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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

`1_g/I=1/11`

B

`I_g/I=2/11`

C

`I_g/I=3/11`

D

`I_g/I=4/11`

Text Solution

Verified by Experts

The correct Answer is:
A

`I_g/I=s/(s+G)=(2.5)/(2.5+25)=2.5/27.5=1/11`
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