Question:
A galvanometer of resistance $100 \Omega$ has 50 divisions on its scale and has sensitivity of $20 \mu \mathrm{A} /$ division. It is to be converted to a voltmeter with three ranges, of $0-2 \mathrm{~V}$, $0-10 \mathrm{~V}$ and $0-20 \mathrm{~V}$. The appropriate circuit to do so is :
Correct Option: , 3
Solution:
(3) $i_{g}=20 \times 50=1000 \mu A=1 \mathrm{~mA}$
Using, $V=i_{g}(G+R)$, we have
$2=10^{-3}\left(100+R_{1}\right)$
$R_{1}=1900 \Omega$
when, $V=10$ volt
$10=10^{-3}\left(100+R_{2}+R_{1}\right)$
$10000=\left(100+R_{2}+1900\right)$
$\therefore R_{2}=8000 \Omega$