FUNDAMENTALS OF ENGINEERING (FE) …

1 1FE: Electric Circuits Gross EE1-1FE: Electric Circuits Gross EE1-2 FUNDAMENTALS OF ENGINEERING (FE) examination ELECTRICALENGINEERINGC harles A. Gross, Professor EmeritusElectrical and Comp ENGINEERING Auburn University Broun Electric Circuits Gross EE1-3EE Review Problems1. dc Circuits2. Complex Numbers3. ac Circuits4. 3-phase CircuitsWe will discuss TransientsControlSignal ProcessingElectronicsDigital SystemsWe may discuss these as time permitsFE: Electric Circuits Gross EE1-41. dc Circuits: Find all voltages, currents, and : Electric Circuits Gross EE1-5 SolutionThe 8 and 7 resistors are in series:18715R R1 and 10 are in parallel:12111011016101 RRRR FE: Electric Circuits Gross EE1-6 Solution:1001010abaabLVIAR 10100 40 60 VVKVL 4440()aVIVL 4 and R2 are in series:4210abRR 4FE: Electric Circuits Gross EE1-7 Solution:10 64bacKCLIIIA 8832()bVIVL 7728()bVIVL 106061010cVIAL FE: Electric Circuits Gross EE1-8 Absorbed 2244 10400aRIW 221010 6360cRIW 22774112bRIW 22884128bRIW Total Absorbed Power 1000W Power Delivered by Source100 101000saVIW In General:PABS= PDEV(Tellegen'sTheorem)5 Summer 200892.

2 Complex Numbers2 Consider250xx 22( 2)4(1)(5)2162124111212x The numbers "1 21" are called complex numbersSummer 200810 The I (j) operatorMath i 112xi ECE Department Define j 112xj We choose ECE notation! Rectangular complex numberreal part of Imimaginary part of ZX jYXeZZYZZ 611 Polar Form Math a complex number|| modulus of argargument of radiansiZReRZZZZ ECE complex number|| magnitude of angangle of degreesZZZZZZZ Summer 200812 The Argand DiagramIt is useful to plot complex numbers in a 2-D cartesian space, creating the so-called Argand Diagram (Jean Argand (1768-1822)). real axis (X) imaginary axis (Y)+1+212 ZXjYj 7 Summer 200813 ConversionsXY Z Polar 221 RetangularPolar tanZXYYX Summer 200814 Example:3434Zj 3Im4 XeZYZ 8 Summer 200815 ConjugateZXjYZ * conjugate of ZZXjYZ (34)* 34 Summer 200816 Addition (think rectangular)Aa jb ABc jd B ()()()( )ABajbcjdac jbd (34) (512)(3 5)(4 12) 88 ABjjjj 9 Summer 200817 Multiplication (think polar)Aa jb ABc jd B ()()()AB ABAB 00000(5 ) ( )(5) (13) ( ) Summer 200818 Division (think polar)Aa jb ABc jd B ()AAABBB 10 Summer 2008 ELEC381019 Multiplication (rectangular) ()AB a jb c jdac bdj adbc 0(34) (512)(15 48)( 36 20)6316 Summer 2008 ELEC381020 Addition (polar)05 Complex numberaddition is the same as "vector addition"!

3 11FE: Electric Circuits Gross EE1-213. ac Circuits 8 mHi(t)+v(t)-Find "everything" in the given circuit.( ) cos(377 )vtt V FE: Electric Circuits Gross EE1-22 Frequency, period( ) cos(377 )vtt V (radian) frequency377/rad s (cyclic) frequency602fHz 12FE: Electric Circuits Gross EE1-23 The ac CircuitTo solve the problem, we convert the circuit into an "ac circuit":, , elements(impedance), sources, (phasors)RLCZviV I :080 RRZRjj :00( )( )010 LLZjLjj 11:0 ( )CCZjjjC FE: Electric Circuits Gross EE1-24 The PhasorTo convert to a 02oMAXVV ()cos()MAXvt Vt 2 MAXVV ( ) cos(377 )vtt For : Electric Circuits Gross EE1-25 The "ac circuit"RCLVVV 100 0 I():100810 ( ) cos( )oitt FE.

4 Electric Circuits Gross EE1-26 Solving for voltages(8)( ) ( ) cos( )ooRRoRVZIVvtt (4)( ) ( ) cos( )ooCCoCVZI jVvtt ( 10)( ) 100 ( ) cos( )ooLLoLVZI jVvtt 14FE: Electric Circuits Gross EE1-27 Absorbed powers* ( )* 8000ooRRSVIj * ( )* 0400ooCCSVIj *SVI P jQ * 100 ( )* 01000ooLLSVIj 800600800 watts;600 var s;|| 1000 TOTRCLTOTTOTTOTTOTS SSSjPQSSVA FE: Electric Circuits Gross EE1-28 Delivered power* 100 0 ( )* 800600ooSSSVIj 800600800 watts600 var sSTOTSTOTSTOTSSjPPQQ In General: PABS= PDEV QABS= QDEV(Tellegen's Theorem)15FE: Electric Circuits Gross EE1-29 The Power Triangle800600Sj P = 800 WQ = 600 varS = 1000 VA = factorcos( ) lagging100 0oV FE: Electric Circuits Gross EE1-30 Leading, Lagging ConceptsVILeading CaseQ < 0 Lagging CaseIVQ > 016FE: Electric Circuits Gross EE1-31A Lagging pf 0 VkV FE: Electric Circuits Gross EE1-32 CurrentsV IIjIA V17FE.

