In the circuit shown here, `E_(1) = E_(2) = E_(3) = 2 V` and `R_(1) = R_(2) = 4 ohms`. The current flowing between point `A` and `B` through battery `E_(2)` is

In the circuit shown in fig. 5.265, E_1 = E_2 = E_3 = 2V and R_1 = R_2 = 4Omega. The current flowing between points A and B through battery E_2 is

In the circuit shown in fig. 5.265, E_1 = E_2 = E_3 = 2V and R_1 = R_2 = 4Omega. The current flowing between points A and B through battery E_2 is

In the following circuit E_(1)=E_(2)=E_(3) 2V and R_(1)=R_(2)=AOmega . The current passing through battery E_(2) between points A and B is

In the circuit as shown in figure, E_(1)=2V , E_(2)=2V, E_(3)=1V, and R=r_(1)=r_(2)=r_(3)Omega . The potential difference between points A and B will be

In the circuit shown in fig. E_(1)=3V, E_(2)= 2V, E_(3)= 1V, R=r_(1)=r_(2)=r_(3)= 1 ohm . Find the potential difference between the points A and B and the currents through each branch.