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Electric Circuits and Machines

by: Melissa Metz

Electric Circuits and Machines EECS 315

Marketplace > Kansas > Elect Engr & Computer Science > EECS 315 > Electric Circuits and Machines
Melissa Metz
GPA 3.74

James Rowland

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James Rowland
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This 12 page Class Notes was uploaded by Melissa Metz on Monday September 7, 2015. The Class Notes belongs to EECS 315 at Kansas taught by James Rowland in Fall. Since its upload, it has received 53 views. For similar materials see /class/186789/eecs-315-kansas in Elect Engr & Computer Science at Kansas.

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Date Created: 09/07/15
EECS 315 Top Concepts in Electric Circuits and Machines DC Circuits An interconnection of resistances with DC voltage sources batteries andor DC current sources having at least one closed path can demonstrate the use of Ohm39s Law and permit the simplest application of the following circuit analysis techniques loop analysis node analysis superposition source nansfomiations equivalent circuits and the combination of series and parallel resistances Kirchho s Laws KVL and KCL Kirchhoth Voltage Law KVL states that the sum of voltage rises is equal to the sum of voltage drops around any closed path Kirchho s Current Law KCL states that the sum of currents entering any node is equal to the sum of the currents leaving that same node Current Divider and Voltage Divider Rules A current entering two parallel resistors divides proportionately to their conductances and a voltage across two resistors divides proportionately to their resistances Dependent Sources Voltage sources or current sources can depend on the voltages or currents elsewhere in the same circuit These are referred to as voltage or current liontrolled voltage or current sources Thevenin39s Equivalent Circuit A 39I39hevenin Equivalent Circuit can be formed with respect to given terminals ab as the series combination ofan independent voltage source having a value of the opencircuit voltage and a resistance A Norton Equivalent Circuit is composed of the parallel combination of an independent current source having a value ofthc short circuit current and the same Thcvenin resistance Capacitance and Inductance The current through a capacitor is equal to the product of the capacitance C and the derivative of its voltage The voltage across an inductor is equal to the product of the inductance L and the derivative of its current Energy Stored The energy stored in an electric eld having a capacitance C is equal to onehalf of the capacitance times the square of the voltage The energy stored in a magnetic eld having an inductance L is equal to onehalf of the inductance times the square of the current Combining Cs and Combining Ls Capacitances combine quotlikequot conductances ie capacitances in parallel sum to form the equivalent capacitance and capacimnces in series surn their inverscs to form the inverse of the equivalent capacitance lnductances combine quotlikequot resistances ie series inductances sum to form the equivalent inductance and parallel inductanccs sum their inverscs to form the inverse of the equivalent inductance AC Circuit Analysis Having sinusoidal voltages and currents as sources AC circuits contain resistances inductances and capacitaan as elements providing impedance dent sources can also be present Using phasors described as the next topic the following AC circuit analysis techniques are applicable loop analysis node analysis superposition source transformations equivalent circuits and the combination of series and parallel resistances Phasors and Impedance Phasors are complex quantities representing voltages and currents that transform AC circuit di erential equations into algebraic equations for ease in performing steadystate analysis Phasors may be expressed as vectors having magnitude and phase polar form or having real and imaginary parts rectangular form Conversion between these two forms is sometimes necessary Average Power and Complex Power in AC Circuits The instantaneous power in ac circuits his an average real and constant part and a time varying sinusoidal component The average power P in Watts is the product of the rootmean square rrns voltage magnitude the rms current magnitude and the power factor de