$20\%$ of the main current passes through the galvanometer. If the resistance of the galvanometer is $G$,then the resistance of the shunt will be

$A$ galvanometer coil has a resistance of $15\; \Omega$ and the meter shows full-scale deflection for a current of $4\; mA$. How will you convert the meter into an ammeter of range $0$ to $6\; A$?

$A$ moving coil galvanometer has $100$ turns and each turn has an area of $2.0 \,cm^2$. The magnetic field produced by the magnet is $0.01 \,T$ and the deflection in the coil is $0.05$ radian when a current of $10 \,mA$ is passed through it. The torsional constant of the suspension wire is $x \times 10^{-5} \,N-m / rad$. The value of $x$ is . . . . . . .

$A$ 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

$A$ moving coil galvanometer is converted into an ammeter reading up to $0.03\,A$ by connecting a shunt of resistance $4r$ across it,and into an ammeter reading up to $0.06\,A$ when a shunt of resistance $r$ is connected across it. What is the maximum current which can be sent through this galvanometer if no shunt is used?

$A$ maximum current of $0.5 \ mA$ can pass through a galvanometer of resistance $15 \ \Omega$. The resistance to be connected in series to the galvanometer to convert it into a voltmeter of range $0-10 \ V$ is (in $Omega$)