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 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 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 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 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 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 in figure E_1=3V, E_2=2V, E_3=1V and R=r_1=r_2=r_3=1Omega Find the potential differece between the points A and B and the currents through each branch.