$A$ galvanometer with $30$ divisions has a current sensitivity of $20 \, \mu A/division$. Its resistance is $25 \, \Omega$. How will you convert it into an ammeter to measure $1 \, A$ current?

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:

The coil of a galvanometer consists of $100$ turns and an effective area of $1 \, cm^2$. The restoring couple per unit radian is $10^{-8} \, N \cdot m/rad$. The magnetic field between the pole pieces is $5 \, T$. The current sensitivity of this galvanometer will be:

Derive the expression for the shunt resistance required to increase the current-measuring capacity of a galvanometer by a factor of $n$.

If an ammeter is to be used in place of a voltmeter,then we must connect with the ammeter a

An ammeter of resistance $20 \Omega$ gives full scale deflection when $1 \text{ mA}$ current flows through it. What is the maximum current that can be measured by connecting $4$ resistors each of $16 \Omega$ in parallel with the meter (in $\text{ mA}$)?