1 ECE 221 Electric Circuit Analysis I Chapter 11 Source Transformations Herbert G. Mayer, PSU Status 10/28/2015

2 Syllabus Goal CVS With R p Removed CCS With R s Removed CVS to CCS Transformation Detailed Sample Conclusion Exercises

3 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

4 CVS With R p Removed

5 Removing the load R p parallel to the CVS has no impact on externally connected loads R L Such loads R L —not drawn here— will be in series with load resistor R Removal of R p decreases the amount of current that the CVS has to produce, to deliver equal voltage to both R p and the series of R and load R L This simplification is one of several obvious source transformations an engineer should look for, before computing unknowns in a circuit

6 CCS With R s Removed

7 Removing the load R s in series with the CCS has no impact on external loads R L Such a load R L —not drawn here— will be parallel to resistor R Removal of R s will certainly decrease the amount of voltage the CCS has to produce, to deliver equal current to both R s in series with R parallel to R L Such a removal is one of several source transformations to simplify computing unknown units in a circuit

8 CVS to CCS Bilateral Transformation

9 CVS to CCS Transformation A given CVS of V s Volt with resistor R in series produces a current i L in loads R and R L, connected externally That current through loads R and R L i L = V s / ( R + R L ) A CCS of i S Ampere with parallel resistor R produces a current i L in an externally connected load R L For the transformation to be correct, these currents must be equal for all loads R L i L = i s * R / ( R + R L ) Setting the two equations for i L equal, we get: i s = V s / R V s = i s * R

10 CVS to CCS Bilateral Transformation

11 A Detailed Sample Transformation

12 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 , 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

13 Detailed Sample Transformation, Step a

14 Detailed Sample Transformation identify all: # of Essential nodes __4__ # of Essential branches __6__

15, Detailed Sample Transformation, Step b

16, Detailed Sample Transformation, Step c

17, Detailed Sample Transformation, Step d

18, Detailed Sample Transformation, Step e

19, Detailed Sample Transformation, Step f

20, Detailed Sample Transformation, Step g

21, Detailed Sample Transformation, Step h

22 Power in 6 V CVS The current through network Step h, in the direction of the 6 V CVS source is: i = ( ) / ( ) [ V / Ω ] i = [ A ] Power in the 6 V CVS, being current * voltage is: P = P 6V = i * V = * 6 P 6V = 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

23 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

24 Exercise 1 To Practice Transformations

25 Exercise 1 Taken from [1], page 112, Example 4.9, part a) Given the circuit on the following page, compute the voltage drop v 0 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

26 Exercise 1

27 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

28 Exercise 1, Simplified Step 1

29 Exercise 1, Cont’d Computation of v 0 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

30 Exercise 1, Simplified Step 3

31 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 v 0 along the equivalent source, and equivalent resistor

32 Exercise 1, Simplified Step 2

33 Exercise 1, Cont’d We can compute v 0 v 0 = 2 A * 10 Ω v 0 = 20 V

34 Exercise 1 Using Ohm’s Law to compute v 0

35 Exercise 1, Using Ohm’s Law

36 Exercise 1, Using Ohm’s Law i 25 = i 3R + i 8A i 8A =8 A i 25 = ( v 0 ) / 25 i 3R =v 0 / ( 100 Ohm || 20 Ohm ) i 3R =v 0 * 3 / 50 (250 - v 0 )/25= 8 + v 0 * 3/50 2* *v 0 = 8*50 + 3*v 0 v 0 = 20 V

37 Exercise 2

38 Exercise 2, Compute Power of V 250 Next compute the power p s delivered (or if sign reversed: absorbed) by the 250 V CVS The current delivered by the CVS is named i s And it equals the sum of i 125 and i 25

39 Exercise 2, Compute Power of V 250

40 Exercise 2, Compute Power of V 250 i s = i i 25 i s = 250/125 + (250 - v 0 )/25 i s = 250/125 + ( )/25 i s = 11.2 A Power p s is i * v p s = 250 * 11.2 = 2,800 W

41 Exercise 3, Compute Power of 8 A CCS Next compute the power p 8A 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 v 8A The voltage drop across the 10 Ω resistor is simply 10 Ω * the current, by definition 8 A We name this voltage drop v 10Ω That is v 10Ω = 80 V

42 Exercise 3, Compute Power of 8 A CCS

43 Exercise 3, Compute Power of 8 A CCS V 0 = v 8A + v 10Ω V 0 = 20 V 20= 8*10 + v 8A v 8A = v 8A = -60 Power p 8A is: p 8A = i 8A * v 8A