If `1 Omega` shunt be connected across a galvanometer of `99Omega` resistance, what fraction of the main current does the flow through the galvanometer?

If a stunt of 1 Omega is connected to a galvanometer of resistance 99 Omega , what fraction of the main current will flow through the galvanometer?

A shunt of resistance 1 Omega is connected with a galvanometer of resistance 100 Omega . What part of the main current will flow through the galvanometer?

A shunt of resistance 1 Omega is connected with a galvanometer of resistance 100 Omega . What part of the main current will flow through the galvanometer?

What should be the value of the shunt to be connected in parallel to a galcanometer of resistance G, so that 1/n part of the main current will pass through the shunt?

What should be the value of the shunt to be connected in the parallel to a galvanometer to resistance G, so that 1/n part of the main current will pass through the shunt?

A shunt of 1 Omega is connected in parallel with a galvanometer of resistance 99 Omega . IF the main current of the circuit be 1A, then what will be the galvanometer current?

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.

When a shunt S is connected across gaivanomenter of resistence G, 1/n of the main current passes through the galvanometer. What is the relation between S and G?

ABCD is a Wheatstone bridge in which the resistanvely 2Omega , 4 Omega , 6Omega and 8Omega . The points A and C are connected to the termicals of a cell of emf 2 v and negligible internal resistance . The points B and D are connected to a galvanometer of resistance 50Omega . Using Kirchhoff's laws find the current flowing through the galvanometer .

A shunt of resistance r_1 is connected in parallel with a galvanometer of resistance G. A steady source of internal resistance r sends current through the galvanometer.Now, removing the shunt r_1 another resistances r_2 is connected in series with the galvanometer. (i) IF the galvanometer current in either case be the same , show that rG=r_1r_2, (ii) if r=0 and r_1=r_2=G/n , then show that the ratio of galvanometer currents in the two cases is (n+1)/n .