The measurement of an unknown resistance R is to be carried out using Wheatstone bridge (see Fig. 2(EP).3). Two students perform an experiment in two way. The first student takes `R_(2)=10 Omega` and `R_(1)=5Omega`. The other student takes `R_(2)=1000Omega` and `R_(1)=500Omega`. In the standard arm, both take `R_(3)=5Omega`. Both find `R=(R_(2))/(R_(1))R_(3)=10Omega` within errors.

The measurement of an unknown resistance R is to be carried out using wheatstone bridge(see Fig. EP 3.10). Two students perform an experiment in two ways. The first student takes R_2 = 10 Omega and R_1 = 5 Omega . The other student takes R_2 = 1000 Omega and R_1 = 500 Omega . In the standard arm, both takes R_3 = 5 Omega . Both find R = R_2/R_1 R_3 = 10 Omega within errors.

The measurement of an unknown resistance R is to be carried out using Wheatstone bridge as given in the figure below. Two students perform an experiment in two ways. The first students takes R_(2) = 10 Omega and R_(1) = 5 Omega . The other student takes R_(2) = 1000 Omega and R_(1) = 500 Omega . In the standard arm, both take R_(3) = 5 Omega . both find R = (R_(2))/(R_(1)) , R_(3) = 10 Omega within errors.

The measurement of an unknown resistance R is to be carried out using Wheatstone's bridge. Two students perform an experiment in two ways. The first student R_2 = 10 Omega and R_1 = 5 Omega the other student takes R_2 = 1000 Omega and R_1 = 500 Omega . In the standard arm ,both take R_3 = 5 Omega . Both find R = [R_2/R_1] R_3 = 10 Omega within errors

Find the current through the resistance R in Fig if R =10 Omega and R=20Omega .

In the given circuit, R_(1) = 10Omega R_(2) = 6Omega and E = 10V, then correct statement is

In the given circuit, R_(1) = 10Omega R_(2) = 6Omega and E = 10V, then correct statement is

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?

In the circuit shown below E_(1) = 4.0 V, R_(1) = 2 Omega, E_(2) = 6.0 V, R_(2) = 4 Omega and R_(3) = 2 Omega . The current I_(1) is

In the circuit shown below E_(1) = 4.0 V, R_(1) = 2 Omega, E_(2) = 6.0 V, R_(2) = 4 Omega and R_(3) = 2 Omega . The current I_(1) is