Given `R_(1) = 5.0 +- 0.2 Omega, and R_(2) = 10.0 +- 0.1 Omega`. What is the total resistance in parallel with possible `%` error?

Given R_(1)=5.0+-0.2Omega, R_(2)=10*0+-0.1Omega. What is total resistance in parallel with possible % error.?

Two resistances are expressed as R_1=(4 ±0.5) Omega and R_2 =(12 ±0.5) Omega . What is the net resistances when these are connected in parallel with percentage error.

Given Resistance R_(1) = (8 +- 0.4) Omega and Resistence, R_(2) = (8 +- 0.6) Omega What is the net resistence when R_(1) and R_(2) are connected in series?

Two resistances are expressed as R_1=(4 +-0.5) Omega and R_2 =(12 +-0.5) Omega . What is the net resistances when these are connected in series.

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 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

You are given two resistance R_(1)=(4.0+-0.1)Omega and R_(2)=(9.1+-0.2)Omega . Calculate their effective resistance when they are connected (i) in series and (ii) in parallel. Also calculate the percentage error in each case.

The vaIue of two resistors are (5.0 pm 0.2)k Omega and (10.0 pm 0.1)k Omega . What is the percentage error in the equivaIent resistance when they are connected in paraIIeI?