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  4. If n cells each of emf ε and internal resistance r are connected in parallel, then the total emf and internal resistances will be

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Q. If $n$ cells each of emf $\varepsilon$ and internal resistance $r$ are connected in parallel, then the total emf and internal resistances will be

Current Electricity

Solution:

In the parallel combination,
$\frac{\varepsilon_{eq}}{r_{eq}}=\frac{\varepsilon_{1}}{r_{1}}+\frac{\varepsilon_{2}}{2}+....+\frac{\varepsilon_{n}}{r_{n}}$
$\frac{1}{r_{eq}}=\frac{1}{r_{1}}+\frac{1}{r_{2}}+....+\frac{1}{r_{n}}$
$(\because \varepsilon_{1}=\varepsilon_{2}=\varepsilon_{3}=.....=\varepsilon_{n}=\varepsilon$ and $r_{1}=r_{2}=r_{3}=....r_{n}=r)$
$\therefore \frac{\varepsilon_{eq}}{r_{eq}}=\frac{\varepsilon}{r}+\frac{\varepsilon}{r}+....+\frac{\varepsilon}{r}=n \frac{\varepsilon}{r}\quad\ldots\left(i\right)$
$\frac{1}{r_{eq}}=\frac{1}{r}+\frac{1}{r}+....+\frac{1}{r}=\frac{n}{r}$
$r_{eq}=r/n\quad\ldots\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$,
$\varepsilon_{eq}=n \frac{\varepsilon}{r} \times r_{eq}$
$=n \times \frac{\varepsilon}{r} \times \frac{r}{n}=\varepsilon$