(a) State the principle of working of a galvonometer.
(b) A galvonometer of resistance G is converted into a voltmeter to measure upto 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 upto 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 upto 2 V. Also find the resistance G of the galvanometer in terms of `R_1 and R_2` .

Explanation

(b) As per question a galvanometer of resistance G is converted into a voltmeter of range V by connecting a series resistance `R_1`, so
`V = I_(g) (G + R_1) " "…..(i)`
And it is converted into a volmeter of range `V/2` by connecting a series resistance `R_2`, so,
`V/2 = I_(g) (G + R_2)" ".......(ii)`
To convert into a voltmeter of range 2 V we should connect a resistance R in series of it. Then,
`2V = I_(g)(G + R) " "......(iii)`
Dividing (i) by (ii), we have
`2 = (G + R_1)/(G + R_2) = G = (R_1 - 2R_2)`
Again dividing (iii) by (i), we get
`2 = (G + R)/(G + R_1) implies R = (G + 2R_1) = (R_1 - 2R_2) = (R_1 - 2R_2) = (3R_1 - 2R_2)`.