EET1122/ET162 Circuit Analysis. Introduction. Acknowledgement OUTLINES. Electrical and Telecommunications Engineering Technology Department

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1 /6 ircuit nalyi ntroduction lectrical and elecommunication ngineering echnology Department cknowledgement want to expre my gratitude to Prentice Hall giving me the permiion to ue intructor material for developing thi module. would like to thank the Department of lectrical and elecommunication ngineering echnology of NY for giving me upport to commence and complete thi module. hope thi module i helpful to enhance our tudent academic performance. Sunghoon Jang Profeor Jang Prepared by textbook baed on ntroduction to ircuit nalyi by obert Boyletad, Prentice Hall, 0 th edition. OUNS ntroduction to lectrical ngineering Brief Hitory Unit of Meaurement Sytem of Unit Operation of a Scientific alculator Significant Figure Key Word: lectrical ngineering, Unit, Power, alculator 6 ircuit nalyi ntroduction Boyletad ntroduction he lectrical/lectronic ngineering he growing enitivity to the technologie on Wall Street i clear evidence that the electrical/electronic indutry i one that will have a weeping impact on future development in a wide range of area that affect our life tyle, general health, and capabilitie. Semiconductor Device nalog & Digital Signal Proceing elecommunication Biomedical ngineering Fiber Optic & Opto-lectronic ntegrated ircuit () Figure. omputer chip on finger. (ourtey of ntel orp.) 6 ircuit nalyi ntroduction Boyletad

2 ntroduction Brief Hitory Unit of Meaurement he numerical value ubtituted into an equation mut have he unit of meaurement pecified by the equation xample FGU. ime chart: (a) long-range; - (b) expanded. 6 ircuit nalyi ntroduction Boyletad 4 mi 580 ft 4000 ft mi min h d mi v mi/ h t h 6 ircuit nalyi ntroduction Boyletad 5 Sytem of Unit Sytem of Unit he nglih ytem i baed on a ingle tandard, the metric i ubdivided into two interrelated tandard: the MKS and the GS. nglih ength: Yard (yd) Ma: Slug Force: Pound emperature: Fahrenheit ( F) nergy: Foot-pound (ft-lb) ime: Second () Metric ength: Meter (m) Ma: Kilogram (kg) Force: Newton (N) emperature: Kelvin (K) nergy: Joule (J) ime: Second () 6 ircuit nalyi ntroduction Boyletad 6 FGU. omparion of

3 Significant Figure, ccuracy, and ounding off n the addition or ubtraction of approximate numb er, the entry with the lowet level of accuracy determine the format of the olution. For the multiplication and diviion of approximate number, the reult ha the ame num ber of ignificant figure a the number with the latet number of ignificant figure. Power of en 0 0 / / / /0, x. - Perform the indicated operation with the following x. - approximate number and round off to the appropriate level of accuracy. a a. + 0 b c d b. 0 e. 40/ f / ircuit nalyi ntroduction Boyletad 8 6 ircuit nalyi ntroduction Boyletad 9 x. - x. -4 x. -5 Baic rithmetic Operation a. (000)(0,000) (0 )( 0 4 ) 0 (+4) 0 7 b. (0.0000)(00) (0-5 )(0 ) 0 (-5+) 0 - When adding or ubtracting number in a power-of-ten format, be ure th at the power of ten i the ame for each number. hen eparate the multiplier, perform the required operation, and apply the ame power of ten to the reult. 00, ( 5 ) a x ) ( ( 4 ( + 4 ) a ,000 b. ) ( ) 0 (6.)(000) + (75)(000) 8. 0 a (96)(0.0000) (8.6)(0.0000) 4 4 ()(4) (96 8.6) 0-5 a. (00) (0 ) 0 0 b. (000) - (0 ) - 0 ()(-) 0-6 c. (0.0) - (0 - ) - 0 (-)(-) ircuit nalyi ntroduction Boyletad 0 6 ircuit nalyi ntroduction Boyletad

4 x. -7 a. (0.000)( ) [()(0.000)] [(7)( )] ( 0-4 )(7 0-6 ) ()(7) (0-4 )(0-6 ) a. (40,000)(0.0006) ( )(6 0-5 ) (.4)(6) (0 5 )(0-5 ) alculator and Order of Operation x. -8 a. b , x. -9 a. (0.0000) ( 0-5 ) ( 0-5 ) () 0-5 b. (90,800,000) ( ) (9.08) (0 7 ) ircuit nalyi ntroduction Boyletad 6 ircuit nalyi ntroduction Boyletad HW -6 Perform the following converion: a..5 min to econd b h to econd c to microecond d. 0.6 m to millimeter e to nanoecond f.,60,000 to day Homework :, 4, 4, 6, 4, 4, 4 6 ircuit nalyi ntroduction Boyletad 4 4

5 /6 ircuit nalyi urrent and oltage lectrical and elecommunication ngineering echnology Department Profeor Jang Prepared by textbook baed on ntroduction to ircuit nalyi by obert Boyletad, Prentice Hall, 0 th edition. OUNS eitance and onductance Ohmmeter urrent and oltage mmeter and oltmeter Key Word: eitance, Ohmmeter, urrent, oltage, mmeter, oltmeter 6 ircuit nalyi urrent and oltage Boyletad ntroduction to eitance he flow of charge through any material encounter an oppoing force imilar in many apect to mechanical friction. hi oppoition, due to the colliion between electron and other atom in the material, which convert electrical energy into another form of energy uch a heat, i called the reitance of the material. h e unit of meaurement of reitance i the ohm (Ω). Figure. eitance ymbol and notation. t a fixed temperature of 0 (room temperature), the reitance i related to the other three factor by l ρ (ohm, Ω) ρ : reitivity of the ample (M-ohm/ft at 0 ) l : the length of the ample (feet) : cro-ectional area of the ample (circular mil (M)) FGU. Factor affecting the reitance of a conductor. 6 ircuit nalyi urrent and oltage Boyletad 6 ircuit nalyi urrent and oltage Boyletad 4

6 eitance: ircular Wire For two wire of the ame phyical ize at the ame temperature, the higher the reitivity (ρ), the more the reitance the longer the length of a conductor, the more the reitance the maller the area of a conductor, the more the reitance the higher the temperature of a conductor, the more the reitance FGU. ae in which >. For each cae, all remaining parameter that control the reitance level are the ame. 6 ircuit nalyi urrent and oltage Boyletad 5 ype of eitor Fixed eitor eitor are made in many form, but all belong in either of two group: fixed or variable. he mot common of the low-wattage, fixed-type reitor i the molded carbon compoition reitor. FGU. Fixed compoition reitor. he relative ize of all fixed and variable reitor change with the power rating, increaing in ize for increaed power rating in order to withtand the higher current and diipation loe. FGU.4 Fixed compoition reitor of different wattage rating. 6 ircuit nalyi urrent and oltage Boyletad 6 ype of eitor ariable eitor ariable reitor have reitance that can be varied by turning a dial, knob, crew, or whatever eem appropriate for the application. olor oding and Standard eitor alue whole variety of reitor are large enough to have their reitance in ohm printed on the caing. However, ome are too mall to have number printed on them, o a ytem of color coding i ued. FGU.6 olor coding of fixed molded compoition reitor. FGU.5 Potentiometer: (a) ymbol: (b) & (c) rheotat connection; (d) rheotat ymbol. 6 ircuit nalyi urrent and oltage Boyletad 7 he firt and econd band repreent the firt and econd digit, repectively. he third band determine the power-often multiplier for the firt two digit. he fourth band i the manufacture tolerance. he fifth band i a reliability factor, which give the percentage of 9 White 0 9 able eitor color 6 ircuit nalyi oltage and urrent Boyletad coding. 8 failure per 000 hour of ue. Band - 0 Black Brown ed Orange 4 Yellow 5 Green 6 Blue 7 iolet 8 Gray Band Band 4 5% Gold 0% Silver 0% No band Band 5 % Brown 0.% ed 0.0% Orange 0.0% Yellow

