To convert a galvanometer into an ammeter,one needs to connect a

$A$ galvanometer with its coil resistance $25\,\Omega$ requires a current of $1\,mA$ for its full-scale deflection. In order to construct an ammeter to read up to a current of $2\,A$,the approximate value of the shunt resistance should be:

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

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

$A$ milliammeter of resistance $40 \Omega$ has a range $0-30 \text{ mA}$. What will be the resistance used in series to convert it into a voltmeter of range $0-15 \text{ V}$ (in $Omega$)?

Two similar galvanometers are converted into an ammeter and a milliammeter. The shunt resistance of the ammeter as compared to the shunt resistance of the milliammeter will be