EE1002 Notes Part1

# Casemeshanalysiswithcontrolledsources

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Unformatted text preview: 2 ) R4 (i1 i2 ) Vs R1i1 i1 Fig. Assigning the branch currents in terms of the mesh currents. Applying the KVL to the meshes, one can get: For mesh ABCDA, algebraic sum of voltage rises: VS R1i1 R2i1 R4 (i1 i2 ) 0 40 (Loop1) EE1002 Introduction to Circuits and Systems For mesh ADA, algebraic sum of voltage rises: R4 (i1 i2 ) R3i2 0 (Loop2) Rearranging the terms we get two independent equations: i1 ( R1 R2 R4 ) i2 R4 Vs i1 R4 i2 ( R3 R4 ) 0 Solving the two equations above, one can get: i1 VS R1 R2 i2 i1 R3 R4 R3 R4 R4 R3 R4 Using the two mesh currents, we can find the current and voltages associated with all the resistors in the circuit. Case : With a voltage source in one of the branches Suppose there were an extra voltage source between in D and D’ as given in the Figure. i1 R2i1 V0 i2 R3i2 (i1 i2 ) R4 (i1 i2 ) Vs R1i1 i1 Applying the KVL to the meshes, one can get: For mesh ABCDD’A, sum of voltage rises: VS R1i1 R2i1 V0 R4 (i1 i2 ) 0 (Loop1) (Loop2) For mesh AD’DA, sum of voltage rises: R4 i1...
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## This document was uploaded on 01/20/2014.

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