Three parallel branches are connected between a common node A and the ground. The first branch consists of a 20 V battery with an internal resistance of 2 . The second branch consists of a 4 V battery with an internal resistance of 2 . The third branch consists of a 4 V battery with an internal resistance of 1 . The positive terminals of all three batteries face node A, and their negative terminals are grounded. The

Step-by-step solution

Let the potential at the common node A be V_A . Using Millman's theorem or Kirchhoff's Current Law, the voltage at node A is given by: V_A = E₁ R₁ + E₂ R₂ + E₃ R₃ 1 R₁ + 1 R₂ + 1 R₃ V_A = 20 2 + 4 2 + 4 1 1 2 + 1 2 + 1 1 V_A = 10 + 2 + 4 0.5 + 0.5 + 1 = 16 2 = 8 V The current I₁ flowing through the first branch is: I₁ = E₁ - V_A R₁ = 20 - 8 2 = 12 2 = 6 A The power dissipated in the internal resistance of the 20 V battery is: P = I₁^2 R₁ = (6)^2 2 = 36 2 = 72 W Answer: 72