7 x. - Find the range in which a reitor having the following color band mut exit to atify the manufacturer tolerance: onductance he quantity of how well the material will conduct electricity i called conductance (S). b. t Band Orange nd Band White rd Band Gold 4 th Band Silver 5 th Band No color G (iemen, S) ±0% a. 8Ω ± 5% (% reliability) Since 5% of 8 4.0, the reitor hould be within a the range of 8Ω ±. 4.0Ω, or between and 86.0Ω. b..9ω ± 0%.9Ω ± 0.9Ω he reitor hould be omewhere between.5 and 4.9Ω. 6 ircuit nalyi urrent and oltage Boyletad 9 G ρ l (S) ndicating that increaing the area or decreaing either the length or the reitivity will increae the onductance. 6 ircuit nalyi urrent and oltage Boyletad 0 x. - What i the relative increae or decreae in conductivity of a conductor if the area i reduced by 0% and the length i increaed by 40%? he reitivity i fixed. G i ρ l i i (iemen, S) with the ubcript i for the initial value. Uing the ubcript n for new value : n i i G n 0. 5 ρ n l n ρi (.4 l i ).4 ρi l i.4gi 0.70 G i Ohmmeter he ohmmeter i an intrument ued to perform the following tak and everal other ueful function.. Meaure the reitance of individual or combined element. Direct open-circuit (high-reitance) and hort-circuit (lowreitance) ituation. heck continuity of network connection and identify wire of a multi-lead cable 4. et ome emiconductor device 6 ircuit nalyi urrent and oltage Boyletad FGU.7 Meauring the reitance of a ingle element. FGU.8 hecking the continuity of a connection.

8 x - n Figure, three conductor of different material are preented. a. Without working out the numerical olution, determine which ection would appear to have the mot reitance. xplain. b. Find the reitance of each ection and compare with the reult of (a) ( 0 ) a. ilver > copper > aluminum l ( 9.9)( ft ) Silver : ρ 9.9 Ω M l ( 0.7)( 0 ft ) opper : ρ.07 Ω 00 M l ( 7)( 50 ft ) lu min um : ρ 0.4 Ω 500 M 6 ircuit nalyi urrent and oltage Boyletad oltage he voltage acro an element i the work (energy) required to move a unit poitive charge from the terminal to the + terminal. he unit of voltage i the volt,. potential difference of volt () exit between two point if joul (J) of energy i exchanged in moving coulomb () of charge between the two point. n general, the potential difference between two point i determined by: W Q voltage () Q coulomb () W potential energy (J) 6 ircuit nalyi urrent and oltage FGU.9 Defining the unit of meaurement for voltage. Boyletad 4 x. -4 Find the potential difference between two point in an electrical ytem if 60 J of energy are expended by a charge of 0 between thee two point. W W 60 J Q 0 x. -5 Determine the energy expended moving a charge of 50 μ through a potential difference of 6. Q ( 50 0 )( 6 ) 6 J 00 µ J 6 ircuit nalyi urrent and oltage Boyletad 5 Fixed (dc) Supplie he terminology dc i an abbreviation for direct current, which encompae the variou electrical ytem in which there i a unidirectional ( one direction ) flow of charge. D oltage Source Dc voltage ource can be divided into three broad categorie: () Batterie (chemical action), () generator (electro-mechanical), and () power upplie (rectification). FGU.0 Symbol for a dc voltage ource. 6 ircuit nalyi urrent and oltage FGU. erminal characteritic: (a) ideal voltage ource; (b) ideal current ource. Boyletad 6 4

9 urrent he electrical effect caued by charge in motion depend on the rate of charge flow. he rate of charge flow i known a the electrical current. With no external force applied, the net flow of charge in a conductor in any direction i zero. f electron ( coulomb) pa through the imaginary plane in Fig..9 in econd, the flow of charge, or current, i aid to be ampere (). 9 h arg e / electron Q e FGU. Baic electrical circuit. he current in ampere can now be calculated uing the following equation: Q ampere () Q coulomb () and Q t (coulomb, ) t t econd () Q t (econd, ) 6 ircuit nalyi urrent and oltage Boyletad 7 6 ircuit nalyi urrent and oltage Boyletad 8 x. -6 he charge flowing through the imaginary urface of Fig. - i 0.6 every 64 m. Determine the current in ampere. Q t mmeter and oltmeter t i important to be able to meaure the current and voltage level of an operating electrical ytem to check it operation, iolate malfunction, and invetigate effect. mmeter are ued to meaure current level while voltmeter are ued to meaure the potential difference between two point. x. -7 Determine the time required for electron to pa through the imaginary urface of Fig.. if the current i 5 m. 6 Q 4 0 electron electron m Q t ircuit nalyi urrent and oltage Boyletad 9 FGU. oltmeter and ammeter connection for an up-cale (+) reading. 6 ircuit nalyi oltage and urrent Boyletad 0 5

10 /6 ircuit nalyi Ohm aw lectrical and elecommunication ngineering echnology Department cknowledgement want to expre my gratitude to Prentice Hall giving me the permiion to ue intructor material for developing thi module. would like to thank the Department of lectrical and elecommunication ngineering echnology of NY for giving me upport to commence and complete thi module. hope thi module i helpful to enhance our tudent academic performance. Sunghoon Jang Profeor Jang Prepared by textbook baed on ntroduction to ircuit nalyi by obert Boyletad, Prentice Hall, 0 th edition. OUNS ntroduction to Ohm aw Plotting Ohm aw Power ntroduction to Ohm aw Ohm law clearly reveal that a fixed reitance, the greater the voltage acro a reitor, the more the current, the more the reitance for the ame voltage, the le the current. ( ampere, ) ( ohm, Ω) ( volt, ) Key Word: Ohm aw, urrent, oltage, Power Figure 4. Baic ircuit. 6 ircuit nalyi Ohm aw Boyletad 6 ircuit nalyi Ohm aw Boyletad

11 x. 4- Determine the current reulting from the application of a 9- battery acro a network with a reitance of. Ω Ω For an iolated reitive element, the polarity of the voltage drop i a hown in Fig. 4.(a) for the indicated current direction. reveral in current will revere the polarity, a hown in Fig. 4.(b). n general, the flow of charge i from a high (+) to a low ( ) potential. x. 4- alculate the reitance of a 60-W bulb if a current of 500 m reult from an applied voltage of Ω FGU 4. Defining polaritie. 6 ircuit nalyi Ohm aw Boyletad 4 6 ircuit nalyi Ohm aw Boyletad 5 x. 4- alculate the current through the -kω reitor of Fig. 4. if the voltage drop acro it i Ω 8 m x. 4-4 alculate the voltage that mut be applied acro the oldering iron of Fig. 4.5 to etablih a current of.5 through the iron if it internal reitance i 80 Ω. (. )( Ω ) FGU 4.4 xample 4.4 FGU 4. xample 4. 6 ircuit nalyi Ohm aw Boyletad 6 Plotting Ohm aw Graph, characteritic, plot play an important role in every technical field a a mode through which the broad picture of the behavior or repone of a ytem can be conveniently diplayed. t i therefore critical to develop the kill neceary both to read data and to plot them in uch a manner that they can be interpreted eaily. For mot et of characteritic of electronic device, the current i repreented by the vertical axi, and the voltage by the horizontal axi, a hown in Fig FGU 4.5 Plotting Ohm law 6 ircuit nalyi Ohm aw Boyletad 7

