Circuit Theory #18 | Supermesh Analysis Example 3 - Three Mesh Currents

Instructor: Dr. Süleyman Burak ÇELİK

Eighteenth lesson in the Circuit Theory series. This lesson solves another supermesh analysis example with three mesh currents and one shared current source. Main ideas: - A current source between two meshes gives a constraint equation - The lower supermesh path bypasses the current-source branch - A second KVL equation comes from the top loop - The three equations are solved together for all mesh currents Worked example: - Current-source constraint: `i3 - i1 = 5` - Lower supermesh KVL: `9i1 - 5i2 + 6i3 = 50` - Top loop KVL: `3i1 - 15i2 + 2i3 = 0` - Results: `i1 = 7/4 A`, `i2 = 5/4 A`, `i3 = 27/4 A` Main takeaway: - Write the current-source constraint first - Choose a KVL path that does not pass through the current source - Use the remaining independent loop equation to complete the system