question_answer

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. Then, [JEE Main 12-4-2019 Afternoon]

A) \[{{R}_{A}}{{R}_{V}}={{G}^{2}}\left( \frac{{{I}_{g}}}{{{I}_{0}}-{{I}_{g}}} \right)\]and\[\frac{{{R}_{A}}}{{{R}_{V}}}={{\left( \frac{{{I}_{0}}-{{I}_{g}}}{{{I}_{g}}} \right)}^{2}}\]

B) \[{{R}_{A}}{{R}_{V}}={{G}^{2}}\]and\[\frac{{{R}_{A}}}{{{R}_{V}}}={{\left( \frac{{{I}_{g}}}{{{I}_{0}}-{{I}_{g}}} \right)}^{2}}\]

C) \[{{R}_{A}}{{R}_{V}}={{G}^{2}}\]and\[\frac{{{R}_{A}}}{{{R}_{V}}}=\frac{{{I}_{g}}}{({{I}_{g}}-{{I}_{g}})}\]

D) \[{{R}_{A}}{{R}_{V}}={{G}^{2}}\left( \frac{{{I}_{0}}-{{I}_{g}}}{{{I}_{g}}} \right)\]and\[\frac{{{R}_{A}}}{{{R}_{V}}}={{\left( \frac{{{I}_{g}}}{{{I}_{0}}-{{I}_{g}}} \right)}^{2}}\]

Correct Answer: B

Solution :

When galvanometer is used as an ammeter shunt is used in parallel with galvanometer.           \[\therefore \]\[{{I}_{g}}G=({{I}_{0}}-{{I}_{g}}){{R}_{A}}\] \[\therefore \]\[{{R}_{A}}=\left( \frac{{{I}_{g}}}{{{I}_{0}}-{{I}_{g}}} \right)G\] When galvanometer is used as a voltmeter, resistance is used in series with galvanometer. \[{{I}_{g}}(G+{{R}_{V}})=V=G{{I}_{0}}\]\[(\text{given}V=G{{I}_{0}})\] \[\therefore \]\[{{R}_{V}}=\frac{({{I}_{0}}-{{I}_{g}})G}{{{I}_{g}}}\] \[\therefore \]\[{{R}_{A}}{{R}_{V}}={{G}^{2}}\And \frac{{{R}_{A}}}{{{R}_{V}}}={{\left( \frac{{{I}_{g}}}{{{I}_{0}}-{{I}_{g}}} \right)}^{2}}\]