When an additional resistance $1980 \ \Omega$ is connected in series with a voltmeter,each scale division has $100$ times larger value. The resistance of the voltmeter is (in $\Omega$)

$20\%$ of the main current passes through the galvanometer. If the resistance of the galvanometer is $G$,then the resistance of the shunt will be

$A$ galvanometer can be converted to a voltmeter of full-scale deflection $V_{0}$ by connecting a series resistance $R_{1}$ and can be converted to an ammeter of full-scale deflection $I_{0}$ by connecting a shunt resistance $R_{2}$. What is the current flowing through the galvanometer at its full-scale deflection?

Assertion: To convert a galvanometer into an ammeter,a small resistance is connected in parallel with it.
Reason: The small resistance increases the combined resistance of the combination.

$A$ galvanometer having a coil resistance of $60\,\Omega$ shows full-scale deflection when a current of $1.0\,A$ passes through it. It can be converted into an ammeter to read currents up to $5.0\,A$ by:

To convert a galvanometer into a voltmeter,one should connect a