ned as the cosine of the phase di erence between the voltage and the current An inductive reactance results in a 13M power factor and a capacitive reactance in a Izadingpower factor The complex power 5 in VoltAmps is de ned as S P j Q where Q in VAR is the reactive power Ideal Transformers An ideal transfonner can step up or step down the voltage and the current from its primary side to its secondary side depending on the number of turns on each side Re ection om one side to the other is a standard analysis technique that simpli es ideal transformer problems Balanced ThreePhase Systems Balanced three phase systems can be connected in Y Y or YA con gurations The Ysource has three sinusoidal voltages equal in magnitude but with each differing by 120 degrees in phase Balanced systems can be analyzed most easily by converting to the YY con guration if YA reinserting the neutral wire and working on a gr ghaxe basis Magnetic Circuits Ironcore transfertriers are examples of magnetic circuits through which lines of ux flow to form magnetic elds with strengths that depend on the number of trans of wires connecting a sinusoidal source and the permeability ofthc magnetic material It is useful to draw the corresponding electric circuit as part of the analysis procedure DC Machines Usually operating in either separately excited or self excited con gurations DC motors are o en eld controlled or annaturecontrolled models Both models feature an electric equation a mechanical rotational equation and an electromechanical equation linking the two parts Armatm e controlled DC motors have a backelectromotive force back emf that is proportional to the motor speed Torque versus speed curves are useful in determining motor selection AC Motors and Generators AC machines operate as synchronous motors or synchronous generators alternators or as induction motors squirrel cage or wound rotor Synchronous machines have AC in one winding and DC in the other stator or rotor and induction machines have AC in both stator and rotor Induction machines are characterized by slip a which is the ratio of the di ercnce in stator and rotor speeds to the stator speed sometimes expressed as a percentage The circuit model for an induction machine incorporates an ideal transformer and re ected impedances are often used Special Purpose Electric Machines Brushless DC motors stepping motors switched reluctance motors universal motors and the singlephase induction machine are regarded as important special purpose electric machines Motor selection depends on desired performance cost nature of the load and thermal issues EECS 315 Finn Eula Elemc Chill Ind Mlchinn MM 20 101 Nun l Farlhcbc hmil xhnwu below L SoMfmthVammznmmmhgm b Samrumemvmmzmcdmumm m lliwhvimdmholhmnlmdimnslyiddlheummm a Inquot 4 in MA LA 9 Va 39v EECSJISHIIIEXI mum Ahayou I Mmi lmmxvdwmbol mhmhu man In mundivider Mot valrgedividet fannull m Wmnrhr u r as quotone Finnllv 39 Condwcc ml 39 duel lpply you mun rawa EECHJIS Flu Sun mum 1 Fe Illa AC drenil mm btlaw um um l h quot quotI0 I0 n minor in urlzs wilh the 20 MN induclnr w mm39 vuiugp Aha mmm 2 WM M 11 mullthnw2ufwhmm mem uqllcncymd nhe equenry inllmdeMSvn of luau Ind cunvnvs are Lb nuthlm quotInn VInd Il divided by Ln J Jo MALquot JD Iw 39 4 an LJ 3100 IDJL I quotI I f Nu CHI 0quot sum huh EECS 315 Flnll En mum Calida the allawhlg 39 r mum Cambium n two quotmarks to aquot cum quot1quot t on onwmunuwauummmdgmmm unnumth min m artwork wall that nuiqu combin lit minuwu quotHF 30 W Ind 60 uF In in pullzl In uric with this plane carnme mm mhzr umm vllm Lanna cambimn39an MEN conducunccs 0 l0 uF Ind 50 ul Sulcd lmyou hive 0m in Drnw Iht Wk ribmm Ind W a unit mm mm Recall mu upwillnbes mnth like sacs n 5 Hum Bun me 5 n s mus Hull zuu hp J at a 5 3 Consider m Y l or m chum for a balanced micephase symcm shnwn L J M quotquot A bclnw Nola mm m am being speci ed 39 m l V K Delmim M M x I in P ram equation wilh V I and Rg I 5min lrcunmethnniul cqmlmn wilh T R I quot Null 1 39 39 39 39 T t 39 39 w 391 v bl I 4 mumn 3 and 1h mom speed toquot nUsing m pmmetcrs pmvidxd whnl sum eld vnlugc V is nquimd In yield a leDr speed mm of 60 mliuns pcr second bWhm motor speed mi ubmincd whn V in set In 140 volu39l gvguvnn lhb molar I cd Ilso lo revmin a R IDA U 1 lg L g LP 3 6 T N 3 B A 5 1 lthlac T 2 It LLon l l w a 1 in MAf L wzsmnmxum I Muquot EECSJISFIMIEHII muse 4 39 39 39 39 39 Nay1 39 n r h me H A L