Two resistance `R_(1)=100pm 3Omega and R_(2)=200pm4Omega` are connected in seriesg. What is their equivalent resistance?

Two resistances R_(1) = 100 +- 3 Omega and R_(2) = 200 +- 4 Omega are connected in series . Find the equivalent resistance of the series combination.

Two resistance (200 pm 4) Omega and (150 pm 3) Omega are connected in series . What is their equivalent resistance ?

Two resistors of resistances R_(1)=(300+-3) Omega and R_(2)=(500+-4) Omega are connected in series. The equivalent resistance of the series combination is

The resistors of resistances R_(1)=100pm3" ohm and R"_(2)=200pm 4ohm are connected in series. The equivalent resistance of the series combination is

Two resistor of resistance R_(1)=(6+-0.09)Omega and R_(2)=(3+-0.09)Omega are connected in parallel the equivalent resistance R with error (in Omega)

Two resistance R_(1) and R_(2) are connected in (i) series and (ii) parallel. What is the equivalent resistance with limit of possible percentage error in each case of R_(1)=5.0+-0.2Omega and R_(2)=10.0+-0.1Omega

Tow resistance R_(1) = 50 pm 2 ohm and R_(2) = 60 pm connected in series , the equivalent resistance of the series combination in

Two resistances R_(1)=(15 pm 0.5) Omega" and "(20 pm 0.7)Omega are connected in parallel. Calculate the total resistance of the combination and maximum percentage error.

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)