$A$ galvanometer has a resistance of $7\,\Omega$ and gives a full-scale deflection for a current of $1.0\,A$. How will you convert it into a voltmeter of range $10\,V$?

The relation between voltage sensitivity $({\sigma _V})$ and current sensitivity $({\sigma _i})$ of a moving coil galvanometer is (Resistance of Galvanometer = $G$)

The coil in a moving coil galvanometer experiences a torque proportional to the current passed through it. If a steady current $i$ is passed through it,the deflection of the coil is found to be $90^o$. Now,the steady current is switched off and a charge $Q$ is suddenly passed through the coil. If the coil has $N$ turns of area $A$ each and its moment of inertia is $I$ about the axis it is going to rotate,then the maximum angle through which it deflects upon passing $Q$ is:

$A$ galvanometer with a resistance of $8 \, \Omega$ is connected to a shunt of $2 \, \Omega$. If the total current is $1 \, A$,how much current in $A$ passes through the shunt?

An ammeter whose resistance is $180\,\Omega$ gives full scale deflection when the current is $2\,mA$. The shunt required to convert it into an ammeter reading $20\,mA$ (in $\Omega$) is

$A$ galvanometer of resistance $40\,\Omega$ gives a deflection of $5\, \text{divisions}$ per $mA$. There are $50\, \text{divisions}$ on the scale. The maximum current that can pass through it when a shunt resistance of $2\,\Omega$ is connected is ................ $mA$.