12 f the reitance of a plot i unknown, it can be determined at any point on the plot ince a traight line indicate a fixed reitance. t any point on the plot, find the reulting current and voltage, and imply ubtitute into following equation: dc he equation tate that by chooing a particular, one can obtain the correponding from the graph, a hown in Fig. 4.6 and 4.7, and then determine the reitance. FGU 4.6 Demontrating on an - plot that the le the reitance, the teeper i the lope FGU 4.7 x. 4-5 Determine the reitance aociated with the curve of Fig. 4.8 uing equation from previou lide, and compare reult. t 6, m, and 6 dc kω m t the int erval between 6 and 8, k m Ω FGU 4.8 xample ircuit nalyi Ohm aw Boyletad 9 Power Power i an indication of how much work can be done in a pecified amount of time, that i, a rate of doing work. Since converted energy i meaured in joule (J) and time in econd (), power i meaured in joule/econd (J/). he electrical unit of meaurement for power i the watt (W), defined by watt (W) joule/econd (J/) W P ( watt, W, or joule / ec ond, J / ) t horepower 746 watt W Q P t t Q ( watt ) where t Q t q. 6 ircuit nalyi Ohm aw Boyletad 0 q. P ( watt ) q. P ( ) (watt) he reult i that the power aborbed by the reitor of Fig. 4.9 can be found directly depend on the information available. Power can be delivered or aborbed a defined by the polarity of the voltage and direction of the current. For all dc voltage ource, power i being delivered by the ource if the current ha the direction appearing in Fig. 4.0 (a). FGU 4.9 Defining the power to a reitive element. 6 ircuit nalyi Ohm aw FGU 4.0 Battery power: (a) upplied; (b) aborbed. Boyletad

13 x. 4-6 Find the power delivered to the dc motor of Fig. 4.. ( 0 )( 5 ) P 600 W 0. 6kW FGU 4. xample 4.6. x. 4-7 What i the power diipated by a 5-Ω reitor if the current i 4? x. 4-8 he - characteritic of a light bulb are powered in Fig. 4.. Note the nonlinearity of the curve, indicating a wide range in reitance of the bulb with applied voltage a defined by the earlier dicuion. f the rated voltage i 0, find the wattage rating of the bulb. lo calculate the reitance of the bulb under rated condition. t 0, P ( 0 )( ) 75 W 0 t 0, 9Ω P (4 ) (5 Ω) 80 W 6 ircuit nalyi Ohm aw Boyletad FGU 4. he nonlinear - characteritic of a 75-W light bulb. 6 ircuit nalyi Ohm aw Boyletad Sometime the power i given and the current or voltage mut be determined. P P or P ( ampere) P P or P ( volt ) HW 4-5 tereo ytem draw.4 at 0. he audio output power i 50 W. a. How much power i lot in the form of heat in the ytem? b. What i the efficiency of the ytem? a. P i (0 )(.4 ) 88 W P i P o + P lot, P lot P i P 0 88 W 50 W 8 W P 0 50W b. η% 00% 00% 7.6% P i 88W x. 4-9 Determine the current through a 5-kΩ reitor when the power diipated by the element i 0 mw. P 0 0 W m 5 0 Ω Homework 4:, 4, 6, 8, 0, 4, 5, 6, 49, 5 6 ircuit nalyi Ohm aw Boyletad 4 6 ircuit nalyi Ohm aw Boyletad 5 4

14 /6 ircuit nalyi Serie ircuit lectrical and elecommunication ngineering echnology Department cknowledgement want to expre my gratitude to Prentice Hall giving me the permiion to ue intructor material for developing thi module. would like to thank the Department of lectrical and elecommunication ngineering echnology of NY for giving me upport to commence and complete thi module. hope thi module i helpful to enhance our tudent academic performance. Sunghoon Jang Profeor Jang Prepared by textbook baed on ntroduction to ircuit nalyi by obert Boyletad, Prentice Hall, 0 th edition. OUNS ntroduction to Serie ircuit Kirchhoff oltage aw oltage Divider ule nterchanging Serie lement Serie ircuit - ntroduction wo type of current are available to the conumer today. One i direct current (dc), in which ideally the flow of charge (current) doe not change in magnitude with time. he other i inuoidal alternating current (ac), in which the flow of charge i continually changing in magnitude with time. (volt) (volt) Serie ircuit Notation deal dc oltage Source v. Non-ideal Source oltage egulation Key Word: Serie ircuit, Kirchhoff oltage aw, oltage Divider ule 6 ircuit nalyi Serie ircuit Boyletad FGU 5. ntroducing the baic component of an electric circuit. 6 ircuit nalyi Serie ircuit Boyletad

15 Serie ircuit circuit conit of any number of element joined at terminal point, providing at leat one cloed path through which charge can flow. wo element are in erie if.hey have only one terminal in common.he common point between the two point i not connected to another current-carrying element. n Fig. 5.(a), the reitor and are in erie becaue they have only point b in common. he current i the ame through erie element. he total reitance of a erie circuit i the um of the reitance level FGU 5. (a) Serie circuit; (b) ituation 6 ircuit nalyi Ohm aw and in Serie which urrent and are not Floyd in erie. 4 he total reitance of a erie circuit i the um of the reitance level. n general, to find the total reitance of N reitor in erie, the following equation i applied: N (ohm, Ω) (ampere, ) P,,, N N (volt, ) P del P del P + P + P + + P N (watt, W) (watt, W) FGU 5. eplacing the erie reitor and of Fig. 5. (a) with the total reitance. he total power delivered to a reitive circuit i equal to the total power diipated by reitive element. 6 ircuit nalyi Serie ircuit Boyletad 5 x. 5- a. Find the total reitance for the erie circuit in Figure 5.4. b. alculate the ource current. c. alculate the voltage,, and. d. alculate the power diipated by,, and. e. Determine the power delivered by the ource, and compare it to the um of the power level of part (b). (a) + + Ω + Ω + 5Ω 8Ω Ω (c) (.5)(Ω) 5 (.5)(Ω).5 (.5)(5Ω).5 (d) P (5)(.5).5W P (.5)(.5) 6.5 W P (.5)(.5).5 W (e) P del (0)(.5) 50W P del P + P + P 50W.5W + 6.5W +.5W FGU ircuit nalyi Ohm aw and Serie urrent Boyletad 6 x. 5- Determine,, and for the circuit of Figure Ω + 4Ω + 7Ω + 7Ω 5Ω 50 5 Ω ()(4Ω) 8 Figure 5.5 x. 5- Given and, calculate and for the circuit of Figure k Ω + 4 k Ω + 6 k Ω k Ω 0 k Ω k Ω ( )( Ω ) 6 ircuit nalyi Serie ircuit Boyletad Figure 5.6

16 oltage Source in Serie oltage ource can be connected in erie, a hown in Fig. 5.7, to increae or decreae the total voltage applied to a ytem. he net voltage i determined imply by umming the ource with the ame polarity and ubtracting the total of the ource with the oppoite polarity FGU 5.7 educing erie dc voltage ource to a ingle ource. 6 ircuit nalyi Serie ircuit Boyletad 8 Kirchhoff oltage aw Kirchhoff voltage law (K) tate that the algebraic um of the potential rie and drop around a cloed loop (or path) i zero. cloed loop i any continuou path that leave a point in one direction and return to that ame point from another direction without leaving the circuit. 0 (Kirchhoff voltage law in ymbolic form) 0 or + rie drop FGU 5.8 pplying Kirchhoff voltage law to a erie dc circuit. 6 ircuit nalyi Serie ircuit Boyletad 9 x. 5-4 For the circuit of Figure 5.9: a. Determine uing Kirchhoff voltage law. b. Determine. c. Find and. a. Kirchhoff voltage law (clockwie direction): or + + and b. 7 Ω c. 8 6Ω 5 5 Ω FGU ircuit nalyi Serie ircuit Boyletad 0 x. 5-5 Find and for the network of Fig For path, tarting at point a in a clockwie direction: and 40 For path, tarting at point a in a clockwie direction: and 0 FGU ircuit nalyi Serie ircuit Boyletad

