$A$ millivoltmeter of $25 \, mV$ range is to be converted into an ammeter of $25 \, A$ range. The value (in $\Omega$) of the necessary shunt will be:

$A$ galvanometer having a resistance $G$ and current $I_g$ flowing in it,produces full-scale deflection. If $S_1$ is the value of the shunt which converts it into an ammeter of range $0-I$ and $S_2$ is the value of the shunt for range $0-8I$,then the ratio $\frac{S_1}{S_2}$ will be:

Give the uses of a galvanometer.

To find the resistance of a galvanometer by the half-deflection method,the experimental data obtained are given in the table below:

$S. No.$	Resistance $R \ (\Omega)$	Deflection $(\theta)$	Shunt $S \ (\Omega)$	Half deflection $(\theta / 2)$	Galvanometer resistance $(G)$
$1$	$3300$	$30$	$80$	$15$	$G_1$
$2$	$5000$	$20$	$80$	$10$	$G_2$

From the above data,the galvanometer resistance $G$ will be near to: (in $\Omega$)

$A$ galvanometer of $10 \,\Omega$ resistance gives full scale deflection with $0.01 \, A$ of current. It is to be converted into an ammeter for measuring $10 \, A$ current. The value of shunt resistance required will be

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: