A multirange current meter can be constructed by using a galvanometer circuit as shown in the figure. We want a current meter that can measure 10mA, 100mA, and 1A using a galvanometer of resistance 10Ω, and that produces maximum deflection for a current of 1mA. Find S1,S2,andS3 that have to be used.

                                     

Hint: The potential drop in the galvanometer and the shunt will be the same.
Step 1:IGG=(I1IG)(S1+S2+S3) for I1=10mA
Step 2: IG(G+S1)=(I2IG)(S2+S3) for I2=100mA
Step 3:IG(G+S1+S2)=(I3IG)(S3) for I3=1A
GivesS1=1W,S2=0.1W
and,S3=0.01W