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 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

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

Two resistance 60.36 Omega and 30.09 Omega are connected in parallel. The equivalent resistance is

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

Two resistor R_(1) = (24 +- 0.5)Omega and R_(2) = (8 +- 0.3)Omega are joined in series, The equivalent resistence is

Two resistor R_(1) = (24 +- 0.5)Omega and R_(2) = (8 +- 0.3)Omega are joined in series, The equivalent resistence is

108 . If two resistors of resistances R_(1)= (4 +- 0.5 )Omega and R_(2) =(16+-0.5)Omega are connected (i) in series and (ii) in parallel, find the equivalent resistance in each case with limits of percentage error.

Two resistors R_1 = (4 pm 0.8) Omega and R_2 (4 pm 0.4) Omega are connected in parallel. The equivalent resistance of their parallel combination will be :

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