Question

A voltmeter has a resistance $G$ and range $V$ . Calculate the resistance to be used in series with it to extend its range to $nV$.

Answer 1

Express the present within the voltmeter before connecting the external resistance within the series. After connecting the external resistance within the series with voltmeter, the range of the voltmeter will increase to $nV$ , where $V$ is that the voltage range of the voltmeter within the first case.

Formula used:
Ohm’s law, $V = IR$

Complete step by step answer:
Potential difference between $A$ and $B$ is given by ${V_{AB}}{\text{ }} = nV$
Current through galvanometer ${I_g} = {\text{ }}\dfrac{V}{G}$
Using ${V_{AB}} = {I_g}\left( {G + R} \right)$
$nV = GV{\text{ }}\left( {G + R} \right)$
$\Rightarrow R = {\text{ }}\left( {n - 1} \right)G$

A galvanometer is often converted into a voltmeter by connecting an outsized resistance serial to the galvanometer. Resistance of a perfect voltmeter is infinite. A voltmeter is usually connected in parallel to the circuit component across which voltage is to be measured.If we've movement with $50mA\,FSD$ , measuring $2.5{\text{ }}V$ full scale, this involves voltmeter resistance of $2.5{\text{ }} \times \dfrac{{{{10}^6}}}{{50}} = 50{\text{ }} \times {10^4}ohms = 500Kohm$ , whereas if a meter with $1{\text{ }}mA{\text{ }}FSD$ is employed for $2000{\text{ }}V$ measurement, its total resistance are going to be $2000{\text{ }} \times \dfrac{{1000}}{1} = 2{\text{ }}Mega\,ohms$ .

Since, the effective resistance of the net-work increases are connected serially, hence the resistance of the voltmeter is quite that of a galvanometer.

Note: It should be noted that resistance of a perfect voltmeter should be infinite. For ammeter, it should be ideally zero. For practical voltmeter, the resistance is in mega ohms.Higher the full-scale voltage, higher is the resistance.