A moving coil galvanometer, having a resistance G,

Question:

A moving coil galvanometer, having a resistance $\mathrm{G}$, 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 $\mathrm{I}_{0}\left(\mathrm{I}_{0}>\mathrm{I}_{g}\right)$ by connecting a shunt resistance $R_{A}$ to it and (ii) into a voltmeter of range 0 to $V$ $\left(\mathrm{V}=\mathrm{GI}_{0}\right)$ by connecting a series resistance $\mathrm{R}_{\mathrm{v}}$ to it. Then,

  1. (1) $R_{A} R_{V}=G^{2}\left(\frac{I_{0}-I_{g}}{I_{g}}\right)$ and $\frac{R_{A}}{R_{V}}=\left(\frac{I_{g}}{I_{0}-I_{g}}\right)^{2}$

  2. (2) $R_{A} R_{V}=G^{2}$ and $\frac{R_{A}}{R_{V}}=\left(\frac{I_{g}}{I_{0}-I_{g}}\right)^{2}$

  3. (3) $R_{A} R_{V}=G^{2}\left(\frac{I_{g}}{I_{0}-I_{g}}\right)$ and $\frac{R_{A}}{R_{V}}=\left(\frac{I_{0}-I_{g}}{I_{g}}\right)^{2}$

  4. (4) $R_{A} R_{V}=G^{2}$ and $\frac{R_{A}}{R_{V}}=\frac{I_{g}}{\left(I_{0}-I_{g}\right)}$


Correct Option: , 2

Solution:

(2) In an ammeter,

$i_{g}=i_{0} \frac{R_{\mathrm{A}}}{R_{\mathrm{A}}+G}$

and for voltmeter,

$V=i_{g}\left(G+R_{V}\right)=G i_{0}$

On solving above equations, we get

$R_{A} R_{V}=G^{2}$

and $\frac{R_{A}}{R_{V}}=\left(\frac{i_{g}}{i_{0}-i_{g}}\right)^{2}$

 

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