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 $\Omega$):

The resistance of a galvanometer is $50\,\Omega$ and the current required to give full-scale deflection is $100\,\mu A$. In order to convert it into an ammeter for reading up to $10\,A$,it is necessary to put a resistance of:

When a shunt resistance of $4r$ is connected to a galvanometer,it becomes an ammeter that can measure $0.03 \, A$. When a shunt resistance of $r$ is connected to the same galvanometer,it becomes an ammeter that can measure $0.06 \, A$. What is the current capacity $(i_g)$ of the galvanometer in $A$?

$A$ galvanometer of resistance $100 \Omega$ gives full scale deflection for a current of $1 \text{ mA}$. It is converted into an ammeter of range $0 - 10 \text{ A}$. The shunt required is: (in $\text{ }\Omega$)

$A$ galvanometer of resistance $100 \Omega$ requires $10 \mu A$ current for full-scale deflection. If a shunt resistance of $1 \Omega$ is connected in parallel to convert it into an ammeter, what is the minimum current required to obtain full-scale deflection (in $\text{ mA}$)?

$A$ voltmeter has a range $V$ and resistance $G \, \Omega$. The value of the series resistance required to convert this voltmeter into a voltmeter of range $nV$ is ....