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Two resistances of $400 \Omega$ and $800 \Omega$ are connected in series with $6 \mathrm{~V}$ battery of negligible internal resistance. A voltmeter of resistance $10000 \Omega$ is used to measure the potential difference across $400 \Omega$. The error in the measurement of potential difference in volts approximately is :
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Verified Answer
The correct answer is:
0.05
$R_1=400 \Omega, R_2=800 \Omega$
$P D$ across $400 \Omega$ resistance
(when voltmeter is not connected)
$V_1=\frac{6}{(400+800)} \times 400$
$=\frac{6 \times 400}{1200}=2 \mathrm{~V}$
when voltmeter is connected
Total resistance of the circuit,
$R=\left(\frac{10000 \times 400}{10000+400}\right)+800$
$=\frac{10000 \times 400}{10400}+800$
$=\frac{40000}{104}+800$
$=\frac{40000+83200}{104}$
$=\frac{123200}{104}=\frac{30800}{26}=\frac{15400}{13}$
New $P D$ across $400 \Omega$ resistance,
$V_2=6-800\left(\frac{6}{15400 / 13}\right)$
$=6-\frac{800 \times 13 \times 6}{15400}$
$=6-4.052=1.95 \mathrm{~V}$
$\therefore \quad$ Error $=V_1-V_2$
$=2-1.95$
$=0.05$
$P D$ across $400 \Omega$ resistance
(when voltmeter is not connected)
$V_1=\frac{6}{(400+800)} \times 400$
$=\frac{6 \times 400}{1200}=2 \mathrm{~V}$
when voltmeter is connected
Total resistance of the circuit,
$R=\left(\frac{10000 \times 400}{10000+400}\right)+800$
$=\frac{10000 \times 400}{10400}+800$
$=\frac{40000}{104}+800$
$=\frac{40000+83200}{104}$
$=\frac{123200}{104}=\frac{30800}{26}=\frac{15400}{13}$
New $P D$ across $400 \Omega$ resistance,
$V_2=6-800\left(\frac{6}{15400 / 13}\right)$
$=6-\frac{800 \times 13 \times 6}{15400}$
$=6-4.052=1.95 \mathrm{~V}$
$\therefore \quad$ Error $=V_1-V_2$
$=2-1.95$
$=0.05$
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