Circuit Theory #19 | Dependent-Source Supermesh Analysis - Find Va

Instructor: Dr. Süleyman Burak ÇELİK

Nineteenth lesson in the Circuit Theory series. This lesson solves a supermesh example that includes a dependent current source. The key step is to connect the dependent source value to the controlling circuit variable before solving the mesh-current equations. Main ideas: - A current source between two meshes gives a mesh-current constraint - A dependent current source gives another constraint, but only after writing it in terms of `Va` - The resistor voltage relation is `Va = -4i1` - The outer supermesh KVL skips the current-source branches Worked example: - Independent source constraint: `i2 - i1 = 12` - Dependent source constraint: `i2 - i3 = 2Va` - Voltage relation: `Va = -4i1` - Outer supermesh KVL: `-4i1 - 12i3 = 24` - Final system gives: `i1 = -1.5 mA`, `i2 = 10.5 mA`, `i3 = -1.5 mA` - Result: `Va = 6 V` Main takeaway: - Do not substitute a dependent source value until the controlling variable is related to the mesh currents - Use consistent units: kOhm times mA gives volts - After the constraints are written, solve the supermesh system normally