When a galvanometer of resistance G iws converted into an ammeter of range IA then the current passing through shunt (S) is

A galvanometer of resistance R_(G) is to be converted into an ammeter, with the help of a shunt of resistance R. If the ratio of the heat dissipated through galvanometer and shunt is 3:4, then

A galvanometer of resistance R_(G) is to be converted into an ammeter, with the help of a shunt of resistance R. If the ratio of the heat dissipated through galvanometer and shunt is 3:4, then

A galvanometer of resistance 50Omega is converted into an ammeter by connecting a low resistance (shunt) of value 1Omega in parallel to the galvanometer, S. If full - scale deflection current of the galvanometer is 10 mA, then the maximum current that can be measured by the ammeter is -

A galvanometer of resistance 50Omega is converted into an ammeter by connecting a low resistance (shunt) of value 1Omega in parallel to the galvanometer, S. If full - scale deflection current of the galvanometer is 10 mA, then the maximum current that can be measured by the ammeter is -

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?