In a circuit,$5$ percent of the total current passes through a galvanometer. If the resistance of the galvanometer is $G$,then the value of the shunt is:

To convert a galvanometer into an ammeter,a shunt of $0.01\,\Omega$ is used. The resistance of the galvanometer coil is $50\,\Omega$ and its maximum deflection current is $20\,mA$. The range of the ammeter is:

$A$ $72 \; \Omega$ galvanometer is shunted by a resistance of $8 \; \Omega$. The percentage of the total current which passes through the galvanometer is $.....$

$A$ 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:

In an ammeter,$4 \%$ of the main current is passing through the galvanometer. If the shunt resistance is $5 \Omega$,then the resistance of the galvanometer will be: (in $\Omega$)

$A$ galvanometer of resistance $100\,\Omega$ has $50$ divisions on its scale and a sensitivity of $20\,\mu A/\text{division}$. It is to be converted into a voltmeter with three ranges: $0-2\,V$,$0-10\,V$,and $0-20\,V$. The appropriate circuit to do so is: