Two resisttors of resistacesR=R_(1)+R_(2)and for 1/R=1/(R_(1))+1/(R_(2))"and"(DeltaR)/(R^(2))=(DeltaR)/(R^(2))+(DeltaR_(1))/(R_(1)^(2))+(DeltaR_(2))/(R_(2)^(2))

Two resisttors of resistacesR=R_(1)+R_(2)and for 1/R=1/(R_(1))+1/(R_(2

1.	Two resisttors of resistacesR=R_(1)+R_(2)and for 1/R=1/(R_(1))+1/(R_(2))"and"(DeltaR)/(R^(2))=(DeltaR)/(R^(2))+(DeltaR_(1))/(R_(1)^(2))+(DeltaR_(2))/(R_(2)^(2))
Answer» Solution :The equivalent resistance of series combination `R=R_(1)+R_(2)=(100+-3)ohm+(200+-4)ohm=300+-7ohm``R.=(R_(1)R_(2))/(R_(1)+R_(2))=200/3=66.7ohm` then , form `1/R.=1/R_(1)+1/R_(2)` we get , `(DeltaR.)/R.^(2)=(DeltaR_(1))/R_(1)^(2)+(DeltaR_(2))/R_(2)^(2)` `=((66.7)/100)^(2)3+((66.7)/200)^(2)4=1.8` Then ,` R.=66.7+-1.8ohm`(here , `DeltaT` is EXPRESSED as `1.8`instead of2 to keep in conformity with the rules of significant FIGURES).