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JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
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,
- A \({R_A}{R_V} = {G^2}\) and \(\,\frac{{{R_A}}}{{{R_V}}} = \frac{{{I_g}}}{{\left( {{I_0} - {I_g}} \right)}}\)
- B \({R_A}{R_V} = {G^2}\) and \(\,\,\frac{{{R_A}}}{{{R_V}}} = {\left( {\frac{{{I_g}}}{{{I_0} - {I_g}}}} \right)^2}\)
- C \({R_A}{R_V} = {G^2}\left( {\frac{{{I_g}}}{{{I_0} - {I_g}}}} \right)\,\) and \(\,\,\frac{{{R_A}}}{{{R_V}}} = {\left( {\frac{{{I_0} - {I_g}}}{{{I_g}}}} \right)^2}\)
- D \({R_A} - {R_V} = {G^2}\left( {\frac{{{I_0} - {I_g}}}{{{I_g}}}} \right)\,\) and \(\,\,\frac{{{R_A}}}{{{R_V}}} = {\left( {\frac{{{I_g}}}{{{I_0} - {I_g}}}} \right)^2}\)
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…
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