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?