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lecture 81 Loop (Mesh) Analysis (3.2) Prof. Phillips February 17, 2003

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lecture 82 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.

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lecture 83 Example: A Summing Circuit The output voltage V of this circuit is proportional to the sum of the two input voltages V 1 and V 2. This circuit could be useful in audio applications or in instrumentation. The output of this circuit would probably be connected to an amplifier.

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lecture 84 Summing Circuit Solution: V out = (V 1 + V 2 )/3 + – V out 1k V1V1 V2V2 +–+– +–+–

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lecture 85 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.

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lecture 86 Mesh 2 1k Identifying the Meshes V1V1 V2V2 Mesh 1 +–+– +–+–

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lecture 87 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.

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lecture 88 1k Assigning Mesh Currents V1V1 V2V2 I1I1 I2I2 +–+– +–+–

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lecture 89 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.

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lecture 810 Voltages from Mesh Currents R I1I1 +– VRVR V R = I 1 R R I1I1 +– VRVR I2I2 V R = (I 1 - I 2 ) R

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lecture 811 KVL Around Mesh 1 -V 1 + I 1 1k + (I 1 - I 2 ) 1k = 0 I 1 1k + (I 1 - I 2 ) 1k = V 1 1k V1V1 V2V2 I1I1 I2I2 +–+– +–+–

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lecture 812 KVL Around Mesh 2 (I 2 - I 1 ) 1k + I 2 1k + V 2 = 0 (I 2 - I 1 ) 1k + I 2 1k = -V 2 1k V1V1 V2V2 I1I1 I2I2 +–+– +–+–

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lecture 813 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.

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lecture 814 Matrix Notation The two equations can be combined into a single matrix/vector equation.

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lecture 815 Solving the Equations Let:V 1 = 7V and V 2 = 4V Results: I 1 = 3.33 mA I 2 = -0.33 mA Finally V out = (I 1 - I 2 ) 1k = 3.66V

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lecture 816 Another Example 1k 2k 12V4mA 2mA I0I0 +–+–

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lecture 817 Mesh 2 Mesh 3 Mesh 1 1. Identify Meshes 1k 2k 12V4mA 2mA I0I0 +–+–

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lecture 818 2. Assign Mesh Currents I1I1 I2I2 I3I3 1k 2k 12V4mA 2mA I0I0 +–+–

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lecture 819 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?

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lecture 820 Current Sources The 4mA current source sets I 2 : I 2 = -4 mA The 2mA current source sets a constraint on I 1 and I 3 : I 1 - I 3 = 2 mA We have two equations and three unknowns. Where is the third equation?

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lecture 821 1k 2k 12V4mA 2mA I0I0 I1I1 I2I2 I3I3 The Supermesh surrounds this source! The Supermesh does not include this source! +–+–

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lecture 822 KVL Around the Supermesh -12V + I 3 2k + (I 3 - I 2 )1k + (I 1 - I 2 )2k = 0 I 3 2k + (I 3 - I 2 )1k + (I 1 - I 2 )2k = 12V

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lecture 823 Matrix Notation The three equations can be combined into a single matrix/vector equation.

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lecture 824 Solve Using MATLAB >> A = [0 1 0; 1 0 -1; 2e3 -1e3-2e3 2e3+1e3]; >> v = [-4e-3; 2e-3; 12]; >> i = inv(A)*v i = 0.0012 -0.0040 -0.0008

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lecture 825 Solution I 1 = 1.2 mA I 2 = -4 mA I 3 = -0.8 mA I 0 = I 1 - I 2 = 5.2 mA

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lecture 826 Class Examples

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