secondquid Dmnimlhcsomphnsormv npclnrfum 39 A It 9Io a l Mnnnwr um um plnc w p a pairs n sixpal molar has Ihnx pol pairs and so form y x 1 A ductmine the fullload speed b For alaurnnle induction molar wilh a WHO speed of L455 mvlmin wm fnqucncy in Hmz yields a slip s om KEGJHl39lnll u musin S 11 3mm shown column In New Thisnnsfnnuci is n impdawn w duy0nhnri i u 39 V u primquot ry 39 n I i 39 A39 394 I when NM quot 39 humans mun Wmmmmvmmumm a 03914 l39 w 5i 4 IOJL V5 9 IDJL msm MERE mm 6 my pal pun Ind w l39n h Dinslip 5 is Mined la be 1h diffumu in tin synchmnoul speed I I i n i h Hummusmmmrym ummmurlmmul syniJamnom speed m F 39 39 39 mm ii i rpm4 in rpm Venus slip I win in pain rm 246 a m Ind 11 b For I RimPole induction mom with l runmid mad at Lass mmiii mi cquuicy iii chz yiclds iiip s 0037 EECS 315 Final Exam EIKlrit Circuit Ind Mlchlnu Deccnbcr 111012 Clam Bach m1 Closed Nata shun Ill Mm Pmrld mm L For lh DC Cu cllil shown bukiw a Soive for he cumnl In using 111 Jugcmugjlinn circuit analysis techniquc b Solve for m cumin In using lhc my iicimiquu 39niii mc md is ul 0 knan as ht node wing quilysix mtlhodi u Dacrmine lhc power being dissipan in um 40 rumor Giv uniu I UJD In 10V 0 m 0 1quot EECS M5 FluI Enm z 2 0396 1 For 1h AC circuil 50 Hertz shown below a an c w L 1 l b camisu of III parallcl combination oilht ion resistor and he 100 m induclor quot Calculuk Lhc power faclm ohth defined in Fun 17 Is 1h pnwcr fncior n lugging or ii leading power faciur39 Which one Explain I LDAJ J 14 a k 7 local 10 a in mm l AlAD EECS 315 FinI Exam Ek ric Cirmln And Human lambr 11 um Hum and low nd clad Nam sn 1 walk Pmllk um I For m DC circuit Shawn below I Solve for m cnrnznl I using my melhod uf ymlr chnicc u Danmine the pawn kip supplied nrv manna by up yoqu source r In 10 V 9 04A mun as us um um Plaza 1 Fa Ike AC drum W shown below a vein m 40 mister in uric with the m mH indmom loo m u ya Ln EECSJISFlnIznn Pqelolb 3 Con ch Ihc YV of Wngm cilcuil m I balanced Iraqthis syilcm shown below Assum nus value Daminc u mm cumin 1 in phuur 139an b Determine m power faclor or m YXoad LI Y luld Kits 15 run um rm 4 on O quot Note 1 L My Detmnin llquot mum plum W v in point can 3 l lut 41M 3 aV ls EECS 315 Final Exam Page 5 of 5 Consider the following two networks a The rst network contains a total of ve inductors Three inductors of 20 mH 30 mH and 40 mll are in parallel In series with this parallel combination are two other inductors with values ofSO mH and 60 mH Stated again you have three inductors in parallel with each other and another two in series with each other and in series with the parallel combination Draw the nehvork diagram and calculate the Quivalent inductance for the entire rst network I The second network contains a total of ve capacitors Three capacitors of 20 11F 30 pF and 40 pF are in parallel ln series with this parallel combination are two Other capacitors with values of50 pF and 60 uF Stated again you have three capacitors in parallel with each other and another two in series with each other and in series with the parallel combination Draw the network diagram and calculate the equivalent capacitance for the entire second network EECS 3l5 Final Exam Page 6 of 6 6 The equcncy of the voltages induced in the rotor conductors of induction motors depends on the rotational speed of the stator eld relative to the rotor and on the number of poles The stator eld rotates at synchronous speed which is 60 times the Frequency in Hertz divided by the number of pole pairs eg a fourpole induction motor has two pole pairs a six pole motor has three pole pairs and so forth The slip s is de ned to be the di erence in the synchronous speed and rotation speed of the rotor divided by the synchronous speed The Slip 5 varies from I when the rotor is stationary to 0 when the rotor turns at synchronous speed a For a Ppole induction motor operating at 60 Hertz with slip 5 determine the rotational speed of the rotor as a function of P and 5 Show a plot of the rotor speed in rpm versus Slip 5 with data points for P 2 4 6 8 IO and 12 For a fourpole induction motor with a fullload speed of 1455 revmin what frequency in Hertz yields a Slip 5 003 9 EECS 315 Final Exam Electric Circuits and Machines December 16 2010 Name Cloud Books 1nd Closed Notes Show Iquot work Provide units l A A doubleloop magnetic circuit shown below has an electmmotive force N1 due to a coil of 500 turns