New version page

UT Arlington EE 2315 - Mesh Current Analysis

Documents in this Course
Load more

This preview shows page 1-2-3 out of 10 pages.

View Full Document
View Full Document

End of preview. Want to read all 10 pages?

Upload your study docs or become a GradeBuddy member to access this document.

View Full Document
Unformatted text preview:

1/9/20111E E 2315Lecture 02 -Mesh Current AnalysisIntroduction to Mesh Current Method• More direct than branch equations• Fewer equations to solve• Express all variables in terms of mesh currents• Solution is set of mesh currents• Solution completely defines the circuit• Most Convenient Method to Model Magnetic Coupling (E E 2446 Topic)1/9/20112Mesh Current Example 1 (1/2)KVL at Mesh 1:KVL at Mesh 2:Using Ohm’s Law:10sacVvv20cbsvvV 2122scbVRIIRI  Mesh Current Example 1 (2/2)Above linear equations can be solved for mesh currents I1and I2.1122saccscbcVRRRIVRRRI   1/9/20113Mesh Current Example 1a (1/2)120 V 64 V8 24 I1I26 KVL at Mesh 1:KVL at Mesh 2:1120 120 6 24III 12120 30 2464 24 32II   Solve:1262.5IAIAMesh Current Example 2 (1/2)KVL @ Mesh 1:KVL @ Mesh 2:But:11210scbVRII RI2210xa ciRIRII  1/9/20114Mesh Current Example 2 (2/2)Solve for I1and I2:112sbc cVRRIRI  120cacRI R R I  Mesh Current Example 2a (1/2)KVL @ Mesh 1:KVL @ Mesh 2:But:12 10 120 24 8III2210410 24xiI II  1/9/20115Mesh Current Example 2a (2/2)12120 32 2402030II   Solve for I1and I2:127.5 5IAI AForced Mesh (1/2)• No KVL equation possible for mesh 2•But I2is known: I2= Is1/9/20116Forced Mesh (2/2)KVL for mesh 1:Substitute and Solve:1120sa cVRIRII  Forced Mesh Example 3a 108 V 5 A6  8 20 I1I2KVL for mesh 1:Substitute and Solve:110 108 6 20 5II 1108 100620I1/9/20117Supermesh Example (1/5)• No KVL possible for meshes 1 or 2• Use Supermesh (dotted loop) for KVLVs1Vs2IsRaRbRcRdReI1I2I3Supermesh Example (2/5)Supermesh KVL:Mesh 3 KVL:Vs1Vs2IsRaRbRcRdReI1I2I3112223 130()()sa b sdcVRIRIVRI I RI I   31 32 30( )( )cd eRI I RI I RI 1/9/20118Supermesh Example (3/5)Also:Vs1Vs2IsRaRbRcRdReI1I2I321 2 1ssIII III  12 1 113 13()()()ss a bsds cVV RIRIIRI I I RI I    Subst for I2:Supermesh Example (4/5)And:Rearranging the equations:Vs1Vs2IsRaRbRcRdReI1I2I331 3 1 30( )( )cdseRII RIII RI1/9/20119Supermesh Example (5/5)Vs1Vs2IsRaRbRcRdReI1I2I312 13s s bds abcdcdVV RRI RRRRIRRI   13ds c d c d eRIRRIRRRI    Supermesh with Numbers (1/3)1/9/201110Supermesh with Numbers (2/3)1340 30 20160 20 40IVIV    1367IAIASupermesh with Numbers


View Full Document
Loading Unlocking...
Login

Join to view Mesh Current Analysis and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Mesh Current Analysis and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?