[" 4.Calcilate equivelent resistance of two "],[" tesistan resistos R,"P" Re in parallel "],[" whete "R_(1)=(6+-0.2)Omega],[" and "R_(2)=(3+-0.1)Omega]

Calculate equivalent resistance of two resistors R_(1) and R_(2) in parallel where, R_(1) = (6+-0.2) ohm and R_2 = (3+-0.1)ohm

Calculate equivalent resistance of two resistors R_(1) and R_(2) in parallel where, R_(1) = (6+-0.2) ohm and R_2 = (3+-0.1)ohm

Two resistance R_(1) and R_(2) are connected in (i) series and (ii) parallel. What is the equivalent resistance with limit of possible percentage error in each case of R_(1)=5.0+-0.2Omega and R_(2)=10.0+-0.1Omega

In the circuit shown in Fig. the emf of the sources is equal to xi = 5.0 V and the resistances are equal to R_(1) = 4.0 Omega and R_(2) = 6.0 Omega . The internal resistance of the source equals R = 1.10 Omega . Find the currents flowing through the resistances R_(1) and R_(2) .

In the circuit shown in Fig. the emf of the sources is equal to xi = 5.0 V and the resistances are equal to R_(1) = 4.0 Omega and R_(2) = 6.0 Omega . The internal resistance of the source equals R = 1.10 Omega . Find the currents flowing through the resistances R_(1) and R_(2) .

In Fig. 27-43, the current in resistance 6 is i_(6)= 2.80 A the resistances are R_(1)=R_(2)=R_(3)=2.00 Omega, R_(4)=16.0 Omega, R_(5)=8.00 Omega and R_(6)=4.00 Omega . Wat is the emf of the ideal battery?

Fig. shows an infinite circuit formed by the repetition of the same link, consisting of resistance R_(1) = 4.0 Omega and R_(2) = 3.0 Omega . Find the resistance of this circuit between points A and B .

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