Consider the circuit shown in the figure, where E ₁ = 3 V , E ₂ = 2 V , E ₃ = 1 V and r₁ = r₂ = r₃ = 1 . Determine the potential difference between points A and B , as well as the current flowing through each branch.

Step-by-step solution

Using Millman's theorem, the equivalent EMF of the parallel combination of the three branches gives the potential difference between nodes A and B. Taking the potential at B to be higher than A (due to the orientation of the batteries): V_B - V_A = E ₁ r₁ + E ₂ r₂ + E ₃ r₃ 1 r₁ + 1 r₂ + 1 r₃ Substituting the given values: V_B - V_A = 3 1 + 2 1 + 1 1 1 1 + 1 1 + 1 1 = 6 3 = 2 V The potential difference between A and B is 2 V . Let the currents flowing from A to B in the three branches be i₁ , i₂ , and i₃ respectivel