A `36 Omega` galvanometer is shunted by resistance of `4Omega`. The percentage of the total current, which passes through the galvanometer is

A 36 Omega galvanometer is shunted by a resistance of 4Omega . The percentage of total current, which passes through the galvanometer is

A galvanometer has a resistance of 98 Omega .it is shunted by 2Omega resistance. The fraction of the total current passing through the galvanometer is

A galvanomter of resistance 50Omega is shunted by a resistance of 5Omega . What fraction of the main current passes through the galvanometer? Through the shunt?

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?

A galvanometer of resistance 36 Omega is changed into an ammeter by using a shunt of 4 Omega . The fraction f of total current passing through the galvanometer is

A resistance of 4 Omega is connected in parallel to a galvanometer of resistance 396 Omega . Find the fraction of the total current passing through the galvanometer and the fraction of the total current passing through the shunt.

We have a galvanometer of resistance 25 Omega . It is shunted by a 2.5 Omega wire. The part of total current that flows through the galvanometer is given as