Two resistance are measured in ohm and is given as:-
`R_(1)=3Omega+-1% &R_(2)=6Omega+-2%`
When they are connected in parallel, the percentage error in equivalent resistance is

Two resistance R1 and R2 connected in parallel, this equivalent resistance is.

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

Two resistances R_(1) = (100 pm 3)Omega, R_(2) = (150 pm 2)Omega , are connected in series. What is their equivalent resistance?

When resistors 1 and 2 are connected in series, the equivalent resistance is 20.0 Omega. When they are connected in parallel, the equivalent resistance is 3.75 Omega. What are (a) the smaller resistance and (b) the larger resistance of these two resistors?

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)

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