Question 355 of 380

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 voltmeter with three ranges, of $0-2V,0-10V$ and $0-20V.$ The appropriate circuit to do so is:

ABCD

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Answer: (D)

Given, the resistance of the galvanometer $G=100 \Omega$
No. of divisions $=50$
Current Sensitivity $=20 \mu A/division$
The maximum current through galvanometer will be
${I}_{max}=50\times 20\times {10}^{-6}={10}^{-3} A$
For ${V}_{1}=2V$
using Ohm's law ${V}_{1}={I}_{max}\left(G+{R}_{1}\right)$
$2={10}^{-3}\times \left(100+{R}_{1}\right)$
$∴{R}_{1}=1900 \Omega$
For ${V}_{2}=10V$
using Ohm's law ${V}_{2}={I}_{max}\left(G+{R}_{1}+{R}_{2}\right)$
$10={10}^{-3}\times \left(100+1900+{R}_{2}\right)$
$∴{R}_{2}=8000 \Omega$
For ${V}_{3}=20V$
using Ohm's law ${V}_{3}={I}_{max}\left(G+{R}_{1}+{R}_{2}+{R}_{3}\right)$
$20={10}^{-3}\left(100+1900+8000+{R}_{3}\right)$
$∴{R}_{3}=10\times {10}^{3} \Omega$