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If only 1% of the total current is to be...

If only 1% of the total current is to be passed through a galvanometer of resistance G, then resistance of the shunt will be

A

`G/(25)Omega`

B

`G/(25)Omega`

C

`G/(49)Omega`

D

`G/(99)Omega`

Text Solution

Verified by Experts

The correct Answer is:
D
`I_g/I=1/(100) therefore I/I_g=100 therefore S=G/(n-1)=G/(99)Omega`
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