A moving coil galvanometer is converted into an ammeter reading upto 0.03 A by connecting a shunt of resistance 4r across it and into an ammeter reading upto 0.06 A when a shunt of resistance r connected across it. What is the maximum current which can be sent through this galvanometer if no shunt is used.
A
0.01 A
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B
0.02 A
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C
0.03 A
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D
0.04 A
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Solution
The correct option is B0.02 A For ammeter, S=ig(i−ig)G⇒igG=(i−ig)S
So igG=(0.03−ig)4r .........(i) and igG=(0.06−ig)r Dividing equation (i) by (ii)1=(0.03−ig)40.06−ig⇒0.06−ig=0.12−4ig ⇒3ig=0.06⇒ig=0.02A