Question

Two resistors $400\Omega \text{and}800\Omega$ are connected in series across a $6\text{V}$ battery. The potential difference measured by a voltmeter of resistance $10\text{k}\Omega \text{across}400\Omega$ resistor is close to:

• A. $2\text{V}$
• B. $1.8\text{V}$
• C. $2.05\text{V}$
• D. $1.95\text{V}$

Answer 1

The correct option is D $1.95\text{V}$

The voltmeter of resistance $10\text{k}\Omega$ will be connected parallel to the resistance of $400\Omega$.

So, their equivalent resistance is
$\frac{1}{R^{\prime }}=\frac{1}{10000}+\frac{1}{400}$

$\Rightarrow \frac{1}{R^{\prime }}=\frac{1+25}{10000}=\frac{26}{10000}$

$\Rightarrow R^{\prime }=\frac{10000}{26}\Omega$

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
$V=I{R}^{\prime }=\left(\frac{6}{\frac{10000}{26}+800}\right)\frac{10000}{26}$

$\Rightarrow V=\frac{150}{77}=1.95\text{V}$

Hence, option $\left(D\right)$ is correct.