A multirange voltmeter can be constructed by using a galvanometer circuit as shown in figure. We want to construct a voltmeter that can measure 2V, 20V and 200V using a galvanometer of resistance `10Omega` and that produces maximum deflection for current of `1mA`. Find `R_1`, `R_2` and `R_3` that have to be used.

Explanation

Given,
Resistance of galvanometer, `R_(G) = 10 Omega`
Current through the galvanometer,`I_(G) = 1 mA = 10^(-3) A`
(i) For a voltameter measuring a voltage of 2 V, resistance `R_(1)` is connected in circuit with the galvanometer.
` :. " " I_(G) xx (R_(G) + R_(1)) = 2`
` rArr 10^(_3) (10 + R_(1)) = 2`
` rArr 10^(-2) + 10^(-3) R_(1) = 2`
` rArr 10^(-3) R_(1) = 2 - 0.01`
` rArr R_(1) = 1,990 Omega`
(ii) For a voltmeter measuring a voltage of 20 V,
resistance `R_(2)`, is connected in circuit with the galvanometer and resistance `R_(1)`.
` :. I_(G) xx (R_(G) + R_(1)+R_(2)) = 20`
` rArr 10^(_3) (10 + 1,990 + R_(2)) = 20`
` rArr 10^(-2) + 1.99 + 10^(_3) R_(2) = 20`
` rArr 10^(-3) R_(2) = 20 - 2`
`rArr R_(2) = 18,000 Omega = 18 kOmega`
(iii) For a voltmeter measuring a voltage of 200 V, resistance `R_(3)` is connected in circuit with the galvanometer and resistances `R_(1)" and "R_(2)`.
`:. I_(G) xx (R_(G) + R_(1) + R_(2) + R_(3)) = 200`
`rArr 10^(-3) (10 + 1,990 + 18,000 + R_(3)) = 200`
` rArr 10^(-3) (20,000 + R_(3)) = 200`
` rArr 10^(-3) R_(3) = 200 - 20 = 180`
` rArr R_(3) = 180000 Omega = 180 k Omega`