$A$ galvanometer of resistance $G$ is shunted by a resistance of $5 \Omega$. To keep the main current in the circuit unchanged,the resistance to be put in series with the same galvanometer is

$A$ galvanometer of resistance $200 \Omega$ is to be converted into an ammeter. The value of shunt resistance which allows $3 \%$ of the main current through the galvanometer is equal to (nearly) (in $\Omega$)

In the circuit diagrams ($A, B, C$ and $D$) shown below,$R$ is a high resistance and $S$ is a resistance of the order of galvanometer resistance $G$. The correct circuit,corresponding to the half deflection method for finding the resistance and figure of merit of the galvanometer,is the circuit labelled as

$A$ multirange voltmeter can be constructed by using a galvanometer circuit as shown in the figure. We want to construct a voltmeter that can measure $2 \ V$,$20 \ V$,and $200 \ V$ using a galvanometer of resistance $10 \ \Omega$ that produces maximum deflection for a current of $1 \ mA$. Find $R_1$,$R_2$,and $R_3$ that have to be used.

$A$ galvanometer can be converted to a voltmeter of full-scale deflection $V_{0}$ by connecting a series resistance $R_{1}$ and can be converted to an ammeter of full-scale deflection $I_{0}$ by connecting a shunt resistance $R_{2}$. What is the current flowing through the galvanometer at its full-scale deflection?

$A$ galvanometer of resistance $100\,\Omega$ has $50$ divisions on its scale and a sensitivity of $20\,\mu A/\text{division}$. It is to be converted into a voltmeter with three ranges: $0-2\,V$,$0-10\,V$,and $0-20\,V$. The appropriate circuit to do so is: