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We have a galvanometer of resistance 25 ...

We have a galvanometer of resistance `25 Omega`. It is shunted by a `2.5 Omega` wire. The part of total current that flows through the galvanometer is given as

A

`(I_(G))/(I)=(1)/(11)`

B

`(I_(G))/(I)=(2)/(11)`

C

`(I_(G))/(I) = (3)/(11)`

D

`(I_(G))/(I) = (4)/(11)`

Text Solution

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The correct Answer is:
A
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