$A$ galvanometer of resistance '$G$' is shunted by a resistance of '$S$' ohm. To keep the main current in the circuit unchanged,the resistance to be put in series with the galvanometer is:

$A$ voltmeter of resistance $R$ is to be used to measure the potential difference across a resistance $r$. In which case will the reading be nearest to the actual potential difference in the absence of the voltmeter?

$A$ galvanometer of resistance $40\,\Omega$ gives a deflection of $5\, \text{divisions}$ per $mA$. There are $50\, \text{divisions}$ on the scale. The maximum current that can pass through it when a shunt resistance of $2\,\Omega$ is connected is ................ $mA$.

$A$ galvanometer of resistance $200 \Omega$ is to be converted into an ammeter. The value of shunt resistance which allows $3 \%$ of the main current through the galvanometer is equal to (nearly) (in $Omega$)

$A$ galvanometer of resistance $25\,\Omega$ gives full-scale deflection for a current of $10\,mA$. It is to be converted into a voltmeter of range $100\,V$ by connecting a resistance $R$ in series with the galvanometer. The value of resistance $R$ in $\Omega$ is:

$A$ galvanometer with a resistance of $12 \,\Omega$ gives full-scale deflection when a current of $3 \, mA$ is passed. It is required to convert it into a voltmeter which can read up to $18 \, V$. The resistance to be connected is ............... $\Omega$.