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Additional Analysis Techniques in Electrical Engineering

Learn about Superposition, Source Transformation, Thevenin and Norton Theorems, and Maximum Power Principle in this chapter of Electrical Engineering.

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Additional Analysis Techniques in Electrical Engineering

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  1. وزارةالتعليم العالي والبحث العلمي الجامعة المستنصرية كلية الهندسة قسم هندسة الحاسوب Ministry of Higher Education and Scientific Research Al-MustansiriyahUniversity Faculty ofEngineering Computer EngineeringDepartment Electrical Eng. Lab 1' class Assistant Lecturer Hazimalkargole 2017 20189

  2. Chapter(4) • Additional AnalysisTechniques • Here we will study four additionalTechniques • Superposition • Sourcetransformation • Thevenin and NortonTheorems • Maximum powerprinciple • 1. Superposition: • Definition: • Whenever a linear circuit is excited by more than one independent source, the total response is the algebraic sum of individualresponses • The idea is to activate one independent source at a timeto get individualresponse. • Then add the individual response to get totalresponse

  3. Note: Dependent source are Neverdeactivated (alwaysactive) When an independent voltage source is deactivated, it is set to zero. replaced by shortcircuit When an independent current source is deactivated, it is set to zero. replaced by opencircuit Example: Use superposition tofind 6Ω i1,i2,i3,i4? 2Ω i3 i1 120V 3Ω 4Ω 12A i2 +- i4

  4. •Activate independent voltage source 120 Vonly 2Ω V 6Ω 1 'i3 'i1 open circuit for currentsource 120V 3Ω 4Ω 'i2 +- 'i4 •Using KCL at V1 (nodalanalysis) 'i1'i2'i30

  5. 120V1V1V1  0 6 3 2 4 20V 1110 6 1 6 3   V1 30V 120V 90 'i  15A 1 1 6 10A 6 V 30 'i  1 2 3 3 1 V 'i i 5 A 3 4 6

  6. * Activate the independent current sourceonly V4 V3 6Ω 2Ω ''i1 ''i3 short circuit for voltagesource 12A 3Ω 4Ω ''i2 ''i4 KCL at V3: "i1"i2"i30 V3 V3 V3 V4  0 6 3 2 V32V33(V3V)0 -6V33V40KK(1)

  7. KCL atV4: "i3"i4120 V3 V4 V4 120 2 4 2 V3  2 V4 V4 48 2 V3  3 V4 48 KK(2) V312V V4 24 V

  8. V 12  2A "i 3 1 6 6 V 12 4 A "i 3 2 3 3 V3 V4 1224   6A "i 3 2  2  6 A V 24 "i 4 4 4 4 i1'i1"i115217A i2'i2"i21046A i3'i3"i35611A i4'i4"i456-1A

  9. Example: 0.4VΦ 5Ω + iΦ + 10V 10Ω V 20Ω 5A Φ V +- o - - 2iΦ Use super positionto findV ? 0 Activate voltage sourceonly: 0.4'V Φ 5Ω X + 'iΦ + 10V 10Ω 'V 20Ω Φ 'V +- o - - 2'iΦ

  10. 'VΦ10(0.4'VΦ)4'VΦ 'VΦ4'VΦ  'VΦ0 Dependent current source isopen 5Ω 'V 1020 25 + 10V 0  +- 20Ω 'V o 'V 8V 0 -

  11. Activate independent current sourceonly: 0.4''VΦ 5Ω Z y + ''iΦ 5A + ''VΦ - 10Ω 20Ω ''V o - 2''iΦ KCL atnode -"V0 "V0 5 20 (y): 0.4"V 0 Φ -4"V0"V08"VΦ0 -5"V08"VΦ0KK(1)

  12. 5"VΦ 0.4"V 0 Φ 10 0.5"VΦ5 "VΦ 10V "V 8"V -8016V 0 Φ 5 5 V0'V0"V0816-8V

  13. Example: Use superposition to find V ? 4A 5 2 100V + V - +- 12 10 Consider the independent sourceonly

  14. 5 2 X i3 10 100Vi1 + 'V - +- 12 i2 Apply KCL at node (x): i1i2i30 100'V'V'V0 5 10 14 22'V'V'V0 5 10 14 ' V 1 1 1 22 'V59.23V 5 10 14

  15. Consider the independent sourceonly. 4A 4A 5 2 5 2 + ''V - 12 10 10 - + ''V 12

  16. 4A 10  3 ix 2 - + ''V Currentdivider iy     12 12 ix4A 2.769A 10 3  12 2 "Vi 109.23V  x 3  V'V"V59.239.2350V

  17. 2.SourceTransformation: A transformation that allow a voltage source inseries with a resistor to be replaced by a current source in parallel with the same resistor or viceversa a How? IL IL a R + VL - + VLRL - IS +- V s b KCT1 We need tofind bothcircuits b KCT2 Is and Vs such that VL and IL is the samein

  18. In KCT 1, Vs I  L R R L In KCT2, R I  I L s R R L For IL to be the same , weneed Vs  RIs

  19. a a R +- IS R VS b b Where V or I  s V RI s s s R

  20. Example: Using source transformation, find the power associated withthe 6 Vsource. 4Ω 6Ω 5Ω 40V +- 6V +- 30Ω 20Ω 10Ω 1. Consider the 40 V source in series with(5 4Ω 6Ω 40 8A 5 6V +- 30Ω 20Ω 5Ω 10Ω

