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=GI_{0}) by connecting a series resistance R_{V} to it. Then, 

  • Option 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}

     

     

     

     

  • Option 2)

    R_{A}R_{V}=G^{2} and \frac{R_{A}}{R_{V}}=\frac{I_{g}}{I_{0}-I_{g}}^{2}

  • Option 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}

  • Option 4)

    \frac{R_{A}}{R_{V}}=G^{2}\frac{R_{A}}{R_{V}}=\frac{I_{g}}{(I_{0}-I_{g})}

 

Answers (1)

 

Conversion of galvanometer into voltameter -

Connected a large Resistance R  in series

- wherein

 

 

(I_{o}-I_{G})R_{A}=I_{G}G             V_{0}=I_{G}(G+R_{A})

                                                        V_{0}=I_{0}G

R_{A}=\frac{I_{G}G}{I_{o}-I_{G}}

I_{0}G=I_{G}(G+R_{V})

\frac{I_{0}G}{I_{G}}-G=R_{V}

R_{V}=\frac{G(I_{0}-I_{G})}{I_{G}}

\frac{R_{A}}{R_{V}}=\frac{I_{G}G}{I_{0}-I_{G}}\times \frac{I_{G}}{G(I_{0}-I_{G})}=\left ( \frac{I_{G}}{I_{0}-I_{G}} \right )^{2}

R_{A}R_{V}=\frac{I_{G}G}{I_{0}-I_{G}}\times \frac{G(I_{0}-I_{G})}{I_{G}}=G^{2}

 


Option 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}

 

 

 

 

Option 2)

R_{A}R_{V}=G^{2} and \frac{R_{A}}{R_{V}}=\frac{I_{g}}{I_{0}-I_{g}}^{2}

Option 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}

Option 4)

\frac{R_{A}}{R_{V}}=G^{2}\frac{R_{A}}{R_{V}}=\frac{I_{g}}{(I_{0}-I_{g})}

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