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

Q: 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

c (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.$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

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

A$\frac{{10}}{9}$
B$0.1$
C$1$
D$10$

Diffcult

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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