Loop Mesh Analysis Dr Holbert January 30 2008
Loop (Mesh) Analysis Dr. Holbert January 30, 2008 Lect 5 EEE 202 1
Loop Analysis • Nodal analysis was developed by applying KCL at each non-reference node • Loop analysis is developed by applying KVL around loops in the circuit • Loop (mesh) analysis results in a system of linear equations which must be solved for unknown currents Lect 5 EEE 202 2
Another Summing Circuit • The output voltage V of this circuit is proportional to the sum of the two input voltages V 1 and V 2 1 k. W + V 1 + – Vout 1 k. W + – V 2 – Lect 5 EEE 202 3
Steps of Mesh Analysis 1. Identify mesh (loops). 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 for the mesh/loop currents. Lect 5 EEE 202 4
1. Identifying the Meshes 1 k. W V 1 + – 1 k. W Mesh 1 Mesh 2 + – V 2 1 k. W Lect 5 EEE 202 5
Steps of Mesh Analysis 1. Identify mesh (loops). 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 for the mesh/loop currents. Lect 5 EEE 202 6
2. Assigning Mesh Currents 1 k. W V 1 Lect 5 + – 1 k. W I 1 I 2 EEE 202 + – V 2 7
Steps of Mesh Analysis 1. Identify mesh (loops). 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 for the mesh/loop currents. Lect 5 EEE 202 8
Voltages from Mesh Currents + VR R + – R I 1 – I 1 V R = I 1 R Lect 5 VR I 2 VR = (I 1 – I 2 ) R EEE 202 9
3. KVL Around Mesh 1 1 k. W V 1 + – 1 k. W I 1 I 2 + – V 2 –V 1 + I 1 1 k. W + (I 1 – I 2) 1 k. W = 0 I 1 1 k. W + (I 1 – I 2) 1 k. W = V 1 Lect 5 EEE 202 10
3. KVL Around Mesh 2 1 k. W V 1 + – 1 k. W I 1 I 2 + – V 2 (I 2 – I 1) 1 k. W + I 2 1 k. W + V 2 = 0 (I 2 – I 1) 1 k. W + I 2 1 k. W = –V 2 Lect 5 EEE 202 11
Steps of Mesh Analysis 1. Identify mesh (loops). 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 for the mesh/loop currents. Lect 5 EEE 202 12
Matrix Notation • The two equations can be combined into a single matrix/vector equation Lect 5 EEE 202 13
4. Solving the Equations Let: V 1 = 7 V and V 2 = 4 V Results: I 1 = 3. 33 m. A I 2 = – 0. 33 m. A Finally Vout = (I 1 – I 2) 1 k. W = 3. 66 V Lect 5 EEE 202 14
Another Example 2 k. W 2 m. A 12 V Lect 5 + – 1 k. W 2 k. W I 0 EEE 202 4 m. A 15
1. Identify Meshes 2 k. W 2 m. A Mesh 3 1 k. W 12 V Lect 5 + – Mesh 1 2 k. W Mesh 2 I 0 EEE 202 4 m. A 16
2. Assign Mesh Currents 2 k. W 2 m. A 12 V Lect 5 + – I 1 I 3 1 k. W 2 k. W I 0 EEE 202 I 2 4 m. A 17
Current Sources • The current sources in this circuit will have whatever voltage is necessary to make the current correct • We can’t use KVL around any mesh because we don’t know the voltage for the current sources • What to do? Lect 5 EEE 202 18
Current Sources • The 4 m. A current source sets I 2: I 2 = – 4 m. A • The 2 m. A current source sets a constraint on I 1 and I 3: I 1 – I 3 = 2 m. A • We have two equations and three unknowns. Where is the third equation? Lect 5 EEE 202 19
2 k. W The Supermesh surrounds this source! 12 V 2 m. A + – I 3 2 k. W I 2 I 1 1 k. W The Supermesh does not include this source! 4 m. A I 0 Lect 5 EEE 202 20
3. KVL Around the Supermesh -12 V + I 3 2 k. W + (I 3 - I 2)1 k. W + (I 1 - I 2)2 k. W = 0 I 3 2 k. W + (I 3 - I 2)1 k. W + (I 1 - I 2)2 k. W = 12 V Lect 5 EEE 202 21
Matrix Notation • The three equations can be combined into a single matrix/vector equation Lect 5 EEE 202 22
4. Solve Using MATLAB >> A = [0 1 0; 1 0 -1; 2 e 3 -1 e 3 -2 e 3+1 e 3]; >> v = [-4 e-3; 2 e-3; 12]; >> i = inv(A)*v i = 0. 0012 -0. 0040 -0. 0008 Lect 5 EEE 202 23
Solution I 1 = 1. 2 m. A I 2 = – 4 m. A I 3 = – 0. 8 m. A I 0 = I 1 – I 2 = 5. 2 m. A Lect 5 EEE 202 24
Advantages of Nodal Analysis • Solves directly for node voltages • Current sources are easy • Voltage sources are either very easy or somewhat difficult • Works best for circuits with few nodes • Works for any circuit Lect 5 EEE 202 25
Advantages of Loop Analysis • Solves directly for some currents • Voltage sources are easy • Current sources are either very easy or somewhat difficult • Works best for circuits with few loops Lect 5 EEE 202 26
Disadvantages of Loop Analysis • Some currents must be computed from loop currents • Does not work with non-planar circuits • Choosing the supermesh may be difficult. • FYI: Spice uses a nodal analysis approach Lect 5 EEE 202 27
Class Examples • Drill Problems P 2 -12, P 2 -14, P 2 -15 Lect 5 EEE 202 28
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