How can a galvanometer be converted into a ammeter and voltmeter?plz explain?

CONVERSION OF GALVANOMETER 
INTO AMMETER

  Since Galvanometer is a very sensitive instrument therefore it can’t measure heavy currents. In order to     convert a Galvanometer into an Ammeter, a very low resistance known as "shunt" resistance is     connected in parallel to Galvanometer. Value of shunt is so adjusted that most of the current passes     through the shunt. In this way a Galvanometer is converted into Ammeter and can measure heavy     currents without fully deflected.

Let resistance of galvanometer = Rg and it gives full-scale deflection when current Ig is passed through     it. Then,

Vg = IgRg -------(i)

Let a shunt of resistance (Rs) is connected in parallel to galvanometer. If total current through the     circuit is I.

Then current through shunt:

Is = (I-Ig)

potential difference across the shunt:

Vs= IsRs

or                Vs = (I – Ig)Rs -------(ii)

But

Vs =Vg (I - Ig)Rs = IgRg

CONVERSION OF GALVANOMETER INTO VOLTMETER

Since Galvanometer is a very sensitive instrument, therefore it can not measure high potential     difference. In order to convert a Galvanometer into voltmeter, a very high resistance known as "series     resistance" is connected in series with the galvanometer.

VALUE OF SERIES RESISTANCE

Let resistance of galvanometer = Rg and resistance Rx (high) is connected in series to it. Then     combined resistance = (Rg + Rx).

If potential between the points to be measured = V and if galvanometer gives full-scale deflection,     when current "Ig" passes through it. Then,

V = Ig (Rg + Rx)  V = IgRg + IgRx  V – IgRg = IgRx  Rx = (V – IgRg)/Ig

Thus Rx can be found.