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JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2019
- Hard
- MCQ
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,
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \({R_A}{R_V} = {G^2}\) and \(\,\,\frac{{{R_A}}}{{{R_V}}} = {\left( {\frac{{{I_g}}}{{{I_0} - {I_g}}}} \right)^2}\)
Step-by-step Solution
Detailed explanation
When galvanometer is used an ammeter shunt is used in parallel with galvanometer.
\(\therefore \quad \mathrm{I}_{g} \mathrm{G}=\left(\mathrm{I}_{0}-\mathrm{I}_{g}\right) \mathrm{R}_{\mathrm{A}}\)
\(\therefore \quad R_{A}=\left(\frac{I_{g}}{I_{0}-I_{g}}\right) G\)
When galvanometer is used as a voltmeter, resistance is used in series with galvanometer.
\(\mathrm{I}_{g}\left(\mathrm{G}+\mathrm{R}_{\mathrm{v}}\right)=\mathrm{V}=\mathrm{GI}_{0}\left(\text { given } \mathrm{V}=\mathrm{GI}_{0}\right)\)
\(\therefore R_{v}=\frac{\left(I_{0}-I_{g}\right) G}{I_{g}}\)
\(\therefore \quad {{\text{R}}_{\text{A}}}{{\text{R}}_{\text{v}}} = {{\text{G}}^2}\& \frac{{{{\text{R}}_{\text{A}}}}}{{{{\text{R}}_{\text{v}}}}} = {\left( {\frac{{{{\text{I}}_g}}}{{{{\text{I}}_0} - {{\text{I}}_g}}}} \right)^2}\)
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