Q. 144.8( 4 Votes )

A multirange volt

Answer :

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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