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MESH ANALYSIS THEORY Loop Analysis 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. EE2 Lect-02 2 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. I1 I2 I3 EE2 Lect-02 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. EE2 Lect-02 4 Identifying the Meshes 1kW 1kW 1kW V1 + + Mesh 1 Mesh 2 V2 – – EE2 Lect-02 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. EE2 Lect-02 6 Assigning Mesh Currents 1kW 1kW 1kW V1 + + I1 I2 V2 – – EE2 Lect-02 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. EE2 Lect-02 8 Voltages from Mesh Currents + VR – + VR – I2 R R I1 I1 VR = I1 R VR = (I1 - I2 ) R EE2 Lect-02 9 KVL Around Mesh 1 1kW 1kW 1kW V1 + + I1 I2 V2 – – I1 1kW + (I1 - I2) 1kW = V1 I1 2kW - I21kW = V1 EE2 Lect-02 10 KVL Around Mesh 2 1kW 1kW 1kW V1 + + I1 I2 V2 – – -(I1 ) 1kW + I2 2kW + = -V2 - I11kW + I2 2kW = -V2 EE2 Lect-02 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. EE2 Lect-02 12 Matrix Notation The two equations can be combined into a single matrix/vector equation. 2kW 1kW I1 V1 1kW 2kW I V 2 2 EE2 Lect-02 13 Solving the Equations Let: V1 = 7V and V2 = 4V Results: I1 = 3.33 mA I2 = -0.33 mA EE2 Lect-02 14 Mesh Analysis: Example 1 Write the mesh equations and solve for the currents I 1, and I2. 4W 2W 7W 6W 10V + _ I1 I2 2V + _ _ 20V + Mesh 1 4I1 + 6(I1 – I2) = 10 - 2 Eq (1) Mesh 2 -6(I1 – I2) + 2I2 + 7I2 = 2 + 20 Eq (2) EE2 Lect-02 15 Mesh Analysis: Example 1, continued. Simplifying Eq (1) and (2) gives, 10I1 – 6I2 = 8 -6I1 + 15I2 = 22 I1 = 2.2105 I2 = 2.3509 EE2 Lect-02 16 Mesh Analysis: Example 2 Solve for the mesh currents in the circuit below. 12V 9W _ _+ + I3 8V 10 W 11 W + _ 6W 4W 3W I1 _ I2 20V + _ 10V _ + The plan: Write KVL, clockwise, for each mesh. Look for a pattern in the final equations. EE2 Lect-02 17 12V 9W _ _+ + I3 8V 10 W 11 W + _ 6W 4W 3W I1 _ I2 20V + _ 10V _ + Mesh 1: 6I1 + 10(I1 – I3) + 4(I1 – I2) = 20 + 10 Eq (1) Mesh 2: -4(I1 – I2) + 11(I2 – I3) + 3I2 = - 10 - 8 Eq (2) Mesh 3: 9I 3 - 11(I2 – I3) - 10(I1 – I2) = 12 + 8 Eq (3) EE2 Lect-02 18 Clearing Equations (1), (2) and (3) gives, Standard Equation form In matrix form: 20I1 – 4I2 – 10I3 = 30 20 4 10 I1 30 4 18 11 I 18 -4I1 + 18I2 – 11I3 = -18 2 10 11 30 I 3 20 -10I1 – 11I2 + 30I3 = 20 WE NOW MAKE AN IMPORTANT OBSERVATION!! EE2 Lect-02 19 Mesh Analysis: By Inspection Consider the following: R11 R12 R13 I1 emfs(1) 20 4 10 I1 30 4 18 11 I 18 R R23 I 2 emfs( 2) 2 21 R22 10 11 30 I 3 20 R31 R32 R33 I 3 emfs( 3) 12V 9W R11 = of resistance around mesh 1, _ _+ + common to mesh 1 current I1. I3 8V 10 W 11 W + _ R22 = of resistance around mesh 2, 6W 4W common to mesh 2 current I2. 