--- title: "Nodal analysis current through 10Ω" subject: "Basic Electrical Engineering" chapter: 2 key: "semester_1/basic_electrical_engineering:nodal-current-10-ohm-25v-50v" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/nodal-current-10-ohm-25v-50v site: BCT Engineering --- # Nodal analysis current through 10Ω Basic Electrical Engineering · Chapter 2: DC circuits - Asked in: 2071 Shrawan, 8 marks - Video: [Nodal Analysis for Circuits Explained](https://www.youtube.com/watch?v=f-sbANgw4fo) — Engineer4Free - 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-current-10-ohm-25v-50v - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2 ## Question Use the node voltage method (nodal) to find the current flowing through 10Ω resistor in the network shown figure below. [Figure: 2 Ω shunt on the left; 5 Ω in series with a 25 V source; 10 Ω and 2 Ω along the top; 4 Ω shunt; 50 V source on the right.] ## Short answer Node 1 and node 2 (bottom ground). Nodal equations: 8V1 - V2 = 50, -2V1 + 17V2 = 500. Solve to get V1=10.07 V, V2=30.60 V. Current through the 10 Ω resistor = (V2 - V1)/10 = 2.05 A, from node 2 to node 1. ## Answer #### Reading the figure Bottom rail = ground. Node 1 is the top of the 2 Ω and of the branch 5 Ω + 25 V (+ on top, so the branch end is at 25 V when the 5 Ω carries no current). Node 2 is the top of the 4 Ω and of the branch 2 Ω + 50 V (+ on top). The 10 Ω joins node 1 and node 2. #### Nodal equations (currents leaving the node) Node 1: V₁/2 + (V₁ − 25)/5 + (V₁ − V₂)/10 = 0 Multiply by 10: 5 V₁ + 2 V₁ − 50 + V₁ − V₂ = 0 ⇒ 8 V₁ − V₂ = 50 ... (1) Node 2: V₂/4 + (V₂ − 50)/2 + (V₂ − V₁)/10 = 0 Multiply by 20: 5 V₂ + 10 V₂ − 500 + 2 V₂ − 2 V₁ = 0 ⇒ −2 V₁ + 17 V₂ = 500 ... (2) #### Solution From (1): V₂ = 8 V₁ − 50. Put in (2): −2 V₁ + 136 V₁ − 850 = 500 134 V₁ = 1350, so V₁ = 10.07 V V₂ = 8 × 10.07 − 50 = 30.60 V #### Current in the 10 Ω I = (V₂ − V₁)/10 = (30.60 − 10.07)/10 = 2.05 A **Answer: I₁₀Ω ≈ 2.05 A, flowing from node 2 to node 1** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The current leaving node 1 through the 5 Ω + 25 V branch is A. (V1 + 25)/5 B. V1/5 C. 25/5 D. (V1 - 25)/5 Answer: D. The battery sets the far end at 25 V, so the drop across 5 Ω is V1 - 25. (Key point: Branch expression) 2. Node 1 simplified is A. 8 V1 + V2 = 50 B. V1 - V2 = 50 C. 8 V1 - V2 = 50 D. 2 V1 - V2 = 25 Answer: C. 5V1 + 2V1 - 50 + V1 - V2 = 0. (Key point: Node 1 equation) 3. Solving gives A. V1 = 30.6 V and V2 = 10.07 V B. V1 = 10.07 V and V2 = 30.60 V C. V1 = 25 V, V2 = 50 V D. V1 = 0 V, V2 = 50 V Answer: B. V2 = 8V1 - 50, and 134 V1 = 1350. (Key point: Solution) 4. The 10 Ω current is 2.05 A flowing A. from node 2 to node 1, because V2 > V1 B. from node 1 to node 2 C. from ground to node 1 D. zero because of the batteries Answer: A. Current flows from higher to lower potential: 30.6 V to 10.07 V. (Key point: Direction) ## Close variants set in other exams - [Nodal analysis current through 10Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-current-through-10-ohm-5a) — Basic Electrical Engineering, chapter 2, asked 2× ## Related in Basic Electrical Engineering - [Nodal analysis, current through 15Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-current-15-ohm-8a-48v) — Basic Electrical Engineering, chapter 2 - [Nodal analysis currents in all resistors](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-currents-23a-2a-10-ohm) — Basic Electrical Engineering, chapter 2 - [Nodal analysis voltage drop on 8Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-8-ohm-voltage-drop-1a-2a) — Basic Electrical Engineering, chapter 2 - [Nodal analysis current I](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-current-i-1a-18v-12v) — Basic Electrical Engineering, chapter 2 - [Nodal analysis Va, Vb, Vc and 2.5Ω current](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-va-vb-vc-2-5-ohm) — Basic Electrical Engineering, chapter 2 - [Nodal analysis, power in 5Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-power-5-ohm-5a-3a-40v-20v) — Basic Electrical Engineering, chapter 2 ## Related in other subjects - [Nodal analysis current in branch a-b](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Anodal-current-branch-ab) — Basic Electrical and Electronics Engineering, chapter 1 - [Nodal analysis current in 6 Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Anodal-analysis-6-ohm-current) — Basic Electrical and Electronics Engineering, chapter 1 - [Nodal analysis: current in each branch](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Anodal-analysis-branch-currents-dependent) — Electrical Circuit Theory, chapter 1 - [Mesh analysis: voltage across 1 Ω resistor](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Amesh-analysis-dependent-source-1ohm) — Electrical Circuit Theory, chapter 1