Circuit Theory #11 | Nodal Analysis Example 4 — Voltage Source to Ground

Instructor: Dr. Süleyman Burak ÇELİK

Eleventh lesson in the Circuit Theory series. A new twist: for the first time, a voltage source appears in a nodal analysis problem. The key insight — when a voltage source has one terminal at ground, the other terminal's voltage is immediately known. The circuit: - 1 mA current source pushes current into node V1 - 1 kOhm, 3 kOhm, and 2 kOhm resistors connect V1 to ground - 6 V source in series with the 2 kOhm fixes the far-end junction at 6 V - Only one unknown node: V1 KCL at V1: V1/1k + V1/3k + (V1 - 6)/2k - 1m = 0 Multiply by 6k: 6V1 + 2V1 + 3(V1 - 6) - 6 = 0 11V1 = 24 Solution: V1 = 24/11 = 2.182 V, so Va = V1 = 2.182 V Take-home: a voltage source to ground reduces the number of unknowns. But what if it sits between two unknown nodes? That is the supernode — coming in the next video.