Question

A galvanometer of resistance $100\Omega$ has $50$ divisions on its scale and has sensitivity of $20\mu \text{A}/\text{division}.$ It is to be converted to a voltmeter with three ranges, of $0-2\text{V},0-10\text{V}$ and $0-20\text{V}$ The appropriate circuit to do is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

The maximum current through the galvanometer is given by,

$i_g=20\mu \text{A}\times 50=1000\mu \text{A}=1\text{mA}$

$\text{Using,}V=i_g(G+R),$

$\text{When}V=2\text{volts}$

$\Rightarrow 2={10}^{-3}(100+R_1)$

$\Rightarrow R_1=1900\Omega$

$\text{When,}V=10\text{volts}$

$\Rightarrow 10={10}^{-3}(100+R_2+R_1)$

$\Rightarrow 10000=(100+R_2+1900)$

$\Rightarrow R_2=8000\Omega$

$\text{When,}V=20\text{volts}$

$\Rightarrow 20={10}^{-3}(100+R_3+R_2+R_1)$

$\Rightarrow 20000=(100+R_3+9900)$

$\Rightarrow R_3=10000\Omega$

Hence, option $\left(C\right)$ is correct.