The resistance of a galvanometer coil is $R$. What is the shunt resistance required to convert it into an ammeter of range $4$ times?

The ratio of heats generated through shunt and galvanometer is $7: 5$ when they are connected to make an ammeter. If the resistance of the galvanometer is $112 \Omega$,then the resistance of the shunt is: (in $\Omega$)

$A$ galvanometer has a current range of $10 \text{ mA}$ and a voltage range of $0.75 \text{ V}$. To convert this galvanometer into an ammeter of range $10 \text{ A}$, what is the shunt resistance?

$A$ moving coil galvanometer has $150$ equal divisions. Its current sensitivity is $10$ divisions per milliampere and voltage sensitivity is $2$ divisions per millivolt. In order that each division reads $1 \, V$,the resistance in $\Omega$ needed to be connected in series with the coil will be

Two identical galvanometers are converted into a voltmeter and a millivoltmeter. As compared to the series resistance of the voltmeter,the series resistance of the millivoltmeter will be

$A$ galvanometer of resistance $G$ can be converted into a voltmeter of range $V$ by connecting a resistance $R$ in series with it. The series resistance required to change its range to $\frac{V}{3}$ is