Question

Two resistances of 400 $\Omega$ and 800 $\Omega$ are connected in series with 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 V
• B. 0.02 V
• C. 0.03 V
• D. 0.05 V

Answer 1

The correct option is D 0.05 V

Before connecting voltmeter potential difference across 400W resistance is $$V_i=\frac{400}{(400+800)}\times 6=2V$$ After connecting voltmeter equivalent resistance between A and B $$=\frac{400\times 10,000}{(400+10,000)}=384.6\Omega$$ 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.