[" 28.If two resistances of values "R_(1)=(2.0+-0.1)" Q and "R_(2)=(12.3+-0.2)" 0are put ( "(" ) in perallel and "(1)" ."5],[" in series,find the error in the equivalent resistance."]

If two resistances of value R _ 1 = ( 2.0 pm 0 .1 ) Omega and R _ 2 = ( 12.3 pm 0.2 ) are connected in (i) parallel and (ii) series then find the error in the estimation of equivalent resistance.

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 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 conductors of resistance R_(1) (50 +- 2) ohm and R_(2) = (100 +- 4) ohm are connected in (a) series and (b) parallel. Find the equivalent resistance.

R_1 and R_2 are two resistances they are connected in series Write down their equivalent resistance in each case.

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 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 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 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?