$A$ galvanometer with $50$ divisions on the scale has a resistance of $25 \Omega$. $A$ current of $2 \times 10^{-4} \text{ A}$ gives a deflection of one scale division. The additional series resistance required to convert it into a voltmeter reading up to $25 \text{ V}$ is $.... \Omega$.

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 $\Omega$):

What is the resistance of an ideal ammeter?

$A$ galvanometer with its coil resistance $25\,\Omega$ requires a current of $1\,mA$ for its full-scale deflection. In order to construct an ammeter to read up to a current of $2\,A$,the approximate value of the shunt resistance should be:

For a moving coil galvanometer,the deflection in the coil is $0.05\,rad$ when a current of $10\,mA$ is passed through it. If the torsional constant of the suspension wire is $4.0 \times 10^{-5}\,Nm\,rad^{-1}$,the magnetic field is $0.01\,T$,and the number of turns in the coil is $200$,the area of each turn (in $cm^2$) is:

In a ballistic galvanometer,the frame on which the coil is wound is non-metallic to