In order to pass $10\,\%$ of the main current through a moving coil galvanometer of $99\,\Omega$,the resistance of the required shunt is ............ $\Omega$.

If the number of turns in the coil of a moving coil galvanometer decreases, the resistance of the galvanometer,

In the adjoining circuit diagram,the readings of the ammeter and voltmeter are $2 \, A$ and $120 \, V$,respectively. If the value of $R$ is $75 \, \Omega$,then the voltmeter resistance will be (in $\Omega$):

$A$ galvanometer has a resistance of $100 \Omega$ and a current of $10 \text{ mA}$ produces full-scale deflection in it. The resistance to be connected to it in series,to convert it into a voltmeter of range $50 \text{ V}$,is: (in $\Omega$)

$A$ 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

$A$ galvanometer gives a full-scale reading of $50 \ mA$,when a potential difference across its terminals is $0.15 \ V$. It can be used as an ammeter of range $0 - 100 \ A$ by connecting a shunt resistance of: (in $\Omega$)