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Series Combination of Cells

A galvanomete...

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?

\frac{V_{0} - I_{0} R_{2}}{R_{1} - R_{2}}

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\frac{V_{0} + I_{0} R_{2}}{R_{1} + R_{2}}

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\frac{V_{0} - I_{0} R_{1}}{R_{1} - R_{2}}

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\frac{V_{0} + I_{0} R_{1}}{R_{1} + R_{2}}

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Solution

The correct option is A $\frac{V_{0} - I_{0} R_{2}}{R_{1} - R_{2}}$
For voltmeter ---(1)
$V = (G + R_{1}) i_{g}$ where $i_{g}$ is the
current through galvanometer
& For Ammeter Shunt S = $R_{2}$
$R_{2} = \frac{G}{\frac{i_{0}}{i_{g}} - 1} = \frac{G i_{g}}{i_{0} - i_{g}} . . . (2)$
Solving (1) & (2)
$V = G i_{g} + R_{1} i_{g} = R_{2} (i_{0} - i_{g}) + R_{1} i_{g} \Rightarrow i_{g} = \frac{V - I_{0} R_{2}}{R_{1} - R_{2}}$

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