17 x. 5-6 Uing Kirchhoff voltage law, determine the unknown voltage for the network of Fig. 5.. x. 5-8 For the circuit of Fig. 5.. a. Determine uing Kirchhoff voltage law. b. Determine. c. Find and and 50 x and FGU 5. x 8 a. Kirchhoff ' voltage law ( clockwie direction ): or x x 6 ircuit nalyi Serie ircuit Boyletad 6 ircuit nalyi Serie ircuit Boyletad b. 7 Ω c Ω 5 Ω FGU 5. oltage Divider ule (D) he voltage acro the reitive element will divide a the magnitude of the reitance level. he voltage divider rule (D) can be derived by analyzing the network of Fig he voltage acro the reitive element of Fig. 5. are provided. Since the reitance level of i 6 time that of, the voltage acro i 6 time that of and /. he fact that the reitance level of i time that of reult in three time the voltage acro. Finally, ince i twice, the voltage acro i twice pplying Ohm law: that of. f the reitance level of all reitor of Fig. 5. are increaed by the ame amount, a hown in Fig. 5.4, the voltage level will all remain the ame. FGU 5. evealing how the voltage will divide acro erie reitive element. FGU 5.4 he ratio of the reitive value determine the voltage diviion of a erie dc circuit. 6 ircuit nalyi Serie ircuit x x FGU 5.5 Developing the voltage divider rule. (voltage divider rule) Boyletad 5 4

18 x. 5-9 Uing the voltage divider rule, determine the voltage and for the erie circuit of Figure ( k Ω )( 45 ) k Ω + 5kΩ + 8kΩ ( 0 Ω)( 45 ) ( 5 0 Ω) ( 8 k Ω )( 45 ) ( 8 0 Ω)( 45 ) 5 0 Ω k Ω FGU ircuit nalyi Serie ircuit Boyletad 6 Notation-oltage Source and Ground Notation will play an increaingly important role on the analyi to follow. Due to it importance we begin to examine the notation ued throughout the indutry. xcept for a few pecial cae, electrical and electronic ytem are grounded for reference and afety purpoe. he ymbol for the ground connection appear in Fig. 5.5 with it defined potential level-zero volt. FGU 5.5 Ground potential. 6 ircuit nalyi Serie ircuit-notation FGU 5.6 hree Boyletad way to ketch the ame erie dc circuit. On large chematic where pace i at a premium and clarity i important, voltage ource may be indicated a hown in Fig. 5.7(a) and 5.8(a) rather than a illutrated in Fig. 5.7(b) and 5.8(b). FGU 5.7 eplacing the pecial notation for dc voltage ource with the tandard ymbol. n addition, potential level may be indicated in Fig. 5.9, to permit a rapid check of the potential level at variou point in a network with repect to ground to enure that the Sytem i operating properly. FGU 5.8 eplacing the notation for a negative dc upply with the tandard notation. FGU 5.9 he expected voltage level at a particular point in a network of the ytem i functioning properly. 6 ircuit nalyi Serie ircuit-notation Boyletad 8 Double-Subcript Notation he fact that voltage i an acro variable and exit between two point ha reulted in a double-cript notation that defined the firt ubcript a the higher potential. n Fig. 5.0(a), the two point that define the voltage acro the reitor are denoted by a and b. Since a i the firt ubcript for ab, point a mut have higher potential than point b if ab i to have a poitive value. f point b i at a higher potential than point a, ab will have a negative value, a indicated in Fig. 5.0(b). he voltage ab i the voltage at point a with repect to point b. 6 ircuit nalyi S FGU 5.0 Defining the ign for double-ubcript notation. 9 erie ircuit-notation Floyd 5

19 Single-Subcript Notation ingle-ubcript notation can be employed that provide the voltage at a point with repect to ground. n Fig. 5., a i the voltage from point a to ground. n thi cae it i obviouly 0 ince it i right acro the ource voltage. he voltage b i the voltage from point b to ground. Becaue it i directly acro the 4-Ω reitor, b 4. ab a b FGU 5. Defining the ue of ingleubcript notation for voltage level. he ingle-ubcript notation a pecifie the voltage at point a with repect to ground (zero volt). f the voltage i le than zero volt, a negative ign mut be aociated with the magnitude of a. 6 ircuit nalyi Serie ircuit-notation Boyletad 0 General omment particularly ueful relationhip can now be etablihed that will have extenive application in the analyi of electronic circuit. For the above notational tandard, the following relationhip exit: ab a b x. 5-4 Find the voltage ab for the condition of Fig. 5.. ab a b FGU 5. xample ircuit nalyi Serie ircuit-notation Boyletad x. 5-5 Find the voltage a for the configuration of Fig. 5.. ab a b a ab + b x. 5-6 Find the voltage ab for the configuration of Fig ab a b 0 ( 5 ) FGU 5.5 he impact of poitive and negative voltage on the total voltage Floydrop. FGU 5. FGU x. 5-7 Find the voltage b, c and ac for the network of Fig Starting at Ground, we proceed through a rie of 0 to reach point a and then pa through a drop in potential of 4 to point b. he reult i that the meter will read b f we then proceed to point c, there i an additional drop of 0, reult in c b he voltage ac can be obtained ac a c 0 ( 4) 4 FGU ircuit nalyi Serie ircuit-notation Boyletad 6

20 x. 5-8 Determine ab, cb and c for the network of Fig here are two way to approach thi problem. he firt i that there i a 54- drop acro the erie reitor and Ω ab (. )( 5 Ω ) 0 cb (. )( 0 Ω ) 4 9 c he other approach i to redraw the network a hown in Fig. 5.7 to clearly etablih the aiding effect of and and then olve the reulting erie circuit Ω 45 Ω and ab cb c FGU 5. FGU 5.8 edrawing the circuit of Fig. 6 ircuit nalyi Ser ie 5.7 ircuit uing dc voltage upply ymbol. Floyd 4 x. 5-9 Uing the voltage divider rule, determine the voltage and for of Fig edrawing the network with tandard battery ymbol will reult in the network of Fig pplying the voltage divider rule, + + ( 4 Ω )( 4 ) 6 4 Ω + Ω ( Ω )( 4 ) 8 4 Ω + Ω FGU 5.40 FGU ircuit nalyi Serie ircuit-notation Boyletad 5 x. 5-0 For the network of Fig. 5.40: a. alculate ab. b. Determine b. c. alculate c. FGU 5.40 c. c ground potential 0 a. oltage divider rule : ab ( Ω )( 0 ) + Ω + Ω + 5 Ω b. oltage divider rule : ( + ) b + ( Ω + 5 Ω )( 0 ) 8 0 Ω or b a ab ab 0 8 deal oltage Source v. Non-ideal oltage Source very ource of voltage, whether a generator, battery, or laboratory upply a hown in Fig. 5.4(a), will have ome internal reitance (know a the nonideal voltage ource). he equivalent circuit of any ource of voltage will therefore appear a hown in Fig. 5.4(b). FGU 5.4 (a) Source of dc voltage; (b) equivalent circuit. 6 ircuit nalyi Serie ircuit-notation Boyletad 6 6 ircuit nalyi Serie ircuit-notation Boyletad 7 7

