--- title: "Nodal analysis, power in 5Ω" subject: "Basic Electrical Engineering" chapter: 2 key: "semester_1/basic_electrical_engineering:nodal-power-5-ohm-40v-20v-5a" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/nodal-power-5-ohm-40v-20v-5a site: BCT Engineering --- # Nodal analysis, power in 5Ω Basic Electrical Engineering · Chapter 2: DC circuits - Asked in: 2070 Chaitra, 6 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-40v-20v-5a - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2 ## Question Determine the power dissipated by 5Ω resistor in the circuit shown in figure below by applying nodal voltage analysis. [Figure: 5 A current source on top; 6 Ω; 20 V source; 3 A current source; 3 Ω; 5 Ω; 40 V source on the left; 4 Ω in the bottom branch.] ## Short answer Using 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. Current through the 5 Ω resistor is 3.625 A, and power dissipated ≈ 65.7 W. ## Answer This circuit is the same as the one in the 2071 Chaitra paper. Use the bottom node of the 40 V source as reference. #### Reading the figure - P: top of the 40 V source (+), V_P = 40 V; left end of the 6 Ω; top of the 3 A source; left end of the 5 A source. - X: right end of the 6 Ω and of the 5 A source; top of the 3 Ω; − 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; right end of the 5 Ω and the 4 Ω (which returns to the reference). The 5 A source takes current from X to P, and the 3 A source takes current from Y to P. #### Nodal equations Supernode (X and Z) with V_Z = V_X + 20: (V_X − 40)/6 + 5 + (V_X − V_Y)/3 + (V_Z − V_Y)/5 + V_Z/4 = 0 Node Y: 3 + (V_Y − V_X)/3 + (V_Y − V_Z)/5 = 0 #### Solution V_X = −15.2 V, V_Y = −13.325 V, V_Z = 4.8 V Current in the 5 Ω: (V_Z − V_Y)/5 = (4.8 + 13.325)/5 = 3.625 A P = 3.625² × 5 = 65.7 W **Answer: Power in the 5 Ω resistor ≈ 65.7 W** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The reference is the bottom of the 40 V source so that A. V_P is unknown B. V_X = 0 C. V_Y = 0 D. V_P = 40 V is known Answer: D. It fixes one node voltage. (Key point: Reference) 2. X and Z are combined because A. a resistor lies between them B. the 5 A source lies between them C. the 20 V source lies between them D. they are at ground Answer: C. This gives V_Z = V_X + 20. (Key point: Supernode) 3. In node Y equation, the 3 A source is written as + 3 because A. it enters Y B. it flows from Y to P, i.e. leaves node Y C. it is a voltage source D. it is in series with 3 Ω Answer: B. Currents leaving the node are positive. (Key point: Source sign at Y) 4. With I = 3.625 A, the power in 5 Ω is A. 65.7 W B. 18.1 W C. 13.1 W D. 131 W Answer: A. P = I^2 R = 13.14 x 5 = 65.7 W. 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