In an ammeter `10%` of main current passes through the galvanometer .If resistance of galvanometer is `G` then resistance of ammeter is :-
(a) `(G)/(5)`
(b) `(G)/(10)`
c) `(G)/(20)`
d)`(G)/(25)`

In a ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G , the resistance of ammeter will be

If 2% of the main current is to be passed through the galvanometer of resistance G , the resistance of shunt required is

To send 10% of the main current through a moving coil galvanometer of resistance 9 Omega , shunt S required is

To send 10% of the main current through a moving coil galvanometer of resistance 99Omega , the shunt required is –

The shunt required for 10% of main current to be sent through the moving coil galvanometer of resistance 99 Omega will be-

In a galvanometer 5% of the total current in the circuit passes through it. If the resistance of the galvanometer is G, the shunt resistance 'S' connected to the galvanometer is

Current passing through galvanometer is 0.002A and resistance of galvanometer is R_(g)=50Omega find out shunt resistance to convert it into ammeter of range 0.5 A

The value of current in the given circuit if the ammeter is a galvanometer with a resistance R_(G)=50Omega is

In the figure, galvanometer G gives maximum deflection when

A galvanometer measures current which passes through it. A galanometer can measure typically carrent of order of mA. To be able to measure currents of the order of amperes of main current, a shunt resistance 'S' is connected in parallel with galvanometer. If resistance of galavonometer is 10Omega and maximum current i_(g) is 10mA then the shunt resistance required so that the main current 'I' can be upto 1A is (in Omega )