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

Explanation

Energy stored in a capacitor, `E=(1)/(2)CV^(2)`
In parallel, `0.25=(1)/(2)(C_(1)+C_(2))(100)^(2)`
In series, `0.045 =(1)/(2)((C_(1)C_(2))/(C_(1)+C_(2)))(100)^(2)`
From (i) `C_(1)+C_(2)=0.25xx2xx10^(-4)`
`C_(1)+C_(2)=5xx10^(-5)`
From (ii) `(C_(1)C_(2))/(C_(1)+C_(2))0.045xx2xx10^(-4)`
`(C_(1)C_(2))/(C_(1)+C_(2))=0.09xx10^(-4)=9xx10^(-6)`
Form (iii) `C_(1)C_(2)=(2xx0.045xx5xx10^(-5))/(10^(4))=4.5xx10^(-10)`
`C_(1)-C_(2)=sqrt((C_(1)+C_(2))^(2)-4C_(1)C_(2))`
`C_(1)-C_(2)=2.64xx10^(-5)`
Solving (ii) and (iv) `C_(1)=38.2 mu F`
`C_(2)=11.8 mu F`
In parallel `Q_(1)=C_(1)V=38.2xx10^(-6)xx100=38.2xx10^(-4)C`
`Q_(2)=C_(2)V=11.8xx10^(-6)xx100=11.2xx10^(-4)C`