1. 1 ECE 221 Electric Circuit Analysis I Chapter 11 Source Transformations Herbert G. Mayer, PSU Status 11/25/2014 For use at Changchun University of Technology CCUT

2. 2 Syllabus Goal Goal CVS With R p Removed CVS With R p Removed CCS With R s Removed CCS With R s Removed CVS to CCS Transformation CVS to CCS Transformation Detailed Sample Detailed Sample Conclusion Conclusion

3. 3 Goal The Node-Voltage and Mesh-Current Methods are powerful tools to compute circuit parameters Cramer’s Rule is especially useful for a large number of unknowns; we practice in ECE 221 just for 3 unknowns Sometimes a circuit can be transformed into another one that is simpler, yet equivalent Generally that will simplify computations We’ll learn a few source transformations here Method 1: remove parallel load from CVS Method 2: remove serial load from CCS Method 3: Transform CVS CCS bilaterally

4. 4 CVS With R p Removed

5. 5 Removing the load R p parallel to the CVS has no impact on externally connected loads R L Such loads R L —not drawn here— will be in series with resistor R Removal of R p decreases the amount of current that the CVS has to produce, to deliver equal voltage to both R p and the series of R plus any load R L This simplification is one of several source transformations an engineer should look for, before computing all unknowns in a circuit

6. 6 CCS With R s Removed

7. 7 Removing the load R s in series with the CCS has no impact on external loads R L Such a load R L —not drawn here— will be parallel to resistor R Removal of R s will certainly decrease the amount of voltage that the CCS has to produce, to deliver equal current to both R s in series with R parallel to load R L This simplification is one of several source transformations to simplify computing unknowns in a circuit

8. 8 CVS to CCS Bilateral Transformation

9. 9 CVS to CCS Transformation A given CVS of V s Volt with resistor R in series produces a current i L in a load, connected externally That current also flows through connected load R L i L = V s / ( R + R L ) A CCS of i S Ampere with parallel resistor R produces a current i L in an externally connected load R L For the transformation to be correct, these currents must be equal for all loads R L i L = i s * R / ( R + R L ) Setting the two equations for i L equal, we get: i s = V s / R V s = i s * R

10. 10 Detailed Sample We’ll use these simplifications in the next example to generate an equivalent circuit that is minimal I.e. eliminate all redundancies from right to left This example is taken from [1], page 110-111, expanded for added detail First we analyze the sample, identifying all # of Essential nodes ____ # of Essential branches ____ Then we compute the power consumed or produced in the 6V CVS

11. 11 Detailed Sample, Step a

12. 12 Detailed Sample identify all: # of Essential nodes __4__ # of Essential branches __6__

13. 13, Detailed Sample, Step b

14. 14, Detailed Sample, Step c

15. 15, Detailed Sample, Step d

16. 16, Detailed Sample, Step e

17. 17, Detailed Sample, Step f

18. 18, Detailed Sample, Step g

19. 19, Detailed Sample, Step h

20. 20 Power in 6 V CVS The current through network h, in the direction of the 6 V CVS source is: i = ( 19.2 - 6 ) / ( 4 + 12 ) [ V / Ω ] i = 0.825 [ A ] Power in the 6 V CVS, being current * voltage is: P = P 6V = i * V = 0.825 * 6 P 6V = 4.95 W That power is absorbed in the 6 V source, it is not being delivered by the 6 V source! It is delivered by the higher voltage CVS of 19.2 V

21. 21 Conclusion Such source transformations are not always possible Exploiting them requires that there be a certain degree of redundancy Frequently that is the case Engineers must check carefully, how much simplification is feasible, and then simplify But no more