$A$ $36 \,\Omega$ galvanometer is shunted by a resistance of $4 \,\Omega$. The percentage of the total current that passes through the galvanometer is (in $\%$)

$A$ moving coil galvanometer has $N$ number of turns in a coil of effective area $A$,and it carries a current $I$. The magnetic field $B$ is radial. The torque acting on the coil is:

$A$ galvanometer having a coil of resistance $30 \ \Omega$ needs $20 \ \text{mA}$ of current for full-scale deflection. If a maximum current of $3 \ \text{A}$ is to be measured using this galvanometer,the resistance of the shunt to be added to the galvanometer should be $\frac{30}{X} \ \Omega$,where $X$ is

$A$ galvanometer of resistance $40 \Omega$ gives a deflection of $10 \text{ divisions per } mA$. There are $50 \text{ divisions}$ on the scale. The maximum current that can pass through the circuit when a shunt resistance of $2 \Omega$ is connected is: (in $\text{ mA}$)

$A$ galvanometer has a resistance of $20\,\Omega$ and reads full-scale when $0.2\,V$ is applied across it. To convert it into a $10\,A$ ammeter,the galvanometer coil should have a

$A$ 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