Question

A moving coil galvanometer, having a resistance G, produces full scale deflection when a current $I_g$ flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to $I_0$
( $I_0$ > $I_g$ ) by connecting a shunt resistance $R_A$ to it and (ii) into a voltmeter of range 0 to V (V = G$I_0$) by connecting a series resistance $R_V$ to it.
A. $R_A{R}_V=G^2\left({\displaystyle \frac{I_g}{I_0-I_g}}\right)\text{ and }{\displaystyle \frac{R_A}{R_V}}=\left({\displaystyle \frac{I_0-I_g}{I_g}}\right)$
B. $R_A{R}_V=G^2\text{ and }{\displaystyle \frac{R_A}{R_V}}={\left({\displaystyle \frac{I_g}{I_0-I_g}}\right)}^2$
C. $R_A{R}_V=G^2\text{ and }{\displaystyle \frac{R_A}{R_V}}=\left({\displaystyle \frac{I_g}{I_0-I_g}}\right)$
D. $R_A{R}_V=G^2\left({\displaystyle \frac{I_0-I_g}{I_g}}\right)\text{ and }{\displaystyle \frac{R_A}{R_V}}={\left({\displaystyle \frac{I_g}{I_0-I_g}}\right)}^2$

Answer 1

Hint: Galvanometer can be converted into ammeter and can be converted to voltmeter. It depends on the type of connection we give to external resistance. If we connect small resistance in parallel to the galvanometer then it would become an ammeter and if we connect big external resistance to the galvanometer in series then it would become voltmeter.
Formula used:
$V=IR$
$R_A={\displaystyle \frac{I_{g}G}{I_0-I_g}}$

Complete answer:
In case of parallel connection of resistors which means that the voltage across the two resistors will be same and current passing through the resistors vary.
We have
$V=IR$
Where V is the voltage and I is the current and R is the resistor.
If it is in a parallel connection then voltage will be the same and current is inversely proportional to voltage.
In case of ammeter shunt resistance($R_A$) will be connected in parallel to galvanometer resistance (G). current passing through total ammeter is ${}^{\prime }{I_0}^{\prime }$and the current passing through galvanometer resistance is $I_g$ so the current passing through shunt resistor is $I_0-I_g$
Voltage across galvanometer resistor will be equal to voltage across shunt resistor, which means
$I_{g}G=R_A(I_0-I_g)$
$\therefore R_A={\displaystyle \frac{I_{g}G}{I_0-I_g}}$ …eq1
After converting galvanometer to voltmeter of value V by connecting resistance $R_V$ in series, we have
$\begin{array}{rl}& V=I_g(R_V+G)\\ & \Rightarrow I_{0}G=I_g(R_V+G)\\ & \therefore R_V=G\left({\displaystyle \frac{I_0-I_g}{I_g}}\right)\end{array}$
If we multiply first equation and above equation we will get
$R_A{R}_V=G^2\left({\displaystyle \frac{I_g}{I_0-I_g}}\right)$
If we divide both we will get
${\displaystyle \frac{R_A}{R_V}}={\left({\displaystyle \frac{I_g}{I_0-I_g}}\right)}^2$

Hence option B will be the answer.

Note:
In case of conversion of galvanometer to ammeter we will attach a shunt resistance in parallel to galvanometer to get the small resistance ammeter and we will connect that ammeter in series with the circuit where we should measure the current where as if we connect external high resistance in series with galvanometer to make high resistance voltmeter and we connect that voltmeter in parallel to the circuit where we want to measure the voltage.