When connecting a shunt in parallel to the internal resistance of an ammeter the maximum range of current intensity that can be indicated is increased. Because the internal resistance is in parallel with shunt resistance the voltage drop is the same on both (U).
with the shunt connected we have
`I_("total") = I_("shunt") + I_("meter")`
The ratio of meter current to shunt current is
`n = I_(meter)/I_(shunt) = (U/R_(meter))*(R_(shunt)/U) = R_(shunt)/R_(meter) =0.1/600 =1.667*10^-4`
`I_(shunt) =I_(meter)/n = (100*10^-6)/(1.667*10^-4) =0.6 A`
This means the total scale of the ammeter with shunt is
`I_("total") = 0.6 +0.0001 = 0.6001 A ~~0.6 A`