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

Explanation

(b) Ist case
`I_(g)` = Maximum current through galvanometer
`R_(1) = V/I_(g) - G ` ….(i)
`2^(nd) "case "`
`R_(2) = V/(2I_(g)) - G ` ….(ii)
`3^(rd) ` case
` R = (2V)/I_(g) - G ` ….(iii)
From eq. (i) and (ii)
`R_(1) - R_(2) = V/I_(g) - G - V/(2I_(g)) + G = V/(2I_(g))`
`2(R_(1) - R_(2)) = V/I_(g)`
Putting the value of `V/I_(g)` in (i) we get
`R_(1) = 2(R_(1) - R_(2)) - G`
` R_(1) = 2R_(1) - 2R_(2) - G`
` G = 2R_(1) - 2R_(2) - R_(1)`
` = R_(1) - 2R_(2)`
Putting the value of `V/I_(g)` and G in eq. (iii) we get :
`R = 2[2(R_(1) - R_(2))] - (R_(1) - 2R_(2))`
` = 4R_(1) - 4R_(2) - R_(1) + 2R_(2)`
` = 3R_(1) - 2R_(2)`