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

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 +- 0.2 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.

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 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)

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.

Two carbon resistances are given by R_(1)=(4pm0.4) ohm and R_(2)=(12pm0.6)Omega . What is their net resistance with percentage error, if they 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.

The values of two resistors are R_(1)=(6+-0.3)kOmega and R_(2)=(10+-0.2)kOmega . The percentage error in the equivalent resistance when they are connected in parallel is