When a galvanometer of resistance G is shunted with a low resistance S, then the effective resistance `R_(eff)` of galvanometer becomes
`R_(eff)=(GS)/(G+S)`
If current is passed through such a galvanometer, then the major amount of current flows through shunt and the rest through galvanometer, i.e., the current divides itself in the inverse ratio of resistance.
Read the above passage and answer the following questions:
(i) Why is the resistance of shunted galvanometer lower than that of shunt?
(ii) A galvanometer of resistance `30Omega` is shunted by a resistance of `3Omega`. What fraction of the main current passes (i) through the galvanometer and (ii) through the shunt?
(iii) What are the basic values you learn from the above study?

Explanation

(i) Shunt is a low resistance connected in parallel with the galvanometer. Therefore, the combined resistance (i.e., resistance of shunted galvanometer) is less than that of shunt.
(ii) Let I be the total current passing through shunted galvanometer. Let `I_g`, `I_s`, be the currents through galvanometer of resistance G and shunt of resultance S respectively.
As current divides itself in the inverse ratio of resistance, therefore, fraction of current passing through galvanometer `=I_g/I=(S)/(G+S)=(3)/(30+3)=1/11`
Fraction of current passing through shunt `=I_s/I=(G)/(G+S)=(30)/(30+3)=10/11`
(iii) From the above study, we find that the current always divides itself in the inverse ratio of resistance. The major part of current flows through shunt, which is the path of low resistance. Similarly in life, if different paths are available to reach a destination, then one selects a path of least resistance to reach there. This would save both, time and energy.