--- title: "Nodal analysis currents and power loss" subject: "Basic Electrical Engineering" chapter: 2 key: "semester_1/basic_electrical_engineering:nodal-analysis-currents-power-loss-1a" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/nodal-analysis-currents-power-loss-1a site: BCT Engineering --- # Nodal analysis currents and power loss Basic Electrical Engineering · Chapter 2: DC circuits - Asked in: 2074 Asoj, 8 marks - Video: [Nodal Analysis problems in Hindi [ Problem 1 ]](https://www.youtube.com/watch?v=ABReC8BUEmk) — Perfect Computer Engineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/nodal-analysis-currents-power-loss-1a - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2 ## Question Using Nodal analysis, determine currents in each branch of the network shown in below figure. Also find the total power loss in the network. [Figure: 1 A current source with 10 Ω in parallel on the left; 10 Ω and 20 Ω in the top branch; 20 Ω and 20 Ω with a 10 V source in the middle; 0.5 A current source on the right.] ## Short answer Using nodal analysis, solve for node voltages (V1 = 6 V, V2 = 2 V, V3 = 6 V). Branch currents are 0.6 A, 0.4 A, 0.6 A, 0.2 A, 0.3 A. Total power loss in resistors is Σ I²R = 15 W. ## Answer #### Reading the figure The bottom rail is ground (0 V). Node 1 has the 1 A source (into the node) and a 10 Ω to ground. Node 1 connects to node 2 through 10 Ω. Node 2 has 20 Ω in series with the 10 V source to ground (assumed + at the bottom, so the branch end sits at −10 V). Node 2 connects to node 3 through 20 Ω. Node 3 has 20 Ω to ground and the 0.5 A source (into the node). #### Nodal equations (V₁, V₂, V₃) Node 1: V₁/10 + (V₁ − V₂)/10 = 1 Node 2: (V₂ − V₁)/10 + (V₂ + 10)/20 + (V₂ − V₃)/20 = 0 Node 3: (V₃ − V₂)/20 + V₃/20 = 0.5 Solving: **V₁ = 6 V, V₂ = 2 V, V₃ = 6 V** #### Branch currents - 10 Ω (node 1 to ground): 6/10 = 0.6 A - 10 Ω (node 1 to node 2): (6 − 2)/10 = 0.4 A - 20 Ω with 10 V (node 2 to ground): (2 + 10)/20 = 0.6 A - 20 Ω (node 3 to node 2): (6 − 2)/20 = 0.2 A (flows from 3 to 2) - 20 Ω (node 3 to ground): 6/20 = 0.3 A #### Total power loss (resistors) P = 0.6² × 10 + 0.4² × 10 + 0.6² × 20 + 0.2² × 20 + 0.3² × 20 P = 3.6 + 1.6 + 7.2 + 0.8 + 1.8 = 15 W **Answer: Total power loss = 15 W** Check: the 1 A source delivers 1 × 6 = 6 W and the 0.5 A source delivers 0.5 × 6 = 3 W. The 0.6 A in the 10 V branch enters the battery at its − terminal and leaves at its + terminal, so the 10 V source also delivers 10 × 0.6 = 6 W. Total supplied = 6 + 3 + 6 = 15 W, equal to the loss. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Nodal analysis writes one equation per unknown node using: A. KVL around each loop B. KCL: currents leaving a node sum to the injected source current C. Thevenin's theorem D. The superposition theorem Answer: B. Node voltages are the unknowns and KCL at each node gives the equations. (Key point: Nodal principle) 2. The solved node voltages V1, V2, V3 are: A. 10 V, 2 V, 6 V B. 6 V, 6 V, 2 V C. 2 V, 6 V, 6 V D. 6 V, 2 V, 6 V Answer: D. Solving the three equations gives V1 = 6 V, V2 = 2 V, V3 = 6 V. (Key point: Node voltages) 3. The current in the 20 ohm branch with the 10 V source (node 2 to ground) is found as: A. (2 + 10)/20 = 0.6 A, since the branch sits at -10 V on the far side B. 2/20 = 0.1 A C. (10 - 2)/20 = 0.4 A D. 10/20 = 0.5 A Answer: A. The source adds to the node voltage, so voltage across the resistor is 2 + 10 = 12 V. (Key point: Branch with 10 V source) 4. The total power loss in the resistors is: A. 9 W B. 15 W C. 6 W D. 30 W Answer: B. 3.6 + 1.6 + 7.2 + 0.8 + 1.8 = 15 W, matching the power delivered by the sources. 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