Explain how a Galvanometer can be converted into Voltmeter.

Explanation

Ammeter : It is a low resistance instrument used to measure electric current in a circuit. An ideal ammeter should have zero resistance. It is always connected in series in an electrical circuit.
Conversion of galvanometer into an ammeter : Consider a galvanometer of resistance G. It can be converted into an ammeter by connecting a low resistance called shunt (S) in series with it.
The value of shunt depends upon the range of current to be measured.
Suppose we want to convert a galvanometer into an ammeter of range`0-I`ampere.
`I_(g) =` Maximum current that can pass through the galvanometer.
n = Number of divisions in the galvanometer scale
k = Figure of merit (or current for deflection of the one division in the galvanometer)
`therefore I_(g) = nk`
the value of S is so adjusted that out of maximum current I, only a part `I_(g)` should flow through the galvanometer and the remaining `(I-I_(g))` flows through the shunt.
As G and S are connected in parallel with each other, so
So `V_(A)-V_(B) =I_(g) G = (I-I_(g))S`
`S = ((I_(g))/(I-I_(g)))G`
Thus, s can be determined.
So by connecting shunt resistance of suitable value in series with a galvanometer, we can get an ammeter of range `0-I` ampere.
Voltmeter : It is a high resistance instrument which is used to measure potential difference across the two points in a circuit.
Resistance of an ideal voltmeter is infinite.
Voltmeter is always connected in parallel in an electrical circuit.
Conversion of galvanometer into a voltmeter : Consider a galvanometer of resistance G.We want to convert it into a voltmeter of range 0 to V volt.

It can be done by connecting a high resistance R in parallel withe the galvanometer.
The value of R depends upon the range of voltage to be measured.
Let `I_(g) =` Current that gives full scale deflection in galvanometer.
n = Number of divisions in the galvanometer scale
k = Figure of merit
`therefore I_(g) =nk`.
The value of R is connected in series with the galvanometer so that a current `I_(g)`flows through the galvanometer when a potential difference V is applied across the points A and B.
As G and R are connected in series, so
`V = I_(g) (G +R)` (By Ohm.s law)
`G + R = (V)/(I_(g))`
If a resistance of the value given by above eqn. is connected in series with the galvanometer, it will work as a voltmeter of the range `(0-V)` volt.