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

Options

A. `(I)/(I_(0)) = (1)/(11)`
B. `(I)/(I_(0)) = (1)/(10)`
C. `(I)/(I_(0)) = (3)/(11)`
D. `(I)/(I_(0)) = (4)/(11)`

Answer

A

Explanation

(a) `(i)/(i_(g)) = (G + S)/(S) implies (i_(g))/(i) = (S)/(G + S) = (2.5)/(27.5) = (1)/(11)`