2 mesh analysis
- 2. Objectives To introduce the mesh – current method. To formulate the mesh-current equations. To solve electric circuits using the mesh-current method.
- 3. Mesh Analysis (Loop Analysis) MeshAnalysis is developed by applying KVL around meshes in the circuit. Loop (mesh) analysis results in a system of linear equations which must be solved for unknown currents. Reduces the number of required equations to the number of meshes Can be done systematically with little thinking As usual, be careful writing mesh equations – follow sign convention.
- 4. Definitions Mesh: Loop that does not enclose other loops Essential Branch: Path between 2 essential nodes (without crossing other essential nodes). How many mesh-currents? A # of essential nodes Ne = 4 # of essential branches Be = 6 + No. of Mesh-currents M = Be –(Ne-1) B C - •Enough equations to get unknowns
- 5. Steps of Mesh Analysis 1. Identify the number of basic meshes. 2. Assign a current to each mesh. 3. Apply KVL around each loop to get an equation in terms of the loop currents. 4. Solve the resulting system of linear equations.
- 6. Identifying the Meshes 1kΩ 1kΩ 1kΩ V1 + + Mesh 1 Mesh 2 V2 – – Assigning Mesh Currents 1kΩ 1kΩ 1kΩ V1 + + I1 I2 V2 – –
- 7. Voltages from Mesh Currents + VR – + VR – R I2 R I1 I1 V R = I1 R VR = (I1 - I2 ) R 1kΩ 1kΩ 1kΩ V1 + + I1 I2 V2 – –
- 8. Mesh-Current Equations R1 Ω R2 Ω R3 Ω V1 + + I1 I2 V2 – – -V1 + I1 R1 + (I1 - I2) R3 = 0 I2 R2 + V2 + (I2-I1) R3 = 0
- 9. Mesh Current Method 6Ω 1. Assign mesh currents 2. Write mesh equations 20 Ω i1 4Ω i1(20+6+4) + (i1-i2)(4+6) = 0 4Ω 6Ω i2(2+4+4) + (i2-i1)(4+6) – 70 = 0 + 3. Solve mesh equations 70V - i2 4Ω 40i1 - 10i2 = 0 2Ω 4Ω -10i1+ 20i2 = 70 ========================= 40i1 - 10i2 = 0 70i2 = 280 Solution: i1 = 1A; i2 = 4A
- 10. Mesh current method Cases Case I: When a current source exists only in one mesh Loop 1 -10 + 4i1 + 6(i1-i2) = 0 Loop 2 i2 = - 5A No need to write a loop equation
- 11. Case II: Super Mesh When a current source exists between two meshes
- 12. Case III: Mesh with Dependent Sources -75 + 5i1 + 20(i1-i2) = 0 10ix + 20(i2-i1) + 4i2 = 0 5Ω 4Ω ix = i1 - i2 + + 75V 20Ω i3 10ix - i1 i2 - -75 + 5i1 + 20(i1-i2) = 0 10(i1-i2) + 20(i2-i1) + 4i2 = 0 i2 = 5A i1 = 7A
- 13. Example Use the mesh-current method to find io Ans. Io = A
- 14. Solution
- 15. Solution