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

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?

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.

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)

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 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