The resistance of a $1\, A$ ammeter is $0.018\,\Omega$. To convert it into a $10\, A$ ammeter,the shunt resistance required will be:

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:

Match the terms given in List-$I$ with suitable options provided in List-$II$ regarding the experiment to calculate the resistance and figure of merit of a galvanometer using the half-deflection method:

List-$I$	List-$II$
$A$. Figure of merit $(k)$	$P$. in series
$B$. Current sensitivity $(SI)$	$Q$. to get power loss minimized
$C$. Deflection is made half by using low resistance	$R$. in shunt/parallel
$D$. High resistance box	$S$. current per unit deflection
	$T$. deflection per unit current

$A$ multirange current meter can be constructed by using a galvanometer circuit as shown in the figure. We want a current meter that can measure $10 \text{ mA}$,$100 \text{ mA}$,and $1 \text{ A}$ using a galvanometer of resistance $10 \text{ } \Omega$ that produces maximum deflection for a current of $1 \text{ mA}$. Find the values of $S_1, S_2$,and $S_3$ that have to be used.

If only $1 \%$ of the total current is passed through a galvanometer of resistance $G$,then the resistance of the shunt is:

$A$ galvanometer coil has a resistance of $12\,\Omega$ and the meter shows full-scale deflection for a current of $3\,mA$. To convert it into a voltmeter of range $0-18\,V$,what resistance should be added and how?