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Learn about source transformations and how they can simplify circuit computations. This chapter covers removing parallel and series loads, and transforming between CVS and CCS circuits. Includes detailed examples and exercises.
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ECE 221Electric Circuit Analysis IChapter 11Source Transformations Herbert G. Mayer, PSU Status 10/28/2015
Syllabus Goal CVS With Rp Removed CCS With Rs Removed CVS to CCS Transformation Detailed Sample Conclusion Exercises
Goal The Node Voltage and Mesh Current Methods are powerful tools to compute circuit parameters Cramer’s Rule is a useful mathematical tool for solving equations with a large number of unknowns Sometimes a circuit can be transformed into another one that is simpler, yet electrically equivalent Generally that will simplify computation We’ll learn about source transformations Method 1: remove parallel load from CVS Method 2: remove serial load from CCS Method 3: Transform CVS CCS bilaterally
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 load resistor R Removal of Rpdecreases 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 obvious source transformations an engineer should look for, before computing unknowns in a circuit
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
CVS to CCS Transformation • A given CVS of Vs Volt with resistor R in series produces a current iL in loads R and RL, connected externally • That current through loads R and 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
Detailed Sample Transformation • 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
Detailed Sample Transformation identify all: # of Essential nodes __4__ # of Essential branches __6__
Power in 6 V CVS • The current through network Step 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
Conclusion about Transformations 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
Exercise 1 • Taken from [1], page 112, Example 4.9, part a) • Given the circuit on the following page, compute the voltage drop v0across the 100 Ω resistor • Solely using source transformations • Do not even resort to KCL or KVL, just simplify and then use Ohm’s Law
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
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
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
Exercise 1, Cont’d • We can compute v0 v0 = 2 A * 10 Ω v0 = 20 V
Exercise 1, Using Ohm’s Law i25 = i3R + i8A i8A = 8 A i25 = ( 250 - v0 ) / 25 i3R = v0 / ( 100 Ohm || 20 Ohm ) i3R = v0 * 3 / 50 (250 - v0)/25 = 8 + v0 * 3/50 2*250 - 2*v0 = 8*50 + 3*v0 v0 = 20 V
Exercise 2, Compute Power of V 250 • Next compute the power psdelivered (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
Exercise 2, Compute Power of V 250 is = 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
Exercise 3, Compute Power of 8 A CCS • Next compute the power p8Adelivered 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 • We name this voltage drop v10Ω • That is v10Ω = 80 V
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: p8A = i8A * v8A
