The potential difference across the $100\,\Omega$ resistance in the following circuit is measured by a voltmeter of $900 \,\Omega$ resistance. The percentage error made in reading the potential difference is
Diffcult
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(c) Before connecting the voltmeter, potential difference across $100\,\Omega $ resistance
${V_i} = \frac{{100}}{{(100 + 10)}} \times V = \frac{{10}}{{11}}\,V$
Finally after connecting voltmeter across $100\,\Omega $ Equivalent resistance $\frac{{100 \times 900}}{{(100 + 900)}} = 90\,\Omega $
Final potential difference
${V_f} = \frac{{90}}{{(90 + 10)}} \times V = \frac{9}{{10}}\,V$
$\%$ error = $\frac{{{V_i} - {V_f}}}{{{V_i}}} \times 100$
$ = \frac{{\frac{{10}}{{11}}\,V - \frac{9}{{10}}\,V}}{{\frac{{10}}{{11}}\,V}} \times 100 = 1.0.$
${V_i} = \frac{{100}}{{(100 + 10)}} \times V = \frac{{10}}{{11}}\,V$
Finally after connecting voltmeter across $100\,\Omega $ Equivalent resistance $\frac{{100 \times 900}}{{(100 + 900)}} = 90\,\Omega $
Final potential difference
${V_f} = \frac{{90}}{{(90 + 10)}} \times V = \frac{9}{{10}}\,V$
$\%$ error = $\frac{{{V_i} - {V_f}}}{{{V_i}}} \times 100$
$ = \frac{{\frac{{10}}{{11}}\,V - \frac{9}{{10}}\,V}}{{\frac{{10}}{{11}}\,V}} \times 100 = 1.0.$
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