Question

Two resistances of $400\Omega$ and $800\Omega$ 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 volts approximately is

• A. 0.1
• B. 0.02
• C. 0.08
• D. 0.05

Answer 1

The correct option is D 0.05

$I=\frac{6}{400+800}=5\times {10}^{-3}A$
$\therefore$ Voltage drop across $400\Omega =5\times {10}^{-3}\times 400=2V$
Because of the presence of the voltmeter having resistance $G=10,000\Omega$ in parallel with $400\Omega$, the effective resistance is
$\frac{400\times 10,000}{10,400}+800=\frac{10,000}{26}\Omega +800=1184.6\Omega$
Current through the circuit will be now : $5.06\times {10}^{-3}A$
which gives us the voltage through 4 ohm resistance as 1.95 V.
Therefore, the error comes out to be 0.05 V