Two resistors $400 \Omega$ and $800 \Omega$ are connected in series across a $6 \mathrm{~V}$ battery. The potential difference measured by a voltmeter of $10 \mathrm{k} \Omega$ across $400 \Omega$ resistor is close to:
Correct Option: , 4
(4) The voltmeter of resistance $10 \mathrm{k} \Omega$ is parallel to the resistance of $400 \Omega$. So, their equivalent resistance is
$\frac{1}{R^{\prime}}=\frac{1}{10 k \Omega}+\frac{1}{400 \Omega}=\frac{1}{10000}+\frac{1}{400}$
$\Rightarrow \frac{1}{R^{\prime}}=\frac{1+25}{10000}=\frac{26}{10000}$
$\Rightarrow R^{\prime}=\frac{10000}{26} \Omega$
Using Ohm's law, current in the circuit
$I=\frac{\text { Voltage }}{\text { Net Resistance }}=\frac{6}{\frac{10000}{26}+800}$
Potential difference measured by voltmeter
$V=I R^{\prime}=\frac{6}{\frac{10000}{26}+800} \times \frac{10000}{26}$
$\Rightarrow V=\frac{150}{77}=1.95$ volt