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$A$ galvanometer coil has a resistance $80 \Omega$ and current for full-scale deflection is $10 \text{ mA}$. $A$ resistance of $920 \Omega$ is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is $0.2 \text{ V}$,the number of divisions on the scale is:

$A$ current of $10^{-3} \ A$ is flowing through a resistance of $1000 \ \Omega$. To measure the correct potential difference across this resistance,a voltmeter must be used whose resistance should be:

$A$ galvanometer of resistance $30 \Omega$ is connected to a battery of emf $2 \text{ V}$ with $1970 \Omega$ resistance in series. $A$ full-scale deflection of $20$ divisions is obtained in the galvanometer. To reduce the deflection to $10$ divisions,the total resistance in series required is: (in $\Omega$)

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

Only $4 \%$ of the total current in the circuit passes through a galvanometer. If the resistance of the galvanometer is $G$,then the shunt resistance connected to the galvanometer is:

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