Derive the expression for the shunt resistance required to increase the current-measuring capacity of a galvanometer by a factor of $n$.

$A$ milliammeter of range $10\, mA$ and resistance $9\, \Omega$ is joined in a circuit as shown. The meter gives full-scale deflection for current $I$ when $A$ and $B$ are used as its terminals,i.e.,current enters at $A$ and leaves at $B$ ($C$ is left isolated). The value of $I$ is

In the experiment to determine the galvanometer resistance by the half-deflection method,the plot of $\frac{1}{\theta}$ vs the resistance $(R)$ of the resistance box is shown in the figure. The figure of merit of the galvanometer is .............. $\times 10^{-1} \text{ A/division}$. [The source has emf $E = 2 \text{ V}$]

The galvanometer has a resistance of $1.8 \Omega$. Calculate the value of shunt to increase the range of galvanometer by $10$ times. (in $\Omega$)

$A$ shunt of resistance $1\,\Omega$ is connected across a galvanometer of $120\,\Omega$ resistance. $A$ total current of $5.5\,A$ is passed through the combination,which gives full-scale deflection in the galvanometer. The current that would give full-scale deflection in the absence of the shunt is nearly ............... $A$.

There are three voltmeters of the same range but of resistances $10000\,\Omega$,$8000\,\Omega$,and $4000\,\Omega$ respectively. The best voltmeter among these is the one whose resistance is ................ $\Omega$.