The reading of a high resistance voltmeter wh | vedclass

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(C) The electromotive force $(E)$ of the cell is the reading of the voltmeter when no current is drawn,so $E = 2.2\, V$.When a resistance $R = 5\,\Ome

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(C) The electromotive force $(E)$ of the cell is the reading of the voltmeter when no current is drawn,so $E = 2.2\, V$.When a resistance $R = 5\,\Omega$ is connected,the terminal voltage $V$ becomes $1.8\, V$.The formula for internal resistance $(r)$ is given by $r = \left( \frac{E}{V} - 1 ight) R$.Substituting the values: $r = \left( \frac{2.2}{1.8} - 1 ight) 	imes 5$.$r = \left( \frac{2.2 - 1.8}{1.8} ight) 	imes 5 = \left( \frac{0.4}{1.8} ight) 	imes 5$.$r = \left( \frac{4}{18} ight) 	imes 5 = \left( \frac{2}{9} ight) 	imes 5 = \frac{10}{9} \approx 1.11\,\Omega$.Rounding to one decimal place,we get $r = 1.1\,\Omega$.

The reading of a high resistance voltmeter when a cell is connected across it is $2.2\, V$. When the terminals of the cell are also connected to a resistance of $5\,\Omega$,the voltmeter reading drops to $1.8\, V$. Find the internal resistance of the cell in $\Omega$.

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