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. \[{{I}_{g}}(G+{{R}_{V}})=V=G{{I}_{0}}\]\[(\text{given}V=G{{I}_{0}})\] \[\therefore \]\[{{R}_{V}}=\frac{({{I}_{0}}-{{I}_{g}})G}{{{I}_{g}}}\] \[\therefore \]\[{{R}_{A}}{{R}_{V}}={{G}^{2}}\And \frac{{{R}_{A}}}{{{R}_{V}}}={{\left( \frac{{{I}_{g}}}{{{I}_{0}}-{{I}_{g}}} \right)}^{2}}\]You need to login to perform this action.
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