# A multirange volt

Given :

Resistance of the galvanometer (G) = 10 ohms

Current measured in the galvanometer (Ig) = 1 mA

We know that potential difference V = IR

R = V/I

For a Voltmeter measuring 2V

G + R1 = V/Ig (As G and R1 are in series)

R1 = V/Ig - G

= (2/10-3)-10

= 1990 Ω

For a Voltmeter measuring 20V

G + R2 + R1 = V/Ig ( As G, R1 and R2 are in series)

R2 = V/Ig - G -R1

= (20/10-3)-10-1990

= 18 KΩ

For a Voltmeter measuring 200V

G + R3 + R2 + R1 = V/Ig

R3 = V/Ig - G - R1-R2

= (200/10-3)-10- 1990- 18000

= 180 KΩ

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