A battery of internal resistances zero is connected to a galvanometer of resistance `80 Omega` and a resistance of `20 Omega` in series . A current flows through the galvanometer If a shunt of `1 Omega` resistance is connected to the galvanometer, show that the current that will now flow through the galvanometer becomes `1/17` of the previous current.

Explanation

The main current before adding shunt to the circuit is
`I=E/(80+20)=E/100`
After adding shunt the main current [Fig.1.20] is
`I'=E/((80times1)/(80+1)+20)=(E times80)/(80+20times81)`
So,`I_G=I'.S/(S+G)=I'.1/(80+1)=1'times1/81=E/(80+20times81)`
`thereforeI_G/I=E/(80+20times81)times100/E=100/(20(4+81))=100/(20times85)=1/17`
i.e.,`I_G=1/17I`