Current sensitivity of a galvanometer is defined as deflection per unit current, $$\frac{\phi}{I}=\frac{NAB}{k}$$
Here, $\phi$ is the angular deflection, k is the torsional constant of the spring, and NAB is the magnetic torque per unit current produced in the galvanometer (MCG).
According to my book: A convenient way for the manufacturer to increase the current sensitivity is to increase the number of turns, N.
My problem: I can't understand the practicality of the current sensitivity. Suppose if we increase N two times the original value of a given galvanometer. It is likely that the resistance will also get doubled, thus the current would also get halved just like in the case of galvanometer turned voltmeter.
So, if I were to measure current using the galvanometer turned ammeter, for a normal current, say 1 A, it would have no significant effect on its angular deflection for if we double the number of turns the resistance would also get doubled and the normal current flowing through the galvanometer will also get halved giving the same value of angular deflection.
Best Answer
When an ammeter is used we usually try to ensure that the resistance of the ammeter is small compared to the resistances of the other elements in the circuit. We try to do this so that including the ammeter in the circuit changes the behaviour of the circuit as little as possible.
So if the resistance of the ammeter is negligible it remains negligible when doubled. However you are correct that there is a limit. If we raise the resistance of the ammeter too far it will start to significantly affect the circuit we are using the ammeter to examine.
Response to comment:
Suppose the ammeter resistance is $r$ and the full scale deflection current is $i$. I won't go through the working, but with a shunt resistor $R$ the full scale deflection current is given by:
$$ I_{fs} = i\left(1 + \frac{r}{R}\right) $$
If we're interested in measuring small currents then we don't use a shunt resistor so we get:
$$ I_{fs} = i $$
and the only way to make our meter more sensitive is to reduce $i$ by using more windings on the coil. I assume this is the point of increasing the windings.
If we're measuring large currents then $r/R \gg 1$ and our equation becomes approximately:
$$ I_{fs} \approx \frac{ir}{R} $$
As you say, $ir$ will be constant because increasing the number of windings decreases $i$ but increases $r$. In this case you're quite correct that there is no point in changing the number of windings.