In the following circuit, the battery has an emf of 2 V and an internal resistance of 2/3 Ω. The power consumption in the entire circuit is _________ W.

Solution

Physics sir · 4 steps

The side wires turn the corners into just two nodes, so this whole square is three 2 Ω paths in parallel.

1. Find the nodes. The left side of the square is a plain wire, so the top-left and bottom-left corners are one node, call it A. The right side is also a plain wire, so the two right corners form node B. The battery connects across A and B.
2. Three paths. The top 2 Ω joins A to B, and so does the bottom 2 Ω. Each diagonal path from a left corner, through the centre, to a right corner is 2 + 2 = 4 Ω. The two 4 Ω diagonals in parallel give 2 Ω. Each diagonal splits at the centre into two equal halves, so the centre sits at the same potential on both. It makes no difference whether the crossing is a junction.
3. Total resistance. Three 2 Ω paths in parallel give 2/3 Ω across A and B. Adding the internal resistance gives Rₜₒₜₐₗ = 2/3 + 2/3 = 4/3 Ω.

   I = E/Rₜₒₜₐₗ = 2/(4/3) = 1.5 A, P = EI = 2 × 1.5 = 3 W

4. So. The power used by the entire circuit, including the battery's internal resistance, is 3 W.

The common slip: 1.5. Counting only the power in the external square, I² × 2/3, gives 1.5 W. The internal resistance also uses power, and the question asks for the entire circuit.

The idea to keep. First merge every pair of points joined by a bare wire into one node, then count the paths between the two nodes. Power in the entire circuit is EI, and that includes the internal resistance.

Answer from the NTA final answer key. Solution by PrepWiser.