Two identical cells of emf 1.5 V each joined in parallel supply energy to an external circuit consisting of two resistances of 7 $\Omega$ each joined in parallel. A very high resistance voltmeter reads the terminal voltage of cells to be 1.4 V. Calculate the internal resistance of each cell.
CBSE OUTSIDE DELHI - SET 1 NORTH 2016
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We have, for a single cell,
$\text{r} = \bigg(\frac{\text{E}}{\text{V}} - 1 \bigg)\text{R}$
$\therefore$ For the parallel combination, as given in the question.
$\frac{\text{r}}{2} = \bigg(\frac{\text{E}}{\text{V} - 1 }\bigg)\frac{\text{R}}{2}$
$\therefore\text{r} = \bigg(\frac{1.5}{1.4} - 1 \bigg)\times\text{7}\Omega$
$ = \frac{0.1}{1.4}\times7\Omega$
$ = 0.5\Omega$
Alternate Answer

$\text{I} =\frac{\text{V}}{(\text{R} / 2 )}$
And $\text{E} = \text{V} - \text{I}(\text{r}/ 2 )$
This gives $\text{I} = \frac{1.4}{7/2}\text{A} = 0.4\text{A}$
$\therefore\frac{\text{r}}{2} = \frac{1.5 - 1.4}{0.4} = 0.25$
$\therefore\text{r} = 0.5\Omega$.
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