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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