3.5 Mesh Current Analysis: Example 3-7
- There are two (left & right) meshes (“window panes”)
- I2 is known to be -3 A. (by inspection or by KCL)
- Write KVLs around meshes whose mesh currents are unknown:
- KVL (left mesh): -2W·I1 - 2W·(I1-I2) - 5 V. = 0 V. which (given I2 = -3 A.) has solution I1 = -2.75 A.