and a currenton amps Neglect fringing Recall that no 41 x 10397 in proper units All bars have crosssectional areas of 1 cm by 1 cm The air gap on the top left bar has a length of001 cm The air gap on the to right bar has a length of002 cm The air gap on the middle bar has a length of003 cm Problem Determine the magnetic ux density B in the me right bar EECS 315 Final Exun Page 1 of 6 2 For the DC circuit shown below solve for the current lo using t1vo of the following circuit analysis methods You must work on this problem until you have obtained the same solution by bod methods that you choose or else there will be major deductions in points Nodevoltage 0 Meshcurrent Superposition Thevenin s circuit 5 A I M 4JL 10V I 9 EECS 315 Final Exam Page 6 of 6 6 Consider a circuit containing an ideal stepup transformer with a turns ratio of 14 This transformer is a stepup transformer because the voltage in the secondary on Lhe right is four times the voltage in the primary on the le ie the turns ratio om the primary to the secondary is 14 The primary circuit consists of a voltage source having a voltage of 150 j0 V ms in series with an impedance of 10 is n The secondary has a single resistance component of 20 9 Determine the real power P in watts being supplied by the voltage source EECS 315 Final Exam Page 3 of 6 3 Two ac loads are placed in parallel with a voltage of200 jO V n39ns across both ofthem One load absorbs 10 kW of real power P and has a power factor of 08 lagging The other parallel load is a resistor of Q 1 Calculate the total complex power being absorbed by the loads b Determine the power factor ofthe combined load 15 it lagging or leading EECS 315 Final Exam Page 4 of 6 4 Consider a balanced abc sequence threephase Y Y or Eyeme circuit having VquotI 200 j0 V rms line impedances of2 jl and a line current in phase a of I 10 j 10 A n39ns There are corresponding voltages line impedances line currents and load impedances in each phase a Calculate the load impedance in Phase 3 b Determine the total real power being absorbed by all three phases ofthe load EECS 315 Final Exam Page 5 ol39 6 5 Consider the equivalent circuit shown below for an induction motor You are given all parameters except you do not know the value of the Slip 5 If the power factor at the stator input terminals is given provide words and formulas to describe the precise steps algorithm you would use to determine the value of st Be very speci c EECS 315 Final Exam Electric Circuits and Machines December 15 2009 Name Closed Book Ind Closed Notes Show Ill work Provide units Formula Sheets I tau of three 8 by 11 front end heck are ellawed I For the DC circuit shown below solve for the voltage VD using 12 of the following circuit analysis methods You must work ul vnu L l by both methods that you choose or else time will be major deduction in pains O Node voltage o Meshcurrent Superposition I Source tansformation AM A A A 9 IL 511 l M V 44 Va 25A lav EECS 315 Final Exam Pnge 2 of 6 2 Consider a balanced abc sequence threephase Y Y or w circuit having Vquot 120 j0 V rms zero line impedances and Y load impedances of30 j 40 2 for each phase 1 Draw the circuit con guration and label all points as a b c n A B C and N h Determine the linetoline voltage VBC at the m Note that the gyital letters B and C indicate that this voltage is at the lom EECS 315 Fine Exam Page 3 of 6 3 The circuit shown below contains an ideal transformer Note This transformer is a step up transformer because the voltage in the secondary on the right is three times the voltage in the primary on the le ie the turns ratio rom the primary to the secondary is 13 Determine the voltage vt as a cosine mction of time t 511 10quot 3 3011 e Al39ltl no to WM 8 1 Valrs IUD H EECS 315 Final Exam Electric Circuits and Mnehln December 19 2008 Name SOLUTIOALS Closed Book Ind Closed Nuts Show all work Provide units 1 For the DC circuit shown below a Solve for the wing Va using m one circuit analysis method Recall that we discussed the following methods Seriespade Nodevoltage Meshcum Superposition Source transformation Thevenin s cixcuit b Determine the power being dissipated in the a resistor Give units 1quot a I gt an 74 1 In M 1 I om 4J1quot in 00V 8 5 Va folu 40 39 5 1 52 Rm 1 7 9 m eke 343103 11 90A 311 sin m 51 quot 154 1 rz2 quot L A aa v 4mm 1 L 5 I iu quotui v gig PM Ifcwanm


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