Question

$A$ galvanometer of resistance $10 \, \Omega$ gives full-scale deflection when $1 \,mA$ current passes through it. The resistance required to convert it into a voltmeter reading upto $2.5 \,V$ is

• A. $24.9 \, \Omega$
• B. $249 \, \Omega$
• C. $2490\, \Omega$
• D. $24900\, \Omega$

Answer 1

Answer: (C)

Here, $I_{g}=1\,mA$
$=1\times10^{-3}\,A$,
$G=10\,\Omega$,
$V=2.5\, V$
From the figure

$V=I_{g}(G+R)$ or $ R=\frac{V}{I_{g}}-G$
Substituting the given values, we get,
$ R=\frac{2.5\,V}{1\times10^{-3}\,A}-10\,\Omega$
$=2500\,\Omega-10\,\Omega$
$=2490\,\Omega$