21 n all the circuit analye to thi point, the ideal voltage ource (no internal reitance) wa ued hown in Fig. 5.4(a). he ideal voltage ource ha no internal reitance and an output voltage of volt with no load or full load. n the practical cae [Fig. 5.4(b)], where we conider the effect of the internal reitance, the output voltage will be volt only when no-load ( 0) condition exit. When a load i connected [Fig. 5.4(c)], the output voltage of the voltage ource will decreae due to the voltage drop acro the internal reitance. FGU 5.4 oltage ource: (a) ideal, int 0 Ω; (b) Determining N; (c) determining int. 6 ircuit nalyi Serie ircuit-notation Boyletad 8 oltage egulation For any upply, ideal condition dictate that for the range of load demand ( ), the terminal voltage remain fixed in magnitude. By definition, the voltage regulation () of a upply between the limit of full-load and no-load condition (Fig. 5.4) i given by the following: oltage regulation ( )% N F F int % 00% 00% For ideal condition, % N and % 0. herefore, the maller the voltage regulation, the le the variation in terminal voltage with change in load. t can be hown with a hort derivation that the voltage regulation i alo given by FGU 5.4 Defining voltage regulation. 6 ircuit nalyi Serie ircuit-notation Boyletad 9 x. 5- alculate the voltage regulation of a upply having the characteritic of Fig FGU 5.44 N F % 00 % F % % 0 % 00 x. 5- Determine the voltage regulation of the upply of Fig int % 00 % Ω 00 % 500 Ω 9. % int 9.48 Ω FGU ircuit nalyi Serie ircuit-notation Boyletad 0 HW 5-4 Determine the value of the unknown reitor in Fig uing the provided voltage level. Figure 5.08 Problem 4. Homework 5:,, 4, 5,7, 0,, 5, 6,,, 4, 6, 0, 4, 4 6 ircuit nalyi Serie ircuit-notation Boyletad 8

22 /6 ircuit nalyi Parallel ircuit lectrical and elecommunication ngineering echnology Department Profeor Jang Prepared by textbook baed on ntroduction to ircuit nalyi by obert Boyletad, Prentice Hall, 0 th edition. OUNS ntroduction to Parallel circuit analyi Parallel lement otal onductance and eitance Parallel circuit analyi and meaurement Kirchhoff urrent aw urrent Divider ule oltage Source in Parallel Key Word: Parallel ircuit, Kirchhoff urrent aw, urrent Divider ule, oltage Source 6 ircuit nalyi Parallel ircuit Boyletad Parallel ircuit ntroduction & lement circuit configuration in which the element have two point in common wo element, branche, or network are in parallel if they have two point in common. Parallel ircuit Parallel lement n Fig. 6., all the element are in parallel becaue they atify the previou criterion. hree configuration provided to demontrate how the parallel network can be drawn. n Fig. 6., for example, element and have terminal a and b in common; they are therefore in parallel. FGU 6. Parallel 6 ircuit nalyi Parallel ircuit Boyletad element. Figure 6. Different way in which three parallel element may appear. 6 ircuit nalyi Parallel ircuit Boyletad 4

23 Parallel ircuit Parallel lement n Fig. 6., element and are in parallel becaue they have terminal a and b in common. he parallel combination of and i then in erie with element due to the common terminal point b. n Fig. 6.4, element and are in erie due to the common point a, but the erie combination of and i in parallel with element a defined by the common terminal connection at a and b. Parallel ircuit otal onductance For parallel element, the total conductance i the um of the individual conductance. G G + G + G + + G N Figure 6. Network in which and are in parallel and i in erie with the parallel combination of and. Figure 6.4 Network in which and are in erie and i in parallel. Figure 6.5 Determining the total conductance of parallel conductance. 6 ircuit nalyi Parallel ircuit Boyletad 6 Parallel ircuit otal eitance Since G /, the total reitance for the network can be determined by direct ubtitution into following equation N x. 6- Determine the total conductance and reitance for the parallel network of Figure 6.7. G G + G + Ω 6 Ω 0. S S 0. 5 S and Ω G 0.5S FGU 6.7 Figure 6.6 Determining the total reitance of parallel reitor. 6 ircuit nalyi Parallel ircuit Boyletad 7 x. 6- Determine the effect on the total conductance and reitance and reitance of the network of Fig. 6.7 if another reitor of 0 Ω were added in parallel with the other element. G 0.5S + 0.5S + 0.S 0.6 S. 667 Ω 0Ω G 0.6S 6 ircuit nalyi Parallel ircuit Boyletad 8

24 x. 6- Determine the total reitance for the network of Fig x. 6-4 a. Find the reitance of the network of Fig b. alculate the total reitance for the network of Fig FGU Ω 4 Ω 5 Ω 05. S S + 0. S 095. S. 05Ω 0.95S 6 ircuit nalyi Parallel ircuit Boyletad 9 Figure 6.9 xample 6-4: three parallel reitor ( a ) ( b ) Figure 6.0 xample 6-4: four parallel reitor of equal value. N N Ω 4 Ω Ω 0.5 Ω 4 he total reitance of parallel reitor i alway le than the value of the mallet reitor. Parallel ircuit otal eitance x. 6-5 epeat xample 6. uing q.(). For two parallel reitor, we write + + () he total reitance of two and parallel reitor i the product + of the two divided by their um. For three parallel reitor, the equation for become () 6 ircuit nalyi Parallel ircuit Boyletad + ( Ω )( 6 Ω ) 8 Ω Ω Ω + 6 Ω 9 6 ircuit nalyi Parallel ircuit Boyletad

25 x. 6-6 epeat xample 6. uing q.(). x. 6-7 alculate the total reitance of the parallel network of Fig. 6.. Figure Ω 4 Ω 5 Ω 05. Ω 05. S S + 0. S 095. S ' (5 Ω )( 4 Ω ) 5 Ω // 4 Ω. Ω 5 Ω + 4 Ω 6 ircuit nalyi Parallel ircuit Boyletad ' (. Ω )( Ω ) // Ω 05. Ω. Ω + Ω Ω Ω 9 Ω 6 Ω 7 Ω 6 Ω ' 6 Ω Ω N '' 4 ( 9 Ω )( 7 Ω ) 648 Ω 8 Ω + 9 Ω + 7 Ω 8 4 ' '' ' '' // + ' '' ( Ω )(8 Ω ) 6 Ω 6. Ω Ω + 8 Ω 0 6 ircuit nalyi Parallel ircuit Boyletad 5 x. 6-8 Determine the value of in Fig. 6. to etablih a total reitance of 9 kω. Figure k Ω 6 kω k Ω k Ω ircuit nalyi Parallel ircuit Boyletad 6 S 4

26 x. 6-9 Determine the value of,, and in Fig. 6. if and and the total reitance i 6 kω. Figure 6. x. 6-0 a. Determine the total reitance of the network of Fig b. What i the effect on the total reitance of the network of Fig.6.4 if additional reitor of the ame value i added, a hown in Fig.6.5? c. What i the effect on the total reitance of the network of Fig.6.4 if very large reitance i added in parallel, a hown in Fig.6.6? d. What i the effect on the total reitance of the network of Fig.6.4 if very mall reitance i added in parallel, a hown in Fig.6.7? k Ω k Ω 8 kω 4 6 ircuit nalyi Parallel ircuit Boyletad 7 0 Ω // 0 Ω 0 Ω 5 Ω Figure Ω // 0 Ω // 0 Ω 0 Ω 0 Ω < 5 Ω Figure ircuit nalyi Parallel ircuit Boyletad 8 Small decreae in Figure 6.6 Figure Ω // 0 Ω // k Ω 5 Ω // k Ω 0 Ω // 0 Ω // 0. Ω 5 Ω // 0. Ω ( 5 Ω )( 000 Ω ) ( 5 Ω )( 0. Ω ) Ω < 5 Ω Ω << 5 Ω 5 Ω Ω 5 Ω + 0. Ω Significant decreae in Parallel ircuit nalyi and Meaurement he network of Fig.6.8 i the implet of parallel circuit. ll the element have terminal a and b in common. he voltage acro parallel element i the ame. FGU 6.8 Parallel network. 6 ircuit nalyi Parallel ircuit Boyletad 9 6 ircuit nalyi Parallel ircuit Boyletad 0 + 5

