For two resistors `R_(1)` and `R_(2)`,connected in parallel,the relative error in their equivalent resistance is (Where `R_(1)=(10.0+-0.1)Omega` and `R_(2)=(20.0+-0.4)Omega`

(1) `0.08`

(2) `0.05`

(3) `0.01`

(4) `0.04`

For two resistors R_1 and R_2 , connected in parallel, find the relative error in their equivalent resistance. if R_1=(50 pm2)Omega and R_2=(100 pm 3) 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 resistors of resistance R_(1) and R_(2) having R_(1) gt R_(2) are connected in parallel. For equivalent resistance R, the correct statement is

Two resistors of resistance R_(1) and R_(2) having R_(1) gt R_(2) are connected in parallel. For equivalent resistance R, the correct statement is

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

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.

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

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

A combination of two resisters R_(1) and R_(4) are placed in an electrical circuit. In an experimental arrangement, the resistance are measured as R_(1)=(100 pm 3)Omega , R_(2)=(200 pm 4)Omega The equivalent resistance, when they are connected in parallel, (1)/(R_(P))=(1)/(R_(1))+(1)/(R_(2)) can be expressed as