5 Electric Circuits Gross EE1-33 PowersV RjXRII 0coscos *5000 RRSVIkWj 500kW0*5001200 1300 S SSj 1300kVA* 01200 varLLSVIj k 1200 varkLIFE: Electric Circuits Gross EE1-34 Add CapacitanceV 125 CIj CjX 8131 RLCII I IIjjA 18FE: Electric Circuits Gross EE1-35 Powers 0coscos 500kW0* 31 SRLCSSSVI S S SSjjSkVA 1200 kvar*0 900kvarCCSVIj 900 kvar kVAFE: Electric Circuits Gross EE1-36 ObservationsBy adding capacitance to a lagging pf (inductive) load, we have significantly reduced the source current., without changing P! ; Before81 ; AfterIf we consider the source in the example to represent an Electric Utility, this reduction in current is of major practical importance, since the utility losses are proportional to the square of the that: low pf, high current; high pf, low current;19FE: Electric Circuits Gross EE1-37 ObservationsThat is, by adding capacitance the utility losses have been reduced by almost a factor of 5!

6 Since this results in significant savings to the utility, it has an incentive to induce its customers to operate at high leads to the Power Factor Correction problem, which is a classic in electric power ENGINEERING and is extremely likely to be on the FE will be using the same numerical data as we did in the previous example. Pretty clever, eh what?FE: Electric Circuits Gross EE1-38 The Power Factor Correction problem An Electric Utility supplies kV to a customer whose load is kV 1300 kVA @ pf = lagging. The utility offers the customer a reduced rate if he will correct ( improve or raise ) his pf to Determine the requisite capacitance. Utility Load pf correctingcapacitance 20FE: Electric Circuits Gross EE1-39PF Correction: the solution1. Draw the load power triangle. 1300 kVA @ pf = lagging.

7 01300 Because the pf is lagging, the load is inductive, and Q is we must add negativeQ to reduce the total, which means we must add : Electric Circuits Gross EE1-40PF Correction: the solution cos31pf 2. We need to modify the source complex power so that the pf rises to lagging. Closing the switch (inserting the capacitors) 50012005001200 SCCSjjQj Q 0 Then 0 Let 1200 Therefore 50031 XCSXSQQSjQSkVA 21FE: Electric Circuits Gross EE1-41PF Correction: the solution1200900 kvarCXQQ 300 kvarXQ 500kW900 kvar1200 kvarThe new source power triangle300 kvarInstall 900 kvar of kV Capacitors 4. Three-phase ac Circuits Although essentially all types of EE s use ac circuit analysis to some degree, the overwhelming majority of applications are in the high energy ( power ) happens that if power levels are above about 10 kW, it is more practical and efficient to arrange ac circuits in a polyphase configuration.

8 Although any number of phases are possible, 3-phase is almost exclusively used in high power applications, since it is the simplest case that achieves most of the advantage of polyphase. It is virtually certain that some 3-phase problems will appear on the FE and PE examinations, which is why 3-phase merits our single-phase ac circuit For a given load, the phase a conductor must have a cross-sectional area A , large enough to carry the requisite current. Since the neutral carries the return current, we needa total of 2A worth of conductors. a is the phase conductor GeneratorLineLoad+an n is the neutral conductor Tripling the capacityWe need a total of A + A +A + 3A = 6A conductors.+++If then 3abcnIII II I aIbInIcI23 But what if the currents are not in phase?Now we only need a total of A + A +A + 0 = 3A conductors!000 Suppose 0120120 abcIIIIII 000 Then 0120120 )( (00) 0nabcnnIIIIIIIII jjjIIjI j A 50% savings!

9 FE: Electric Circuits Gross EE1-46 The 3-Phase Situation24FE: Electric Circuits Gross EE1-47"Balanced" voltage means equal in magnitude, 120oseparated in phasemax00max00max( )cos()2cos()( )cos(120 )2cos(120 )( )cos(120 )2cos(120 )anbncnvt VtVtvt VtVtvt VtVt 0000120120anbncnVVVVVV FE: Electric Circuits Gross EE1-48 The Line VoltagesBy KVL abanbnVVV 000130120 1033022abVVVV jjV 003903 150bccaVVVV 3abbccaLVVVVV 25FE: Electric Circuits Gross EE1-49 When a power engineer says the primary distribution voltage is 12 kV he/she An FE: Electric Circuits Gross EE1-50 All balanced three-phase problems can be solved by focusing on a-phase, solving the single-phase (a-n) problem, and using 3-phase symmetry to deal with b-n and c-n values!

10 BcAn Important involves judicious use of the factors 3, 3, and 120 !To 26FE: Electric Circuits Gross EE1-51bcRecall the pf Correction ProblemAn Electric Utility supplies kV to a single-phase customer whose load is kV 1300 kVA @ pf = lagging. 0 VkV 104 018167 RLIIIA 1300kVA500kW1200 kvarFE: Electric Circuits Gross EE1-52 The pf Correction Problem in the 3-phase caseAn Electric Utility supplies to a 3-phasecustomer whose load is 3900 kVA @ pf = lagging. 3900kVA1500kW3600 kvar3 times bigger!018167A 0kV 104 anV27FE: Electric Circuits Gross EE1-If we want all the V s, I s, and S s0001816718118718153abcIAIAIA 500120050012005001200abcSjkVASjkVASjkVA FE: Electric Circuits Gross EE1-54PF Correction: the 3ph solution1500kW2700 kvar3600 kvar900 kvarInstall 2700 kvar of Capacitance.