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

$A$ galvanometer coil has a resistance of $15\; \Omega$ and the meter shows full-scale deflection for a current of $4\; mA$. How will you convert the meter into an ammeter of range $0$ to $6\; A$?

$A$ galvanometer having $30$ divisions has a current sensitivity of $0.0625 \frac{\text{div}}{\mu A}$. If it is converted into a voltmeter to read a maximum of $6 \text{ V}$,then the resistance of that voltmeter is:

The resistance of a galvanometer coil is $R$. What is the shunt resistance required to convert it into an ammeter of range $4$ times?

$A$ circuit contains an ammeter,a battery of $30\,V$ and a resistance of $40.8\,\Omega$ all connected in series. If the ammeter has a coil of resistance $480\,\Omega$ and a shunt of $20\,\Omega$,the reading in the ammeter will be .................. $A$.

$A$ 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?