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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/i_0)=1/11`

B

`(i/i_0)=1/10`

C

`(i/i_0)=1/9`

D

`(i/i_0)=2/11`

Text Solution

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The correct Answer is:
A
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