$A$ galvanometer having internal resistance $10 \Omega$ requires $0.01 \ A$ for a full scale deflection. To convert this galvanometer to a voltmeter of full-scale deflection at $120 \ V$, we need to connect a resistance of

The graph which represents the relation between the total resistance $R$ of a multi-range moving coil voltmeter and its full-scale deflection $V$ is

$A$ multirange current meter can be constructed by using a galvanometer circuit as shown in the figure. We want a current meter that can measure $10 \text{ mA}$,$100 \text{ mA}$,and $1 \text{ A}$ using a galvanometer of resistance $10 \text{ } \Omega$ that produces maximum deflection for a current of $1 \text{ mA}$. Find the values of $S_1, S_2$,and $S_3$ that have to be used.

$A$ galvanometer coil has a resistance $80 \Omega$ and current for full-scale deflection is $10 \text{ mA}$. $A$ resistance of $920 \Omega$ is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is $0.2 \text{ V}$,the number of divisions on the scale is:

Explain the solution to the difficulties that arise when using a galvanometer directly as an ammeter.

$A$ 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 $\text{ }\Omega$)