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

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