A galvanometer of resistance $100\,\Omega $ has $50\, divisions$ on its scale and has sensitivity of $20\,\mu A / division$. It is to be converted to a voltmeter with three ranges, of $0-2\, V$, $0-10\, V$ and $0-20\, V$. The appropriate circuit to do so is
JEE MAIN 2019, Diffcult
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$20 \times 50 \times 10^{-6}=10^{-3}\, \mathrm{Amp}$
$\mathrm{v}_{1}=\frac{2}{10^{-3}}=100+\mathrm{R}_{1}$
$1900=\mathrm{R}_{1}$
$\mathrm{V}_{2}=\frac{10}{10^{-3}}=\left(2000+\mathrm{R}_{2}\right)$
$\mathrm{R}_{2}=8000$
$\mathrm{V}_{3}=\frac{20}{10^{-3}}=10 \times 10^{3}+\mathrm{R}_{3}=10 \times 10^{3} \mathrm{R}_{3}$
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