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ECE 102 Engineering Computation Chapter 20 Source Transformations Dr. Herbert G. Mayer, PSU Status 10/2/2015 For use at CCUT Fall 2015 1 Syllabus Goal CVS With Rp Removed CCS With Rs Removed CVS to CCS Transformation Detailed Sample Conclusion Exercises 2 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 3 CVS With Rp Removed 4 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 5 CCS With Rs Removed 6 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 7 CVS to CCS Bilateral Transformation 8 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 9 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 10 Detailed Sample, Step a 11 Detailed Sample identify all: # of Essential nodes __4__ # of Essential branches __6__ 12 Detailed Sample, Step b , 13 Detailed Sample, Step c , 14 Detailed Sample, Step d , 15 Detailed Sample, Step e , 16 Detailed Sample, Step f , 17 Detailed Sample, Step g , 18 Detailed Sample, Step h , 19 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 20 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 21 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 22 Exercise 1 23 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 24 Exercise 1, Simplified Step 1 25 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 26 Exercise 1, Simplified Step 3 27 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 28 Exercise 1, Simplified Step 2 29 Exercise 1, Cont’d We can compute v0 v0 = 2 A * 10 Ω v0 = 20 V 30 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 31 Exercise 2, Compute Power of V 250 32 Exercise 2, Compute Power of V 250 is is is is = = = = i125 + i25 250/125 + (250 - v0)/25 250/125 + (250 - 20)/25 11.2 A Power ps is i * v ps = 250 * 11.2 = 2,800 W 33 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 34 Exercise 3, Compute Power of 8 A CCS 35 Exercise 3, Compute Power of 8 A CCS V0 V0 20 v8A v8A = = = = = v8A + v10Ω 20 V 8*10 + v8A 20 - 80 -60 Power p8A is i8A * v8A p8A = i8A*v8A 36