Given, the resistance of the galvanometer G=100Ω
No. of divisions =50
Current Sensitivity =20μA/division
The maximum current through galvanometer will be
Imax=50×20×10−6=10−3A
For V1=2V
using Ohm's law V1=Imax(G+R1)
2=10−3×(100+R1)
∴R1=1900Ω
For V2=10V
using Ohm's law V2=Imax(G+R1+R2)
10=10−3×(100+1900+R2)
∴R2=8000Ω
For V3=20V
using Ohm's law V3=Imax(G+R1+R2+R3)
20=10−3(100+1900+8000+R3)
∴R3=10×103Ω