A galvanometer of resistance G is converted into a voltmenter to measrue upto V volts by connecting a resistance `R_(1)` in series with the coil. If a resistance `R_(2)` is connected series with it, then it can measure upto 1//2 volts. Find the resistance, in terms of `R_(1)` and `R_(2)`, required to be connected to convert in into a voltmeter that can read upto 2V. Also find the resistance G of the galvanometer in terms of `R_(1)` and `R_(2)`

Explanation

Let G be the resistance of galvanometer and `l_(g)` be the current through galvanometer for full scale deflection. If this galvanometer is converted into voltmenter to measure upto V volts, then
`R_(1)=(Vl2)/(l_(g))-G=(V)/(2l_(g))-G`
Substracting (ii) from (i), we get
`R_(1)-R_(2)=(V)/(l_(g))-(V)/(2l_(g))=(V)/(2l_(g))`
or `(V)/(l_(g))=2(R_(1)-R_(2))`
Multiply (ii) with 2 and substracting from (i), we get
`R_(1)-2R_(2)=-G+2G=G, i.e., G=R_(1)-2R_(2)`
Let `R_(3)` resistance be connected in series to convert the galvanometer into a voltmenter of range 0 to 2V then
`R_(3)=(2V)/(l_(g))-G=2xx2(R_(1)-R_(2))-(R_(1)-2R_(2))=3R_(1)-2R_(2)`