$A$ galvanometer has a resistance of $80 \Omega$ and it is shunted with a resistance of $20 \Omega$. If $20 \%$ of the main current flows through the galvanometer, what is the value of the main current (in $\text{ A}$)?

In the moving coil galvanometer, if the number of turns in the coil is doubled, then the current sensitivity . . . . . . and the voltage sensitivity . . . . . . .

Current sensitivities of two galvanometers $G_1$ and $G_2$ of resistances $100 \ \Omega$ and $50 \ \Omega$ are $10^8 \ \text{div/A}$ and $0.5 \times 10^5 \ \text{div/A}$ respectively. The galvanometer in which the voltage sensitivity is more is

$A$ galvanometer has $30$ divisions and a current sensitivity of $20 \, \mu A$ per division. Its resistance is $25 \, \Omega$. How will you convert this galvanometer into a voltmeter of $1 \, V$ range? (Find the series resistance 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:

The deflection in a moving coil galvanometer is reduced to half when it is shunted with a $X \ \Omega$ coil. The relation between $X$ and the resistance of the galvanometer $G$ is: