Circuit Theory #09 | Nodal Analysis Example 2 — Voltage Across Each Source

Instructor: Dr. Süleyman Burak ÇELİK

Ninth lesson in the Circuit Theory series. A second worked nodal-analysis example that pushes the same four-step method through two new twists. What's new vs Lesson #08: • A current source sits **between** two non-reference nodes (5 A flowing from V₁ toward V₂), not between a node and ground • One node has **two** current sources at once • The question asks for the **voltage across each current source**, not just the node voltages The circuit: • Two unknown nodes V₁, V₂ — reference at the bottom • 2Ω from V₁ to ground, 4Ω from V₁ to V₂, 4Ω from V₂ to ground • 3 A current source enters V₁; 5 A flows from V₁ to V₂; 6 A enters V₂ Solution: • KCL at V₁: V₁/2 + (V₁−V₂)/4 − 3 + 5 = 0 → 3V₁ − V₂ = −8 • KCL at V₂: V₂/4 + (V₂−V₁)/4 − 5 − 6 = 0 → −V₁ + 2V₂ = 44 • Substitution: V₁ = 5.6 V, V₂ = 24.8 V • KVL gives the source voltages: V₃A = 5.6 V, V₅A = 19.2 V, V₆A = 24.8 V Take-home: once the node voltages are known, KVL gives the voltage across **any** element — just take the difference. The plus terminal of a current source is on the side where the current exits. Covers content from TR lecture d17 (Düğüm Analizi Örnek 2).