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ECE 221Electric Circuit Analysis I

Chapter 11Source Transformations

Herbert G. Mayer, PSUStatus 11/25/2014

For use at Changchun University of Technology CCUT

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Syllabus

Goal CVS With Rp Removed CCS With Rs Removed CVS to CCS Transformation Detailed Sample Conclusion

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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

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CVS With Rp Removed

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CVS With Rp Removed Removing the load Rp parallel to the CVS has

no impact on externally connected loads RL

Such loads RL—not drawn here— will be in series with resistor R

Removal of Rp decreases the amount of current that the CVS has to produce, to deliver equal voltage to both Rp and the series of R plus any load RL

This simplification is one of several source transformations an engineer should look for, before computing all unknowns in a circuit

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CCS With Rs Removed

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CCS With Rs Removed Removing the load Rs in series with the CCS

has no impact on external loads RL

Such a load RL—not drawn here— will be parallel to resistor R

Removal of Rs will certainly decrease the amount of voltage that the CCS has to produce, to deliver equal current to both Rs in series with R parallel to load RL

This simplification is one of several source transformations to simplify computing unknowns in a circuit

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CVS to CCS Bilateral Transformation

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CVS to CCS Transformation A given CVS of Vs Volt with resistor R in series

produces a current iL in a load, connected externally That current also flows through connected load RL

iL = Vs / ( R + RL ) A CCS of iS Ampere with parallel resistor R produces

a current iL in an externally connected load RL

For the transformation to be correct, these currents must be equal for all loads RL

iL = is * R / ( R + RL ) Setting the two equations for iL equal, we get:

is = Vs / RVs = is * R

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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

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Detailed Sample, Step a

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Detailed Sample

identify all:# of Essential nodes __4__

# of Essential branches __6__

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,Detailed Sample, Step b

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,Detailed Sample, Step c

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,Detailed Sample, Step d

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,Detailed Sample, Step e

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,Detailed Sample, Step f

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,Detailed Sample, Step g

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,Detailed Sample, Step h

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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 = P6V = i * V = 0.825 * 6 P6V = 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

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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