--- title: "Nodal analysis, power in 5Ω" subject: "Basic Electrical Engineering" chapter: 2 key: "semester_1/basic_electrical_engineering:nodal-power-5-ohm-5a-3a-40v-20v" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/nodal-power-5-ohm-5a-3a-40v-20v site: BCT Engineering --- # Nodal analysis, power in 5Ω Basic Electrical Engineering · Chapter 2: DC circuits - Asked in: 2071 Chaitra, 8 marks - Video: [Nodal Analysis : Determine v1, v2, and the power dissipated in all the resistors in the circuit of](https://www.youtube.com/watch?v=WFx8TkJvJUo) — Electrical and Electronics Engineering - 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-power-5-ohm-5a-3a-40v-20v - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2 ## Question Apply nodal analysis to calculate power dissipated by 5Ω resistor in the circuit shown in figure below: [Figure: 5 A current source and 6 Ω on top; 20 V source on the top right; 40 V source on the left; 3 A current source with 3 Ω shunt; 5 Ω on the right; 4 Ω in the bottom branch.] ## Short answer Using nodal analysis with a supernode for the 20 V source, node voltages are V_X = −15.2 V, V_Y = −13.325 V, V_Z = 4.8 V. The 5 Ω current is (V_Z − V_Y)/5 = 3.625 A, so power dissipated ≈ 65.7 W. ## Answer #### Reading the figure Take the negative terminal of the 40 V source as the reference (0 V). Nodes: - P: top of the 40 V source (+), so V_P = 40 V. Also the left end of the 6 Ω, the top of the 3 A source and the left end of the 5 A source. - X: right end of the 6 Ω, right end of the 5 A source, top of the 3 Ω, and the − terminal of the 20 V source. - Y: bottom of the 3 Ω, bottom of the 3 A source, left end of the 5 Ω. - Z: + terminal of the 20 V source (V_Z = V_X + 20), right end of the 5 Ω and of the 4 Ω. The 4 Ω returns to the reference. Source directions: 5 A flows from X to P; 3 A flows from Y to P. #### Nodal equations X and Z are joined by the 20 V source, so use a supernode with V_Z = V_X + 20. Node Y (currents leaving Y): 3 + (V_Y − V_X)/3 + (V_Y − V_Z)/5 = 0 Supernode X + Z (currents leaving): (V_X − 40)/6 + 5 + (V_X − V_Y)/3 + (V_Z − V_Y)/5 + V_Z/4 = 0 Substituting V_Z = V_X + 20 and solving: V_X = −15.2 V, V_Y = −13.325 V, V_Z = 4.8 V #### Current and power in the 5 Ω I₅ = (V_Z − V_Y)/5 = (4.8 + 13.325)/5 = 3.625 A (from Z to Y) P = I₅² × 5 = 3.625² × 5 = 65.7 W **Answer: P₅Ω ≈ 65.7 W** Figure notes: the 40 V source has its + terminal on top, the 20 V source has its + terminal on the right (Z side), the 5 A source points from X to P and the 3 A source points up (from Y to P). The vertical wire joining the 5 A and 3 A sources crosses the wire from the 40 V source to the 6 Ω; I took this crossing as a junction (otherwise the 5 A and 3 A sources would not satisfy KCL). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Taking the negative terminal of the 40 V source as 0 V gives V_P = A. 0 V B. unknown C. 20 V D. 40 V, a known value Answer: D. The battery fixes P at 40 V, so it is not an unknown. (Key point: Reference choice) 2. X and Z form a supernode because A. a resistor joins them B. a current source joins them C. the 20 V source joins them D. they are both grounded Answer: C. A voltage source between two non-reference nodes needs a supernode. (Key point: Supernode) 3. The 20 V source gives A. V_Z = V_X - 20 B. V_Z = V_X + 20 C. V_Z = 20 D. V_X = V_Z + 20 Answer: B. Z is the + terminal, so it is 20 V above X. (Key point: Constraint) 4. With V_Z = 4.8 V and V_Y = -13.325 V, P in the 5 Ω is A. 3.625^2 x 5 = 65.7 W B. 3.625 x 5 = 18 W C. 4.8^2/5 = 4.6 W D. 13.3^2/5 = 35 W Answer: A. I = (4.8 + 13.325)/5 = 3.625 A, so P = I^2 R = 65.7 W. 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