epsilon }_1\right)$i) and internal resistances $r_1$ and $r_2$ respectively are connected in parallel as shown in figure

• A. The equivalent emf ${\epsilon }_{eq}$ of the two cells is between ${\epsilon }_1$ and ${\epsilon }_2$, i.e. ${\epsilon }_1<{\epsilon }_{eq}<{\epsilon }_2$.
• B. The equivalent emf ${\epsilon }_{eq}$ is smaller than ${\epsilon }_1$
• C. The ${\epsilon }_{eq}$ is given by ${\epsilon }_{eq}={\epsilon }_1+{\epsilon }_2$ always
• D. ${\epsilon }_{eq}$ is independent of internal resistances $r_1$ and $r_2$

Answer 1

Answer: (A)

A. The equivalent emf ${\epsilon }_{eq}$ of the two cells is between ${\epsilon }_1$ and ${\epsilon }_2$,
i.e., ${\epsilon }_1<{\epsilon }_{eq}<{\epsilon }_2$
The cells of emf ${\epsilon }_1$ and ${\epsilon }_2$ with internal resistance $r_1$ and $r_2$ respectively.
In parallel combination,
the relation between equivalent emf and equivalent resistance is,
$\frac{{\epsilon }_{eq}}{r_{eq}}=\frac{{\epsilon }_1}{r_1}+\frac{{\epsilon }_2}{r_2}$
The resistance in parallel combination is,
$r_{eq}=\frac{r_1\cdot r_2}{r_1+r_2}$
From the above equation,
${\epsilon }_{eq}=\frac{{\epsilon }_1{r}_2+{\epsilon }_2{r}_1}{r_1+r_2}$
In question,
${\epsilon }_2>{\epsilon }_1$
Then The equivalent emf ${\epsilon }_{eq}$ of two cell is between ${\epsilon }_1$ and ${\epsilon }_2$.