A voltmeter has resistance of G ohm and range of V volt. The value of resistance used in series to convert it into a voltmeter of range nV volt is:

Question

A voltmeter has resistance of G ohm and range of V volt. The value of resistance used in series to convert it into a voltmeter of range nV volt is: (A)  n G  (B)  (n-1) G  (C)  (G/n)  (D)  (G/n-1)

Tardigrade Admin · Accepted Answer

Let G is resistance of galvanometer and ig the current through it. From Ohms law ig=(V/G+R) ⇒ G+R=(V/ig) ⇒ R=(V/ig)-G Given, V=n V, ig=(V/G) ∴ R=(n V/V / G)-G =n G-G=(n-1) G

Q. A voltmeter has resistance of $G$ ohm and range of $V$ volt. The value of resistance used in series to convert it into a voltmeter of range $nV$ volt is:

$n G$

$(n - 1) G$

$\frac{G}{n}$

$\frac{G}{n - 1}$

Let $G$ is resistance of galvanometer and $i_{g}$ the current through it.
From Ohms law
$i_{g} = \frac{V}{G + R}$
$\Rightarrow G + R = \frac{V}{i _{g}}$
$\Rightarrow R = \frac{V}{i _{g}} - G$
Given, $V = nV,$
$i_{g} = \frac{V}{G}$
$∴ R = \frac{nV}{V / G} - G$
$= n G - G = (n - 1) G$

Q. A voltmeter has resistance of G ohm and range of V volt. The value of resistance used in series to convert it into a voltmeter of range nV volt is:

nG

(n−1)G

nG​

n−1G​