Two resistors of resistances `R_1 = 100 pm 3` ohm and `R_2 = 200 pm 4 ` ohm are connected (a) in series , (b) in parrallel. 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 (DeltaR')/(R^(,2)) = (DeltaR_1)/(R_1^2) + (DeltaR_2)/(R_2^2)`

Explanation

(a)The equivalent resistance of series combination
`R= R_(1)+R_(2)= (100 pm 3) ` ohm + `(200 pm 4)` phm
= `300 pm 7` ohm
(b ) The equivalent resistance of parallel combination
`R= (R_(1)R_(2))/(R_(1)+R_(2))=(200)/(3)=66.7` ohm
Then from `(1)/(R) = (1)/(R_(1))+(1)/(R_(2))` we get `(DeltaR)/(R^(2))=(DeltaR_(1))/(R_(1)^(2))+(DeltaR_(2))/(R_(2)^(2))`
`DeltaR=(R^(2))(DeltaR_(1))/(R_(1)^(2))+(R^(2))(DeltaR_(2))/(R_(2)^(2))`
`=((66.7)/(100))^(2)3+((66.7)/(200))^(2)4=1.8`