Two resistors of resistances R_(1)=100pm3 ohm and R_(2)=200pm4 ohm are connected (a) 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 (DeltaR')/(R'^(2)) = (DeltaR_(1))/(R_(1)^(2))+(DeltaR_(2))/(R_(2))

Two resistors of resistances R_(1)=100pm3 ohm and R_(2)=200pm4 ohm are

1.	Two resistors of resistances R_(1)=100pm3 ohm and R_(2)=200pm4 ohm are connected (a) 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 (DeltaR')/(R'^(2)) = (DeltaR_(1))/(R_(1)^(2))+(DeltaR_(2))/(R_(2))
Answer» Solution :(a)The equivalent resistance of series COMBINATION `R = R_(1)+R_(2) = (100 pm 3)` OHM + `(200 pm4)` ohm ` = 300pm7` 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 Then, R. = `66.7pm1.8` ohm (Here, `DeltaR` is expressed as 1.8 INSTEAD of 2 to keep in confirmity with the rules of significant figures.) Error in case of a measured quantity raised to a power Suppose Z = `A^(2)`, Then, `DeltaZ//Z=(DeltaA//A)+(DeltaA//A)=2(DeltaA//A)`. Hence, the relative error in `A^(2)` is two times the error in A. In general , if `Z=A^(p)B^(q)//C^(r)` Then, `DeltaZ//Z = p(DeltaA//A)+q(DeltaB//B)+r(DeltaC//C)`.