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A moving coil galvanometer is converted into an ammeter reading up to 0.03 A by connecting a shunt of resistance 4r across it and into an ammeter reading up to 0.06 A when a shunt of resistance r is connected across it. What is the maximum current which can be sent through this galvanometer if no shunt is used

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0.01 A

0.02 A

0.03 A

0.04 A

answer is B.

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Detailed Solution

$\frac{i_{g}}{2} = \frac{S}{G + S} \Rightarrow i_{g} G = (i - i_{g}) S$

$i_{g} G = (0.03 - i_{g}) 4 r$ ......(i)

and $i_{g} G = (0.06 - i_{g})$ .........(ii)

From (i) and (ii)

$0.12 - 4 i_{g} = 0.06 - i_{g}$

$3 i_{g} = 0.06$

$i_{g} = 0.02 A$

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