Question

A battery of $9$ volt is connected to a $3K\Omega$ resistor as shown in the figure. The potential difference between two points $(A,B)$ of the resistor is measured using (i) $20K\Omega$ voltmeter and (ii) $1K\Omega$ voltmeter. Which voltmeter will have higher reading? Explain it by calculations.

Answer 1

Current in the given circuit will be $I=\frac{V}{R}=\frac{9}{3000}=0.003A$. Potential difference across $3K\Omega$ will be equal to $0.003A\times 3000\Omega =9V$. The voltmeter must be connected parallel to the resistor. This is necessary because in parallel it experiences the same potential difference across the resistor. So the final reading on the voltmeter will be $9V$. The resistances of voltmeters will not play any role.