27 For ingle ource parallel network, the ource current ( ) i equal to the um of the individual branch current. he power diipated by the reitor and delivered by the ource can be determined from. P P P x. 6- For the parallel network of Fig a. alculate. b. Determine. c. alculate and, and demontrate that +. d. Determine the power to each reitive load. e. Determine the power delivered by the ource, and compare it to the total power diipated by the reitive element. FGU 6.9 a. ( 9 Ω )( 8 Ω ) + 9 Ω + 8 Ω 6 Ω 6 Ω 7 b Ω 6 ircuit nalyi Parallel ircuit Boyletad 6 ircuit nalyi Parallel ircuit Boyletad c. 7 9 Ω Ω ( check ) x. 6- Given the information provided in Fig.6.0. a. Determine. b. alculate. c. Find &. d. Determine P. ( a ) Ω 0 Ω 0 Ω d. P ( 7 )( ) 8 W P ( 7 )( 5. ) W e. P ( 7 )( 4. 5 ) 5. W P + P (8 W )( W ) 5. W FGU 6.0 (b) (4)(0Ω) 40 6 ircuit nalyi Parallel ircuit Boyletad 6 ircuit nalyi Parallel ircuit ( d 0 Ω 40 ( c) 0 4 Ω 40 0 Ω ( ) ( ) W ) Boyletad P 0Ω 80 6

28 Kirchhoff urrent aw Kirchhoff current law (K) tate that the algebraic um of the current entering and leaving an area, ytem, or junction i zero. he um of the current entering an area, ytem, or junction mut equal to the um of the current leaving the area, ytem, or junction. entering leaving x. 6- Determine the current and 4 of Fig. 6. uing Kirchhoff current law. FGU 6. t a entering : t b entering : 4 leaving leaving FGU 6. ntroduction to K: FGU 6. Demontrating K: 6 ircuit nalyi Parallel ircuit Boyletad 6 x. 6-4 Determine the current,, 4, and 5 for the network of Fig x. 6-5 Determine the current and 5 of Fig. 6.5 through application of Kirchhoff current law. FGU 6.4 t c : entering leaving 4 4 t a t b t d : entering leaving : entering leaving : entering leaving FGU 6.5 t node a, t node b, ircuit nalyi Parallel ircuit Boyletad 7 6 ircuit nalyi Parallel ircuit Boyletad 8 7

29 x. 6-6 Find the magnitude and direction of the current, 4, 6, and 7 for the network of Fig.6.6. urrent Divider ule (D) For two parallel element of equal value, the current will divide equally. For parallel element with different value, the maller the reitance, the greater the hare of input current. For parallel element with different value, the current will plit with a ratio equal to the invere of their reitor value. FGU ircuit nalyi Parallel ircuit Boyletad 9 FGU 6.7 Demontrating how current will divide between unequal reitor. 6 ircuit nalyi Parallel ircuit Boyletad 0 For network in which only the reitor value are given along with the input current, the current divider rule hould be applied to determine the variou branch current. t can be derived uing the network of Fig FGU 6.8 Deriving the current divider rule. and x x 6 ircuit nalyi Parallel ircuit Boyletad x x For the particular cae of two parallel reitor, a hown in Fig and + + FGU 6.9 Developing an equation for current diviion between two parallel reitor. + n word, for two parallel branche, the current through either branch i equal to the product of the other parallel reitor and the input current divided by the um of the two parallel reitance. 6 ircuit nalyi Parallel ircuit Boyletad 8

30 x. 6-7 Determine the current for the network of Fig.6.0 uing the current divider rule. x. 6-8 Find the current for the network of Fig.6.. FGU ( 6 ) ( 4 k Ω )( 6 ) 4 k Ω + 8 k Ω 6 ircuit nalyi Parallel ircuit Boyletad FGU 6. 6 Ω 4 m m Ω 4 Ω 48 Ω 6 ircuit nalyi Parallel ircuit Boyletad 4 x. 6-9 Determine the magnitude of the current,, and for network of Fig. 6.. x. 6-9 Determine the reitance to effect the diviion of current in Fig ( 4 Ω )( ) ( Ω )( ) 4 Ω + 4Ω Figure 6. Figure 6. 8 Ω + 4Ω 7 Ω ( 7 m m ) m 6 ircuit nalyi Parallel ircuit Floyd 5 6 ircuit nalyi Parallel ircuit Boyletad 6 + ( + ) Ω Ω ( ) 9

31 urrent eek the path of leat reitance.. More current pae through the maller of two parallel reitor.. he current entering any number of parallel reitor divide into thee reitor a the invere ratio of their ohmic value. hi relationhip i depicted in Fig.6.. oltage Source in Parallel oltage ource are placed in parallel a hown in Fig. 6.4 only if they have ame voltage rating. Figure 6.4 Parallel voltage ource. 6 ircuit nalyifigure Parallel ircuit 6. urrent diviion through Floyd parallel branche. 7 6 ircuit nalyi Parallel ircuit Boyletad 8 f two batterie of different terminal voltage were placed in parallel both would be left ineffective or damaged becaue the terminal voltage of the larger battery would try to drop rapidly to that of the lower upply. onider two lead-acid car batterie of different terminal voltage placed in parallel, a hown in Fig HW 6-9 Baed olely on the reitor value, determine all the current for the configuration in Fig Do not ue Ohm law. 6 int + int 00. Ω Ω Ω Figure 6.99 Problem 9. Figure 6.5 Parallel batterie of different terminal voltage. 6 ircuit nalyi Parallel ircuit Boyletad 9 Homework 6:, 4, 7, 0, 8, 0,, 8, 9 6 ircuit nalyi Parallel ircuit Boyletad 40 0

32 /6 ircuit nalyi Serie and Parallel Network lectrical and elecommunication ngineering echnology Department cknowledgement want to expre my gratitude to Prentice Hall giving me the permiion to ue intructor material for developing thi module. would like to thank the Department of lectrical and elecommunication ngineering echnology of NY for giving me upport to commence and complete thi module. hope thi module i helpful to enhance our tudent academic performance. Sunghoon Jang Profeor Jang Prepared by textbook baed on ntroduction to ircuit nalyi by obert Boyletad, Prentice Hall, 0 th edition. OUNS ntroduction to Serie-Parallel Network educe and eturn pproach Block Diagram pproach Decriptive xample adder Network Serie-Parallel Network educe and eturn pproach Serie-parallel network are network that contain both erie and parallel circuit configuration For many ingle-ource, erie-parallel network, the analyi i one that work back to the ource, determine the ource current, and then find it way to the deired unknown. Key Word: Serie-Parallel Network, Block Diagram, adder Network 6 ircuit nalyi Serie and Parallel Network Boyletad FGU 7. ntroducing the reduce and return approach. 6 ircuit nalyi Serie and parallel network Boyletad

33 FGU 7. ntroducing the block diagram approach. Serie-Parallel Network Block Diagram pproach he block diagram approach will be employed throughout to emphaize the fact that combination of element, not imply ingle reitive element, can be in erie or parallel. 6 ircuit nalyi Serie and parallel network Boyletad 4 FGU 7. x. 7- f each block of Fig.7. were a ingle reitive element, the network of Fig. 7.4 might reult. ( ) ( ) ( ) ( ) m m k k k m m k k k B Ω Ω + Ω Ω Ω + Ω FGU ircuit nalyi Serie and parallel network Boyletad 4 N B B // // Ω Ω Ω Ω + + Ω Ω x. 7- t i alo poible that the block, B, and of Fig. 7. contain the element and configuration in Fig Working with each region: Ω + Ω Ω Ω : 4 // : 4 : 4,5 5 4 // N B B FGU 7.5 FGU ircuit nalyi Boyletad 6 B B 0.5 ( )( ) ( )( ) B B B B 8 4 Ω Ω 6 ircuit nalyi Serie and parallel network Boyletad 7 FGU 7.6

