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ECE 102Engineering Computation

Chapter 20Source Transformations

Dr. Herbert G. Mayer, PSUDr. Herbert G. Mayer, PSUStatus 9/2/2015Status 9/2/2015

For use at CCUT Fall 2015For use at CCUT Fall 2015

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Syllabus

GoalGoal CVS With RCVS With Rpp Removed Removed CCS With RCCS With Rss Removed Removed CVS to CCS TransformationCVS to CCS Transformation Detailed SampleDetailed Sample ConclusionConclusion ExercisesExercises

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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 Sometimes a circuit can be transformed into

another one that is simpler, yet electrically 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 and load RL

This simplification is one of several source transformations an engineer should look for, before computing 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 the CCS has to produce, to deliver equal current to both Rs in series with R parallel to RL

Such a removal is one of several source transformations to simplify computing unknown units 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 / R

Vs = 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, and then we can simplify

Engineers must check carefully, how much simplification is feasible, and then simplify

But no more

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Exercise 1

Taken from [1], page 112, Example 4.9, part a)

Given the circuit on the following page, compute the voltage drop v0 across the 100 Ω resistor

Solely using source transformations

Do not even resort to KCL or KVL, just simplify and then use Ohm’s Law

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Exercise 1

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Exercise 1

We know that the circuit does not change, when we remove a resistor parallel to a CVS

Only the power delivered by the CVS will change

So we can remove the 125 Ω resistor We also know that the circuit does

not change, when we remove a resistor in series with a CCS

Only the overall power delivered by the CCS will change

So we can remove the 10 Ω resistor

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Exercise 1, Simplified Step 1

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Exercise 1, Cont’d

Computation of v0 does not change with these 2 simplifications

If we substitute the 250 V CVS with an equivalent CCS, we have 2 parallel CCS

These 2 CCSs can be combined

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Exercise 1, Simplified Step 3

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Exercise 1, Cont’d

Combine 2 parallel CCS of 10 A and -8 A

And combine 3 parallel resistors: 25 || 100 || 20 Ω = 10 Ω

Yielding an equivalent circuit that is simpler, and shows the desired voltage drop v0 along the equivalent source, and equivalent resistor

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Exercise 1, Simplified Step 2

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Exercise 1, Cont’d

We can compute v0

v0 = 2 A * 10 Ω

v0 = 20 V

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Exercise 2, Compute Power of V 250

Next compute the power ps delivered (or if sign reversed: absorbed) by the 250 V CVS

The current delivered by the CVS is named is

And it equals the sum of i125 and i25

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Exercise 2, Compute Power of V 250

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Exercise 2, Compute Power of V 250is = i125 + i25

is = 250/125 + (250 - v0)/25

is = 250/125 + (250 - 20)/25

is = 11.2 A

Power ps is i * v

ps = 250 * 11.2 = 2,800 W

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Exercise 3, Compute Power of 8 A CCS

Next compute the power p8A delivered by the 8 A CCS

First we find the voltage drop across the 8 A CCS, from the top essential node toward the 10 Ω resistor, named v8A

The voltage drop across the 10 Ω resistor is simply 10 Ω * the current, by definition 8 A, named v10Ω

That is v10Ω = 80 V

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Exercise 3, Compute Power of 8 A CCS

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Exercise 3, Compute Power of 8 A CCS

V0 = v8A + v10Ω

V0 = 20 V

20 = 8*10 + v8A

v8A = 20 - 80

v8A = -60

Power p8A is i8A * v8A

p8A = i8A*v8A