In the adjoining circuit, the $e.m.f.$ of the cell is $2\, volt$ and the internal resistance is negligible. The resistance of the voltmeter is $80 \,ohm$. The reading of the voltmeter will be ............. $volt$
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(c) Total resistance of the circuit $ = \frac{{80}}{2} + 20 = 60\,\Omega $
$ \Rightarrow $ Main current $i = \frac{2}{{60}} = \frac{1}{{30}}A$
Combination of voltmeter and $80$ $\Omega$ resistance is connected in series with $20$ $\Omega$, so current through $20$ $\Omega$ and this combination will be same $ = \frac{1}{{30}}A$.
Since the resistance of voltmeter is also $80$ $\Omega$, so this current is equally distributed in $80$ $\Omega$ resistance and voltmeter (i.e.$\frac{1}{{60}}A$ through each)
$P.D.$ across $80$ $\Omega$ resistance $ = \frac{1}{{60}} \times 80 = 1.33\,V$
$ \Rightarrow $ Main current $i = \frac{2}{{60}} = \frac{1}{{30}}A$
Combination of voltmeter and $80$ $\Omega$ resistance is connected in series with $20$ $\Omega$, so current through $20$ $\Omega$ and this combination will be same $ = \frac{1}{{30}}A$.
Since the resistance of voltmeter is also $80$ $\Omega$, so this current is equally distributed in $80$ $\Omega$ resistance and voltmeter (i.e.$\frac{1}{{60}}A$ through each)
$P.D.$ across $80$ $\Omega$ resistance $ = \frac{1}{{60}} \times 80 = 1.33\,V$
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