In a circuit, three parallel branches are connected between two nodes A and B. Branch 1 contains a 3 , resistor in series with a 10 V battery, with the battery's positive terminal connected to node A. Branch 2 contains a 6 , resistor in series with a 4 V battery, with the battery's positive terminal connected to node A. Branch 3 contains a variable resistor R_L . The value of R_L is adjusted such that it dissipates m

Step-by-step solution

To find the power dissipated in R_L , we first determine the Thevenin equivalent circuit across nodes A and B with R_L removed. Let node B be the reference node ( 0 V ). The Thevenin voltage V_ th (open-circuit voltage across A and B) can be found using Millman's theorem: V_ th = E₁ R₁ + E₂ R₂ 1 R₁ + 1 R₂ Given E₁ = 10 V , R₁ = 3 , and E₂ = 4 V , R₂ = 6 , (both batteries have their positive terminals at A). V_ th = 10 3 + 4 6 1 3 + 1 6 = 20 + 4 6 2 + 1 6 = 24 3 = 8 V The Thevenin resistance R_ th is the equivalent