The maximum permissible current through a galvanometer (G) of resistance 10ohmams is 0.05A. A resistance (R) is used to convert G into a voltameter with a maximum reading of 100v . find the value of R and how it is connected to G an?

Answer 1

full scale voltage=.05*10=.5volts.
so for 100 volt, 99.5 volts=.05(10+RinSeries) solve for the series resistance.

99.5/.05=10+R=1990
R=1980 ohms

Answer 2

R=1980