Basic Laws of Electric Circuits Mesh Analysis Lesson
Basic Laws of Electric Circuits Mesh Analysis Lesson 7
Basic Circuits Mesh Analysis: Basic Concepts: In formulating mesh analysis we assign a mesh current to each mesh. Mesh currents are sort of fictitious in that a particular mesh current does not define the current in each branch of the mesh to which it is assigned.
Basic Circuits Mesh Analysis: Basic Concepts: Figure 7. 2: A circuit for illustrating mesh analysis. Around mesh 1: Eq 7. 1
Basic Circuits Mesh Analysis: Basic Concepts: Eq 7. 2 Eq 7. 3 Eq 7. 4
Basic Circuits Mesh Analysis: Basic Concepts: We are left with 2 equations: From (7. 1) and (7. 4) we have, Eq 7. 5 Eq 7. 6 We can easily solve these equations for I 1 and I 2.
Basic Circuits Mesh Analysis: Basic Concepts: The previous equations can be written in matrix form as: Eq (7. 7) Eq (7. 8)
Basic Circuits Mesh Analysis: Example 7. 1. Write the mesh equations and solve for the currents I 1, and I 2. Figure 7. 2: Circuit for Example 7. 1. Mesh 1 Mesh 2 4 I 1 + 6(I 1 – I 2) = 10 - 2 6(I 2 – I 1) + 2 I 2 + 7 I 2 = 2 + 20 Eq (7. 9) Eq (7. 10)
Basic Circuits Mesh Analysis: Example 7. 1, continued. Simplifying Eq (7. 9) and (7. 10) gives, 10 I 1 – 6 I 2 = 8 -6 I 1 + 15 I 2 = 22 » % A MATLAB Solution » » R = [10 -6; -6 15]; » » V = [8; 22]; » » I = inv(R)*V I = 2. 2105 2. 3509 Eq (7. 11) Eq (7. 12) I 1 = 2. 2105 I 2 = 2. 3509
Basic Circuits Mesh Analysis: Example 7. 2 Solve for the mesh currents in the circuit below. Figure 7. 3: Circuit for Example 7. 2. The plan: Write KVL, clockwise, for each mesh. Look for a pattern in the final equations.
Basic Circuits Mesh Analysis: Example 7. 2 Mesh 1: 6 I 1 + 10(I 1 – I 3) + 4(I 1 – I 2) = 20 + 10 Eq (7. 13) Mesh 2: 4(I 2 – I 1) + 11(I 2 – I 3) + 3 I 2 = - 10 - 8 Eq (7. 14) Mesh 3: 9 I 3 + 11(I 3 – I 2) + 10(I 3 – I 1) = 12 + 8 Eq (7. 15)
Basic Circuits Mesh Analysis: Example 7. 2 Clearing Equations (7. 13), (7. 14) and (7. 15) gives, Standard Equation form In matrix form: 20 I 1 – 4 I 2 – 10 I 3 = 30 -4 I 1 + 18 I 2 – 11 I 3 = -18 -10 I 1 – 11 I 2 + 30 I 3 = 20 WE NOW MAKE AN IMPORTANT OBSERVATION!!
Basic Circuits Mesh Analysis: Standard form for mesh equations Consider the following: R 11 = of resistance around mesh 1, common to mesh 1 current I 1. R 22 = of resistance around mesh 2, common to mesh 2 current I 2. R 33 = of resistance around mesh 3, common to mesh 3 current I 3.
Basic Circuits Mesh Analysis: Standard form for mesh equations R 12 = R 21 = - resistance common between mesh 1 and 2 when I 1 and I 2 are opposite through R 1, R 2. R 13 = R 31 = - resistance common between mesh 1 and 3 when I 1 and I 3 are opposite through R 1, R 3. R 23 = R 32 = - resistance common between mesh 2 and 3 when I 2 and I 3 are opposite through R 2, R 3. = sum of emf around mesh 1 in the direction of I 1. = sum of emf around mesh 2 in the direction of I 2. = sum of emf around mesh 3 in the direction of I 3.
Basic Circuits Mesh Analysis: Example 7. 3 - Direct method. Use the direct method to write the mesh equations for the following. Figure 7. 4: Circuit diagram for Example 7. 3. Eq (7. 13)
Basic Circuits Mesh Analysis: With current sources in the circuit Example 7. 4: Consider the following: Figure 7. 5: Circuit diagram for Example 7. 4. Use the direct method to write the mesh equations.
Basic Circuits Mesh Analysis: With current sources in the circuit This case is explained by using an example. Example 7. 4: Find the three mesh currents in the circuit below. Figure 7. 5: Circuit for Example 7. 4. When a current source is present, it will be directly related to one or more of the mesh current. In this case I 2 = -4 A.
Basic Circuits Mesh Analysis: With current sources in the circuit Example 7. 4: Continued. An easy way to handle this case is to remove the current source as shown below. Next, write the mesh equations for the remaining meshes. Note that I 2 is retained for writing the equations through the 5 and 20 resistors.
Basic Circuits Mesh Analysis: With current sources in the circuit Example 7. 4: Continued. Equation for mesh 1: 10 I 1 + (I 1 -I 2)5 = 10 or 15 I 1 – 5 I 2 = 10 Equations for mesh 2: 2 I 3 + (I 3 -I 2)20 = 20 or - 20 I 2 + 22 I 3 = 20 Constraint Equation I 2 = - 4 A
Basic Circuits Mesh Analysis: With current sources in the circuit Example 7. 4: Continued. Express the previous equations in Matrix form: I 1 = -0. 667 A I 2 = - 4 A I 3 = - 2. 73 A
circuits End of Lesson 7 Mesh Analysis
- Slides: 20