What will be the resistance of the shunt when $5 \%$ of the main current is passed through a galvanometer of resistance $G$?

When a resistance of $200\Omega$ is connected in series with a galvanometer of resistance $G$, its range is $V$. To triple its range, a resistance of $2000\Omega$ is connected in series. The value of $G$ is (in $\Omega$)

$A$ 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 $\text{ A}$)?

How can we convert a galvanometer into a voltmeter?

Explain the equation of a shunt.

$A$ galvanometer coil has a resistance of $10 \ \Omega$ and the meter shows full-scale deflection for $3 \ \text{mA}$. The value of the shunt required to convert this meter into an ammeter of range $0$ to $10 \ \text{A}$ is . . . . . . $\Omega$.