34 B x. 7- nother poible variation of Fig. 7. appear in Fig // Ω + ( 9 Ω )( 6Ω) 9 Ω + 6Ω 54 Ω.6 Ω 5 4//5 4 Ω + Ω 6 Ω ( 9Ω)( Ω) 4 Ω + 9 Ω + Ω FGU Ω ( 6Ω)( Ω) 6 ircuit nalyi Serie and parallel network 8 B //.6Ω + Ω 5.6 Ω FGU 7.7.6Ω + 6 Ω + Ω pplying the current divider rule yield By Ohm' law, B + ( Ω )( ) By Kirchhoff ' current law, B B B B B ( )(.6Ω) Ω + 6Ω 0.8 ( )( Ω) 6 + ( 6 Ω )( ) 6 Ω + 9 Ω...8 Serie-Parallel Network - Decriptive xample x. 7-4 Find the current 4 and the voltage for the network of Fig D B // Ω // 6Ω Ω D + D Ω ( Ω )( ) 4 Ω + 4Ω FGU 7.9 FGU 7.0 FGU 7. 6 ircuit nalyi Serie and parallel network Boyletad 0 x. 7-5 Find the indicated current and the voltage for the network of Fig. 7.. // B N // // 4 //5 FGU 7. 6 Ω Ω ( Ω )( Ω) 6 Ω. Ω Ω + Ω 5 ( 8 Ω )( Ω) 96 Ω 4.8 Ω 8 Ω + Ω 0 FGU 7. 6 ircuit nalyi Serie and parallel network Boyletad

35 // // + 4 // 5. Ω Ω 6 Ω 4 6 Ω 4 x. 7-6 a. Find the voltage,, and ab for the network of Fig b. alculate the ource current. // // 4//5 ( 4 )(.Ω) 4.8 ( 4 )( 4.8Ω) 9. FGU Ω ircuit nalyi Serie and parallel network Boyletad 6 Ω 5 4 a. pplying the voltage divider rule yield pplying Kirchhoff ' voltage law around ( 5 Ω )( ) the indicated loop of Fig Ω + Ω + ab 0 ( 6 Ω )( ) ab Ω + Ω 4 FGU 7.4 FGU ircuit nalyi Serie and parallel network Boyletad x. 7-7 For the network of Fig. 7.6, determine the voltage and and current. b. By Ohm ' law, Ω Ω pplying Kirchhoff current law, ircuit nalyi Serie and parallel network Boyletad 4 6 FGU 7.6 pplying K to the loop pplying K to node a yield Ω 6 Ω Ω FGU ircuit nalyi Serie and parallel network Boyletad 5 4

36 x. 7-9 alculate the indicated current and voltage of Fig.7.7. x. 7-0 hi example demontrate the power of Kirchhoff voltage law by determining the voltage,, and for the network of Fig kω 5 (,,) // //(8,9) //(8,9) FGU m + k Ω + kω 4 k Ω 7 //(8,9) ( 4.5k Ω )( k Ω + kω m 4.5k Ω m +4.5 m 7.5 m 6 ircuit nalyi Serie and parallel network Boyletad 6 ) FGU 7.7. FGU 7.8. For the path, For the path, For the path, ( 7 ) 5 6 ircuit nalyi Serie and parallel network Boyletad 7 Serie-Parallel Network adder Network three-ection ladder appear in Fig FGU 7.0. FGU 7.9. adder network. 5 Ω + Ω 8 Ω Ω FGU ircuit nalyi Serie and parallel network Boyletad 8 6 ircuit nalyi Serie and parallel network Boyletad 9 5

37 0 5 HW 7-6 For the ladder network in Fig. 7.86: a Determine. b. alculate. c. Find 8. Figure 7.86 Problem 6. 6 ( 6 Ω ) 6 ( 5 ) 0 6 Ω + Ω 9 ( )( 0 Ω ) Homework 7:, 4, 7,, 5, 5, 6 6 ircuit nalyi Serie and parallel network Boyletad 0 6 ircuit nalyi Serie and parallel network Boyletad 6

38 //6 ircuit nalyi Method of nalyi lectrical and elecommunication ngineering echnology Department cknowledgement want to expre my gratitude to Prentice Hall giving me the permiion to ue intructor material for developing thi module. would like to thank the Department of lectrical and elecommunication ngineering echnology of NY for giving me upport to commence and complete thi module. hope thi module i helpful to enhance our tudent academic performance. Sunghoon Jang Profeor Jang Prepared by textbook baed on ntroduction to ircuit nalyi by obert Boyletad, Prentice Hall, 0 th edition. OUNS ntroduction to Method nalyi urrent Source Source onverion urrent Source in Serie Branch urrent nalyi Meh & Super Meh nalyi Nodal & Super Nodal nalyi ntroduction to Method of nalyi he circuit decribed in the previou chapter had only one ource or two or more ource in erie or parallel preent. he tep-by-tep procedure outlined in thoe chapter cannot be applied if the ource are not in erie or parallel. Method of analyi have been developed that allow u to approach, in a ytematic manner, a network with any number of ource in any arrangement. Branch-current analyi, meh analyi, and nodal analyi will be dicued in detail in thi chapter. Key Word: urrent Source, Source onverion, Branch urrent, Meh nalyi, Nodal nalyi 6 ircuit nalyi Method of nalyi Boyletad 6 ircuit nalyi Method of nalyi Boyletad

39 urrent Source he interet in the current ource i due primarily to emiconductor device uch a the tranitor. n the phyical model (equivalent circuit) of a tranitor ued in the analyi of tranitor network, there appear a current ource a indicated in Fig. 8.. current ource determine the current in the branch in which it i located and the magnitude and polarity of the voltage acro a current ource are a function of the network to which it i applied. x. 8- Find the ource voltage and the current for the circuit of Fig. 7.. FGU 8. urrent ource within the tranitor equivalent circuit. FGU 8. 0 m (0 m)(0 kω) 00 6 ircuit nalyi Method of nalyi Boyletad 4 6 ircuit nalyi Method of nalyi Boyletad 5 x. 8- Find the voltage and the current and for the network of Fig. 8.. x. 8- Determine the current and voltage for the network of Fig Ω FGU 8. pplying Kirchhoff ' current law : ircuit nalyi Method of nalyi Boyletad 6 FGU 8.4 he voltage i ( )( Ω ) 4 U in g the current divider rule : + and, applying Kirchhoff ' voltage law, ( Ω )( 6 ) ( Ω ) + ( Ω ) 6 ircuit nalyi Method of nalyi Boyletad 7

40 Source onverion ll ource-whether they are voltage or current-have ome internal reitance in the relative poition hown in Fig. 8.5 and 8.6. For the voltage ource, if 0 Ω or i o mall compared to any erie reitor that it can be ignored, then we have an ideal voltage ource. For the current ource, if Ω or i large enough compared to other parallel element that it can be ignored, then we have an ideal current ource. he equivalent ource, a far a terminal a and b are concerned, are repeated in Fig. 8.7 with the equation for converting in either direction. Note, a jut indicated, that the reitor i the ame in each ource; only it poition change. he current of the current ource or the voltage of the voltage ource i determined uing Ohm law and the parameter of the other configuration. FGU 8.5 FGU ircuit nalyi Method of nalyi Boyletad 8 FGU 8.6 Source converion 6 ircuit nalyi Method of nalyi Boyletad x. 8-4 a. onvert the voltage ource of Fig. 8.8 (a) to a current ource, and calculate the current through the 4-Ω load for each ource. b. eplace the 4-Ω load with a -kω load, and calculate the current for the voltage ource. c. eplace the calculation of part (b) auming that the voltage ource i ideal ( 0 Ω) becaue i o much larger than. thi one of thoe ituation where auming that the ource i ideal i an appropriate approximation? 6 a. Fig. 88(. a ): + Ω + 4Ω ( Ω)( ) Fig. 88(. b ): + Ω + 4Ω b Ω + k Ω c. 6 6 m m k Ω Ye, >> ( voltage ource ) 6 ircuit nalyi Method of nalyi FGU 8.8 Boyletad 0 x. 8-5 a. onvert the current ource of Fig. 8.9(a) to a voltage ource, and find the load current for each ource. b. eplace the 6-kΩ load with a 0-kΩ load, and calculate the current for the current ource. c. eplace the calculation of part (b) auming that the vcurrent ource i ideal ( Ω) becaue i o much maller than. thi one of thoe ituation where auming that the ource i ideal i an appropriate approximation? a. Fig. 89. ( a ): + ( k Ω )( 9 m ) k Ω + 6 k Ω m Fig. 89. ( b ): FGU ( k Ω )( 9 m ) b.. + k Ω + 0 Ω k Ω + 6 k Ω c. 9 m m m Ye, >> ( current ource ) 6 ircuit nalyi Method of nalyi Boyletad

