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

Step-by-step solution

As shown in figure a shunt S is connected in parallel with the galvanometer to convert it into the ammeter, let G is the resistant of the galvanometer. Where I is the current in the circuit as shown, according to question 0 · 2 % current passing through the galvanometer, i.e., I G   =   0 · 2 %   o f   I   =   0 · 2 100 I =   I 500 ; so, the current through the shunt is I = I G   +   I S   ⇒   I S   =   I - I G ⇒ I S