L13 - 1)) mesh equations. 5A source gives us current in...

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Lecture 13 Sept. 24, 2008 Finishing Mesh Current Analysis Sumit Chaudhary Office: 2124 Coover [email protected]
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Reading and HW Read 4.6 and 4.7 HW –4 .46 Exam 1: Oct 1 st (stuff done till Friday of this week) KVL, KCL, equivalent resistance, node Voltage, mesh Current, Superposition, Source
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Setting Up Mesh Current equations 0 ) ( 0 ) ( 2 3 2 3 1 1 = + + = + + R i R i i v R i i R i v b a b b a a Instead of Setting up in mesh form Solve for mesh currents i a and i b . Then express branch current in terms of mesh currents
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Mesh current method with dependent sources φ i i i i i 15 ) ( 4 ) ( 20 0 2 3 1 3 + + = 3 1 i i i = Constraint Equation Dependent voltage source is no special thing for mesh current method. Just count it as drop or rise and write another constraint equation
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Mesh with dependent sources (AP 4.8)
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Mesh current method with special cases (current sources) We defined i a , i b , i c and v 5 branches with current unknown and 4 nodes. So we need ( 5–(4
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Unformatted text preview: 1)) mesh equations. 5A source gives us current in that branch. But we need an unknown voltage across it for the mesh current equations. ) ( 2 4 50 6 ) ( 3 100 b c c a b a i i v i i v i i − + − = − + + − = Mesh current method with special cases (current sources) ‐ SUPERMESH ) ( 2 4 50 6 ) ( 3 100 b c c a b a i i v i i v i i − + − = − + + − = c b a i i i 6 5 9 50 + − = Eliminating v Constraint Equation a c i i − = 5 Similar approach if the current source was dependent Supermesh (Assessment Prb 4.11) Supermesh (Assessment Prob 4.12) Supernodes vs. Supermeshes Supernode 4 100 50 5 3 2 1 2 = − + + − v v v v φ i v v 10 2 3 + = 5 50 2 − = v i 6 4 50 ) ( 2 ) ( 3 100 = + + + − + − + − a c b c b a i i i i i i a c i i − = 5 Constraint Equations...
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This note was uploaded on 02/08/2009 for the course EE 201 taught by Professor Kruempel during the Fall '06 term at Iowa State.

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L13 - 1)) mesh equations. 5A source gives us current in...

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