In a galvanometer,$5 \%$ of the total current in the circuit passes through it. If the resistance of the galvanometer is $G$,the shunt resistance $S$ connected to the galvanometer is

$A$ galvanometer of resistance $50 \Omega$ is connected to a battery of $3 \text{ V}$ along with a resistance of $2950 \Omega$ in series. $A$ full-scale deflection of $30$ divisions is obtained in the galvanometer. In order to reduce this deflection to $20$ divisions,the resistance in series should be: (in $\Omega$)

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 $\Omega$):

$A$ milliammeter of resistance $40 \Omega$ has a range $0-30 \text{ mA}$. What will be the resistance used in series to convert it into a voltmeter of range $0-15 \text{ V}$ (in $\Omega$)?

The resistance of an ideal voltmeter is . . . . . . .

$A$ moving coil galvanometer has $100$ 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 Ohm's needed to be connected in series with the coil will be