In a moving coil galvanometer,the deflection of the coil $\theta$ is related to the electrical current $i$ by the relation

$A$ student is provided with a variable voltage source $V$,a test resistor $R_T=10\,\Omega$,two identical galvanometers $G_1$ and $G_2$,and two additional resistors,$R_1=10\,M\,\Omega$ and $R_2=0.001\,\Omega$. For conducting an experiment to verify Ohm's law,the most suitable circuit is:

To convert a galvanometer into an ammeter,we connect . . . . . . .

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

The resistance of a galvanometer coil is $R$. What is the shunt resistance required to convert it into an ammeter of range $4$ times?

$A$ galvanometer coil has a resistance of $15\; \Omega$ and the meter shows full-scale deflection for a current of $4\; mA$. How will you convert the meter into an ammeter of range $0$ to $6\; A$?