Resistances of two resistors is measured as (20+-0.2)Omega and "(40+-0.2)Omega .The maximum absolute error in their series combination will be

Minimum resistance obtained from two resistors 3Omega and 2 Omega is

If three resistors of resistance 2Omega, 4Omega and 5 Omega are connected in parallel then the total resistance of the combination will be

The resistors of resistances 2 Omega, 4 Omega and 8 Omega are connected in parallel, then the equivalent resistance of the combination will be

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

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

Two resistors R_1 = (4 pm 0.8) Omega and R_2 (4 pm 0.4) Omega are connected in parallel. The equivalent resistance of their parallel combination will be :

Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistance of 2 Omega in series with a combination of two resistors ( 4 Omega each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2 Omega resistor be the same as that across the parallel combination of 4 Omega resistors ? Give reason.