The shunt resistance required to allow 4...

The shunt resistance required to allow 4% of the main current through the galvanometer of resistance `48 Omega` is

A

`1Omega`

B

`2Omega`

C

`3Omega`

D

`4Omega`

Text Solution

Verified by Experts

The correct Answer is:

B

Required shunt `S=(i_(g))/((i-i_(g)))G`
`=((4)/(100)xxi)/((i-(4)/(100)xxi))=((i)/(25))/((24)/(25)i)cc48`
`S=2Omega`

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