Question

A galvanometer can be converted to a voltmeter of full-scale deflection \(V_{0}\) by connecting a series resistance \(R_{1}\) and can be converted to an ammeter of full-scale deflection \(I_{0}\) by connecting a shunt resistance \(R_{2}\). What is the current flowing through the galvanometer at its full-scale deflection ?
(A) \(\frac{\mathrm{V}_{0}-\mathrm{I}_{0} \mathrm{R}_{2}}{\mathrm{R}_{1}-\mathrm{R}_{2}}\)
(B) \(\frac{V_{0}+I_{0} R_{2}}{R_{1}+R_{2}}\)
(C) \(\frac{V_{0}-l_{0} R_{2}}{R_{2}-R_{1}}\)
(D) \(\frac{V_{0}+I_{0} R_{1}}{R_{1}+R_{2}}\)

Answer 1

Ans: (A)
 \(R_{1}=\frac{V_{0}}{I_{g}}-G \ldots \ldots \ldots .\) (1)
$$
\mathrm{R}_{2}=\frac{\mathrm{Ig}_{g}}{\mathrm{I}_{0}-\mathrm{I}_{g}} \ldots \ldots \ldots \ldots(2)
$$
Eliminating \(\mathrm{G}\) we get
$$
\mathrm{I}_{g}=\frac{\mathrm{V}_{0}-\mathrm{I}_{0} \mathrm{R}_{2}}{\mathrm{R}_{1}-\mathrm{R}_{2}}
$$

Answer 2

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Answer 3

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