$A$ moving coil galvanometer is converted into an ammeter of range $0$ to $5 \, mA$. The galvanometer resistance is $90 \, \Omega$ and the shunt resistance has a value of $10 \, \Omega$. If there are $50$ divisions in the galvanometer-turned-ammeter on either side of zero, its current sensitivity is

$A$ moving coil galvanometer of resistance $100 \ \Omega$ shows full-scale deflection when a current of $100 \ \mu A$ passes through it. If it is intended to show full-scale deflection when a current of $1 \ mA$ passes through it,the value of shunt resistance in ohms to be connected to the galvanometer is:

In the given figure,an ammeter $A$ consists of a $240 \Omega$ coil connected in parallel to a $10 \Omega$ shunt. The reading of the ammeter is . . . . . . $mA$.

Explain the solution to the difficulties that arise when using a galvanometer directly as an ammeter.

The resistance of a galvanometer is $50\,\Omega$ and the current required to give full-scale deflection is $100\,\mu A$. In order to convert it into an ammeter for reading up to $10\,A$,it is necessary to put a resistance of:

Moving coil galvanometers $M_{1}$ and $M_{2}$ have resistance,number of turns,area of coil,and magnetic field as follows:
$R_{1}=10 \Omega, R_{2}=14 \Omega, N_{1}=30, N_{2}=42$
$A_{1}=3.6 \times 10^{-3} \ m^{2}, A_{2}=1.8 \times 10^{-2} \ m^{2}, B_{1}=0.25 \ T, B_{2}=0.50 \ T$
(Spring constants are same for both galvanometers).
The ratio of $(i)$ current sensitivity and (ii) voltage sensitivity for galvanometer ($M_{2}$ to $M_{1}$) is respectively: