In a galvanometer,$5 \%$ of the total current in the circuit passes through it. If the resistance of the galvanometer is $G$,the shunt resistance $S$ connected to the galvanometer is

$A$ galvanometer having a coil resistance $100 \; \Omega$ gives a full-scale deflection when a current of $1 \; mA$ is passed through it. What is the value of the resistance in $k\Omega$ which can convert this galvanometer into a voltmeter giving full-scale deflection for a potential difference of $10 \; V$?

Two identical galvanometers are converted into a voltmeter and a millivoltmeter. As compared to the series resistance of the voltmeter,the series resistance of the millivoltmeter will be

Why is the total resistance of a voltmeter kept very high?

$A$ galvanometer whose resistance is $120 \, \Omega$ gives full-scale deflection with a current of $0.05 \, A$. To enable it to read a maximum current of $10 \, A$,a shunt resistance is added in parallel with it. The resistance of the ammeter so formed is .............. $\Omega$.

If only $2 \%$ of the main current is to be passed through a galvanometer of resistance $G$,then the resistance of the shunt connected to it will be: