Post on 08-Jan-2018
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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