Two resistances of $400\,\Omega $ and $800\,\Omega $ are connected in series with a $6\,volt $ battery of negligible internal resistance. A voltmeter of resistance $10,000\,\Omega $ is used to measure the potential difference across $400\,\Omega $ . The error in the measurement of potential difference (in volt) approximately is
Diffcult
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The current
$\mathrm{I}=\frac{6}{400+800}=\frac{6}{1200}=\frac{1}{200}=5 \times 10^{-3} \mathrm{\,A}$
$\therefore $ Voltage drop across $400 \Omega=5 \times 10^{-3} \times 400$
$=2000 \times 10^{-3}=2 \mathrm{\,V}$
Because of the presence of the voltmeter having resistance $\mathrm{G}=10,000\, \mathrm{G}$ in parallel with $400\, \Omega$ the effective resistance is
${\frac{400 \times 10,000}{10,400}=\frac{10,000}{26} \Omega}$
$\therefore $ Voltage measured $=\frac{10,000}{26} \times 5 \times 10^{-3}=\frac{50}{26} \mathrm{\,V}$
$\therefore $ Relative error in the measurement
$=\frac{2-(50 / 26)}{2}=\frac{1}{26}=0.04 \mathrm{\,volt}$
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