A galvanometer together with an unknown resistance in series is connected to two identical batteries each of $1.5\, V$. When the batteries are connected in series, the galvanometer records a current of $1\,A$, and when batteries are in parallel the current is $0.6\,A$. What is the internal resistance of the battery ?
Diffcult
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Case $I$ $:-$ $I_{g}=\frac{E_{1}+E_{2}}{R_{g}+R+2 r} \Rightarrow 1=\frac{3}{R_{g}+R+\frac{r}{2}}$
$\Rightarrow \mathrm{R}_{\mathrm{g}}+\mathrm{R}+2 \mathrm{r}=3$ ............$(1)$
Case $II: - {E_{eq}} = E = 1.5{\rm{\,V}}$
$\mathrm{I}_{\mathrm{g}}=\frac{\mathrm{E}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{g}}+\mathrm{R}+\frac{\mathrm{r}}{2}} \quad \Rightarrow 0.6=\frac{1.5}{\mathrm{R}_{\mathrm{g}}+\mathrm{R}+\frac{\mathrm{r}}{2}}$
$\Rightarrow \mathrm{R}_{\mathrm{g}}+\mathrm{R}+\frac{\mathrm{r}}{2}=\frac{1.5}{0.6}=2.5$ .........$(2)$
from eq $(1)$ and $(2)$
$\frac{3 \mathrm{r}}{2}=0.5 \quad \Rightarrow \quad \mathrm{r}=\frac{1}{3}\, \Omega$
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