(a) Briefly explain how a galvanometer is converted into an ammeter.
(b) A galvanometer coil has a resistance of `15 Omega` and it shows full scale deflection for a current of 4 mA. Convert it into an ammeter of range 0 to 6 A.

Explanation

(a) A galovanometer can be converted into an ammeter by connecting a low resistance called shunt parallel to the galvanometer as galvanometer and shunt are in parallel
`V_(s)=V_(g)`
`I_(s)S=I_(g)G" ….(1)"`
`I=I_(s)+I_(g)" ….(2"`
From equation (1) and (2)
`S=(I_(g)G)/((I-I_(g))`
Total effective resistance of ammeter
`R_(eff)=(GS)/(G+S)`
(b) `I_(s)=I-I_(g)=6000-4=5996 mA`
Voltage of shunt = Voltage of galvanometer
`V_(s)=V_(G)`
`I_(s)S=I_(g)G`
`5996xxs=4xx15" "S=(4xx15)/(5996)=0.010 Omega`
to covert the galvanometer into ammeter low Resistance (Shunt) used in parallel will be `=0.010 Omega.`