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A moving coil galvanometer is converted ...

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

A

`0.01A`

B

`0.02A`

C

`0.03A`

D

`0.04A`

Text Solution

Verified by Experts

The correct Answer is:
B
`I_(g)=((S_1)/(S_1+G))I=((S_2)/(S_2+G))`
From this `G=2r`.
`I_(g) = (4r)/(4r+2r)xx0.03=0.02A`.
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