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. `(l_(g))/(l)=4/11`
B. `(l_(g))/(l)=3/11`
C. `(l_(g))/(l)=2/11`
D. `(l_(g))/(l)=-1/11`

Answer

D

Explanation

Current through shunt is given by difference of total current in the circuit to current thorogh galvanomter
If l=total current in the circuit
G= resistance of the galvanometer
S=resistance of the shunt
`l_(g),l_(s)`=current throught gavanometer and shunt
respectively then `l_(g)=l(S)/(G+S)`

Hence part of current that flows throiugh galvanometer is given by
`(l_(g))/(l)=(S)/(G+S)=(2.5)/(2.5+2.5)=(2.5)/(27.5)=(1)/(11)`