Two resistors of resistances `R_(1)=100 pm 3` ohm and `R_(2)=200 pm 4` ohm are connected (a) in series, (b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel combination. Use for (a) the relation `R=R_(1)+R_(2)` and for (b) `1/(R')=1/R_(1)+1/R_(2)` and `(Delta R')/R'^(2)=(Delta R_(1))/R_(1)^(2)+(Delta R_(2))/R_(2)^(2)`

Explanation

(a) For the series combination, equivalent resistance is `R= R_(1)+R_(2)`
`=(100 pm 3) Omega+(200 pm 4)Omega`
`=(100+200)pm (3+4)`
`(300 pm 7)Omega`
(b) For the parallel combination, equivalent resistance of the combination is :
`(1)/(R.)=(1)/(R_1)+(1)/(R_2)" ""..........."(i)`
`implies R.= (R_(1)+R_(2))/(R_(1)+R_(2))=(100xx200)/(100+200)=(200)/(3)=66.7 Omega`
On differentiating equation (i) we get
Also, `(triangleR.)/(R.^2)=(triangleR_1)/(R_(1)^(2))+(triangleR_2)/(R_(2)^(2))`
`implies triangleR. =(R.^2)(triangleR_1)/(R_(1)^(2))+(R.^2)(triangleR_2)/(R_(2)^(2))`
`implies AR.= ((R.)/(R_1))^(2) triangleR_(1)+((R.)/(R_2))^(2) triangleR_(2)`
`=((66.7)/(100))^(2)xx3+((66.7)/(200))^(2)xx4`
`=1.8 Omega`
`R. =66.7 pm 1.8 Omega`.