Question

Two batteries of emf ${ϵ}_1$ and ${ϵ}_2({ϵ}_2>{ϵ}_1)$ and internal resistances $r_1$ and $r_2$ respectively are connected in parallel as shown in figure.

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

Answer 1

Answer: (A)

The equivalent emf ${ϵ}_{eq}$ of the two cells is between ${ϵ}_1$ and ${ϵ}_2$, i.e. ${ϵ}_1<{ϵ}_{eq}<{ϵ}_2$

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,

$r_{eq}=\frac{r_1.r_2}{r_1+r_2}$

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 ${\epsilon }_2$.