Question

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 volts, approximately, is

• A. 0.01
• B. 0.02
• C. 0.03
• D. 0.05

Answer 1

Answer: (D)

Before connecting voltmeter, potential difference across $400 \,\Omega$ resistance is

$V_{i}=\frac{400}{(400+800)} \times 6=2 \,V$
After connecting voltmeter, equivalent resistance
between $A$ and $B=\frac{400 \times 10000}{(400+10000)}=384.6\, \Omega$
Hence, potential difference measured by voltmeter
$V_{f}=\frac{384.6}{(384.6+800)} \times 6=1.95 \,V$
Error in measurement $=V_{i}-V_{f}$
$=2-1.95=0.05 \,V$