  21. 2. Take 5 6Ω 4Ω 8A 6V 5 //20  4Ω +- 30Ω 10Ω 3. Consider 8A in parallel with(4 4Ω 6Ω 4A (8A)(4)= 32V 6V +- +- 30Ω 10Ω

  22. 4. Take (4+6+10) inseries 4Ω 20 6V 32V +- +- 30Ω 5 Consider 32 V in series with(20 4Ω 32 1.6A 20 6V +- 20Ω 30Ω

  23. 6. Take (30//20 4Ω 1.6A 30 //20 12Ω 6V +- 7Consider 1.6A in parallel with (15 4Ω 12 i 1.6 A(12) =19.2V 6V +- +-

  24. i19.260.825A  P vi6(0.825) 6V 412 P6v  4.95 W(absorbing) Example: Use source transformation to findV0 25Ω 5Ω 250V + Vo - 8A +- 100Ω 15Ω

  25.  Take (5//15)//100 = 6.66 25Ω 250V + Vo - 8A +- 16.66Ω 2. Consider (250 V) in series with (25 250  10A 25 + Vo - 8A 16.66Ω 25Ω

  26. 3. Findequivalent + Vo - i1082A R16.66//25 10Ω V0iR(2A)(10Ω)20V

  27. Example: Use source transformation to findV0 1.6Ω 60V 20Ω +- + 36A Vo - 8Ω 120V 6Ω 5Ω +- 1.6Ω 120  6A 60 12A + 36A 5 20 8Ω V 20Ω o 6Ω 5Ω -

  28. 1.6Ω i1 i2 20 // 5 //6  2.4Ω R2 i = 36 + 6 -12 = 30A + Vo - 8Ω R3 R 1 R1 2.4 i i  (30)  6A 2 R R R 2.41.68 1 2 3 V0i2R36A8Ω48V

  29. Example : Use source transformation to findV0? 1ohm 1ohm 1ohm + Vs 10V + 2A 1ohm 1ohm - Vo -- Consider (10V) in series with (11ohm 1ohm + 1 ohmVo 10A 2A 1ohm 1ohm --

  30. Take(1//1)=0.5 1ohm 1ohm + Vo 2A 1/2ohm 10A 1ohm -- Consider (10A) in parallel with (0.5 1/2 ohm 1ohm 1ohm + + Vo 2A 5V 1ohm - --

  31. Take 0.5Ω in series with 1Ω 1.5ohm 1ohm + + 5V - 2A Vo 1ohm -- Consider 5V in series with 1.5 1ohm + Vo 2A 1.5ohm 10/3A 1ohm --

  32. 1ohm Add the currentsources + 4/3A Vo 1.5ohm 1ohm -- 7. Take (4/3A) in parallel with (3/2 1.5ohm 1ohm + + Vo 2V 1ohm 1  - V0111.5 2  4/7V --

  33. Thevenin and NortonTheorems TheveninTheorem: A portion of the circuit at a pair of nodes can be replaced by a voltage source Voc in series with a resistor RTH, where Voc is the open circuit voltage and RTH is the Thevenin’s equivalent resistance obtained by considering the open circuit with all independent sources madezero a RTH a circuit VOC RL +- RL b b

  34. Norton Theorem: A portion of the circuit at pair of nodes can be replaced by a current source Isc in a parallel with aresistor RTH. Isc is the short circuit current at the terminals, and RTH is theThevenin’s equivalentresistance a a ISC circuit RL RL RTH b Here we will consider ( 3 ) cases: b Circuit containing only independentsources. Circuit containing only dependentsources. Circuit containing both independent and dependentsources.

  35. Case (1): Circuit containing only independentsources: • Procedure of Thevenin’sTheorm: • Find the open circuit voltage at the terminals ,Voc. • Find the Thevenin’s equivalent resistance, RTH at the terminals when all independent sources arezero: • Replacing independent voltage sources by shortcircuit • Replacing independent current sources by opencircuit • Reconnect the load to the Thevenin equivalentcircuit • Procedure of Norton’sTheorm: • Find the short circuit current attheterminals, Isc. • Find Thevenin’s equivalent resistance, RTH (asbefore). • Reconnect the load to Norton’s equivalentcircuit.

  36. RTH VOC ISC +- RL RL RTH Example: Use Thevenin’s and Norton Theorms to findV0 +- - + + Vo - 6V 12V 2kΩ 2kΩ 4kΩ Using TheveninTheorm:

  37. First findVOC: +- - + 6V 12V + Voc -  V i (4k)4V + 4kΩ - i1 2kΩ 6V i  1 mA 1 4kΩ 1 2k4k Voc12V4V8V

  38. Second, findRTH RTH 2kΩ 4kΩ RTH = 2k//4k = 4/3 k Thevenin equivalent circuitis RTH 2 kΩ V  V 0 oc 2kR TH + Vo - VOC 2k 8V +-  2 kΩ 10 k 3 V0 4.8V

  39. +- Using NortonTheorm First findIsc - + 6V 12V ISC 2kΩ 4kΩ X +- i1 4kΩ i 6V + V2k - - +- 12V + 2kΩ i2 12V

  40. i 12V3m A 1 4k KVL around outerloop: 126V2k0 V2k  6V V2k 63mA i 2 2k 2k KCL at x: i1i2i0 3m3mi0 i6m A Isc 6 mA

  41. RTH is the same asbefore: Io RTH + Vo - ISC 2k 4 k 3 R (6m)2.4mA ) I TH (I 0 sc 4 R  2k k2k 3 TH V0I0(2k)(2.4m)(2k)4.8V

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