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A galvanometer can be converted to a voltmeter of full-scale deflection $\mathrm{V}_{0}$ by connecting a series resistance $\mathrm{R}_{1}$ and can be converted to an ammeter of full-scale deflection $I_{0}$ by connecting a shunt resistance $\mathrm{R}_{2}$. What is the current flowing through the galvanometer at its full-scale deflection?
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Verified Answer
The correct answer is:
$\frac{\mathrm{V}_{0}-\mathrm{I}_{0} \mathrm{R}_{2}}{\mathrm{R}_{1}-\mathrm{R}_{2}}$
Hint:
$R_{1}=\frac{V_{0}}{I_{g}}-G_{\ldots} \ldots \ldots . .(1)$
$\mathrm{R}_{2}=\frac{\mathrm{I}_{\mathrm{g}} \mathrm{G}}{\mathrm{I}_{0}-\mathrm{I}_{\mathrm{g}}} \ldots \ldots \ldots . .(2)$
Eliminating G we get
$I_{g}=\frac{V_{0}-I_{0} R_{2}}{R_{1}-R_{2}}$
$R_{1}=\frac{V_{0}}{I_{g}}-G_{\ldots} \ldots \ldots . .(1)$
$\mathrm{R}_{2}=\frac{\mathrm{I}_{\mathrm{g}} \mathrm{G}}{\mathrm{I}_{0}-\mathrm{I}_{\mathrm{g}}} \ldots \ldots \ldots . .(2)$
Eliminating G we get
$I_{g}=\frac{V_{0}-I_{0} R_{2}}{R_{1}-R_{2}}$
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