3W I1 _ I2 20V + _ 10V _ + R33 = of resistance around mesh 3, common to mesh 3 current I3. EE2 Lect-02 20 Mesh Analysis: By Inspection R12 = R21 = - resistance common between mesh 1 and 2 when I1 and I2 are opposite through R1,R2. R13 = R31 = - resistance common between mesh 1 and 3 when I1 and I3 are opposite through R1,R3. R23 = R32 = - resistance common between mesh 2 and 3 when I2 and I3 are opposite through R2,R3. emfs(1) = sum of emf around mesh 1 in the direction of I1. emfs( 2) = sum of emf around mesh 2 in the direction of I2. emfs( 3) = sum of emf around mesh 3 in the direction of I3. EE2 Lect-02 21 Mesh Analysis: Example 3 – By Inspection Use the direct method to write the mesh equations for the following. 20 W 30 W 12 W _ 8W 10 W 20V + _ I1 I2 + 15V I3 + _ _ 10V 10 W + 30V 30 10 0 I 1 10 10 50 10 I 25 2 0 10 30 I 3 15 EE2 Lect-02 22 Mesh Analysis: Example 4 2kW 2mA 1kW + 12V 2kW 4mA – I0 EE2 Lect-02 23 1. Identify Meshes 2kW 2mA Mesh 3 1kW + 2kW 12V Mesh 1 Mesh 2 4mA – I0 EE2 Lect-02 24 2. Assign Mesh Currents 2kW 2mA I3 1kW + 2kW 12V I1 I2 4mA – I0 EE2 Lect-02 25 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 the loop because we don’t know the voltage. What to do? EE2 Lect-02 26 Current Sources The 4mA current source sets I2: I2 = -4 mA The 2mA current source sets a constraint on I1 and I3: I1 - I3 = 2 mA We have two equations and three unknowns. Where is the third equation? EE2 Lect-02 27 The 2kW Supermesh The Supermesh does not surrounds 2mA I3 include this this source! 1kW source! + 2kW 12V I1 I2 4mA – I0 EE2 Lect-02 28 KVL Around the Supermesh -12V + I3 2kW + (I3 – I2)1kW + (I1 - I2)2kW = 0 I3 2kW + (I3 - I2)1kW + (I1 - I2)2kW = 12V I3 3kW - I23kW + I12kW = 12V EE2 Lect-02 29 Matrix Notation The three equations can be combined into a single matrix/vector equation. 0 1 0 I 1 4mA 1 0 1 I 2 2mA 2kW 3kW 3kW I 3 12V EE2 Lect-02 30 Solution I1 = 1.2 mA I2 = -4 mA I3 = -0.8 mA EE2 Lect-02 31 Mesh Analysis: Example 5 20V 2W _ + 10 W I3 20 W 10V + _ I1 5W I2 4A 15 W EE2 Lect-02 32 This case is explained by using an example. 20V 2W _ + 10 W I3 20 W 10V + _ I1 5W I2 4A 15 W When a current source is present, it will be directly related to one or more of the mesh current. In this case I 2 = -4A. EE2 Lect-02 33 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. 20V 2W _ + 10 W I3 20 W 10V + _ I1 5W I2 15 W Note that I 2 is retained for writing the equations through the 5 W and 20 W resistors. EE2 Lect-02 34 20V 2W _ + 10 W I3 20 W Equation for mesh 1: 10V + _ I1 5W I2 10I1 + (I1-I2)5 = 10 15 W or 15I1 – 5I2 = 10 Equations for mesh 2: 2I3 - (I2-I3)20 = 20 Constraint Equation or I2 = - 4A - 20I2 + 22I3 = 20 EE2 Lect-02 35 15 5 0 I1 10 0 20 22 I 20 2 0 1 0 I 3 4 I1 = -0.667 A I2 = - 4 A I3 = - 2.73 A EE2 Lect-02 36

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