Question

A galvanometer coil has a resistance of `15Omega` and the meter shows full scale deflection for a current of `4mA`. How will you convert the meter into an ammeter of range 0 to 6A?

Answer 1

Here, `G=15Omega`, `I_g=4mA=4xx10^-3A`, `S=?`, `I=6A`
From `S=(I_g.G)/(I-I_g)=(4xx10^-3xx15)/(6-4xx10^-3)=(6xx10^-2)/(5.996)=10^-2ohm`
This low resistance, called shunt resistance, is to be connected in parallel with the galvanometer.