When a shunt of $4\,\Omega$ is connected to a galvanometer,the deflection reduces to $1/5$ of its initial value. If an additional shunt of $2\,\Omega$ is connected in parallel,what will be the new deflection?

Explain the equation of a shunt.

$A$ galvanometer gives full scale deflection with $0.006 \ A$ current. By connecting it to a $4990 \ \Omega$ resistance,it can be converted into a voltmeter of range $0-30 \ V$. If connected to a $\frac{2n}{249} \ \Omega$ resistance,it becomes an ammeter of range $0-1.5 \ A$. The value of $n$ is:

$A$ 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:

$A$ 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?