41 x. 8-6 eplace the parallel current ource of Fig. 8.0 and 8. to a ingle current ource. x. 8-7 educe the network of Fig. 8. to a ingle current ource, and calculate the current through. FGU 8.0 FGU 8. FGU // 8 Ω //4 Ω 6 Ω + ( 6 Ω )( 0 ) 6 Ω + 4 Ω 6 ircuit nalyi Method of nalyi Boyletad 6 ircuit nalyi Method of nalyi Boyletad x. 8-8 Determine the current in the network of Fig. 8.. urrent Source in Serie he current through any branch of a network can be only ingle-valued. For the ituation indicated at point a in Fig. 8.4, we find by application of Kirchoff current law that the current leaving that point i greater than entering-an impoible ituation. herefore, FGU 8. ( 4 )( Ω ) Ω Ω + Ω 6 ircuit nalyi Method of nalyi Boyletad 4 urrent ource of different current rating are not connected in parallel. FGU ircuit nalyi Method of nalyi Boyletad 5 4

42 Branch-urrent nalyi We will now conider the firt in a erie of method for olving network with two or more ource. FGU 8.5 Determining the number of independent cloed loop.. ign ditinct current of arbitrary direction to each branch of the network.. ndicate the polaritie for each reitor a determined by the aumed current direction.. pply Kirchhoff voltage law around each cloed, independent loop of the network. 4. pply Kirchhoff current law at the minimum number of node that will include all the branch current of the network. 5. Solve the reulting imultaneou linear equation for aumed branch current. FGU 8.6 Determining the number of application of Kirchhoff current law required. 6 ircuit nalyi Method of nalyi Boyletad 6 6 ircuit nalyi Method of nalyi Boyletad 7 x. 8-9 pply the branch-current method to the network of Fig FGU 8.7 loop : loop : + 0 and loop : + ( Ω ) + ( 4 Ω ) 0 loop : ( 4 Ω ) ( Ω ) pplying + loop : + ( Ω ) + ( 4 Ω )( + ) 0 loop : ( 4 Ω )( + ) ( Ω ) ( ) ( 6 ) ( ) ( 6 ) ircuit nalyi Method of nalyi Meh nalyi he econd method of analyi to be decribed i called meh analyi. he term meh i derived from the imilaritie in appearance between the cloed loop of a network and wire meh fence. he meh-analyi approach imply eliminate the need to ubtitute the reult of Kirchhoff current law into the equation derived from Kirchhoff voltage law. he ytematic approach outlined below hould be followed when applying thi method.. ign a ditinct current in the clockwie direction to each independent, cloed loop of the network. t i not abolutely neceary to chooe the clockwie direction for each loop current.. ndicate the polaritie with each loop for each reitor a determined by the aumed current direction of loop current for that loop.. pply Kirchhoff voltage law around each cloed loop in the clockwie direction. 4. Solve the reulting imultaneou linear equation for aumed branch current. 6 ircuit nalyi Method of nalyi Boyletad 9 5

43 x. 8-0 onider the ame baic network a in xample 8.9 of the preceding dection, now appearing in Fig.8.8. loop : + ( Ω ) + ( 4 Ω )( ) 0 loop : + ( 4 Ω )( ) + ( Ω ) and loop : loop : FGU ( ) and 6 ircuit nalyi Method of nalyi Boyletad 0 x. 8- Find the current through each branch of the network of Fig.8.9. FGU 8.9 he current in the 6Ω reitor and 0 ource for loop i loop : 5 + ( Ω ) + ( 6 Ω )( ) loop : 0 + ( 6 Ω )( ) + ( Ω ) 0 and loop : loop : ( ) and 6 ircuit nalyi Method of nalyi Boyletad x. 8- Find the branch current of the network of Fig.8.0. x. 8- Uing meh analyi, determine the current of the network of Fig.8.. loop : 6 + ( Ω ) ( 4 Ω )( ) 0 loop : ( 4 Ω )( ) 4 + ( 6 Ω ) + 0 o that loop : loop : FGU 8.0 he current in the 4Ω reitor and 4 ource for loop i.8 ( 0.77) ( ) and ircuit nalyi Method of nalyi Boyletad FGU ( 6 Ω ) + ( 4 Ω ) + ( Ω ) 0 or 0 + pplying ( + 4 ) + 8 or ircuit nalyi Method of nalyi Boyletad 6

44 Nodal nalyi x. 8-4 pply nodal analyi to the network of Fig.8.. We will employ Kirchhoff current law to develop a method referred to a nodal analyi. node i defined a a junction of two or more branche. Since a point of zero potential or ground i ued a a reference, the remaining node of the network will all have a fixed a fixed potential relative to thi reference. For a network of N node, therefore, there will exit (N ) node.. Determine the number of node within the network.. Pick a reference node, and label each remaining node with a ubcripted value of voltage:,, and o on.. pply Kirchhoff current law at each node except the reference. ume that all unknown current leave the node for each application of Kirchhoff current law. 4. Solve the reulting equation for the nodal voltage. 6 ircuit nalyi Method of nalyi Boyletad 4 FGU 8. 4 Node : Ω Ω or Ω Ω 6 ircuit nalyi Method of nalyi 5 x. 8-5 pply nodal analyi to the network of Fig.8.. FGU ( 80 ) ( ) ` ` ` 64 Node : + 8 Ω 4 Ω + 0 Node : Ω 0 Ω o that Ω Ω Ω 6 ircuit nalyi Method of nalyi Boyletad 6 6 ircuit nalyi Method of nalyi Boyletad 7 7

45 x. 8-6 Determine the nodal voltage for the network of Fig.8.4. FGU 8.4 Node : Ω Ω 0 Node : Ω 6 Ω o that ( ) 6 7 ` ` ` 6 ( 6 ) Ω 6 Ω 6 6 Ω 6 ircuit nalyi Method of nalyi Boyletad 8 6 ircuit nalyi Method of nalyi Boyletad 9 x. 8-7 Determine the nodal voltage and Fig.8.5 uing the concept of a upernode. Supernode : o that FGU 8.5 Node,: Ω 4 Ω 0 Ω Ω ( 40 ) ( 5 ) ircuit nalyi Method of nalyi Boyletad 0 HW 8- Uing meh analyi, determine the current for the network in Fig ompare your anwer to the olution of Problem 8. Figure 8.9 Problem. Homework 8:, 4, 6, 7, 8, 9,,, 5, 6 ircuit nalyi Method of nalyi Boyletad 8

46 /6 ircuit nalyi Method of nalyi lectrical and elecommunication ngineering echnology Department cknowledgement want to expre my gratitude to Prentice Hall giving me the permiion to ue intructor material for developing thi module. would like to thank the Department of lectrical and elecommunication ngineering echnology of NY for giving me upport to commence and complete thi module. hope thi module i helpful to enhance our tudent academic performance. Sunghoon Jang Profeor Jang Prepared by textbook baed on ntroduction to ircuit nalyi by obert Boyletad, Prentice Hall, 0 th edition. OUNS ntroduction to Network heorem ntroduction to Network heorem Superpoition hevenin heorem Norton heorem hi chapter will introduce the important fundamental theorem of network analyi. ncluded are the uperpoition, hevenin, Norton, and maximum power tranfer theorem. We will conider a number of area of application for each. through undertanding of each theorem will be applied repeatedly in the material to follow. Maximum Power ranfer heorem Key Word: Network heorem, Superpoition, hevenin, Norton, Maximum Power 6 ircuit nalyi Network heorem Boyletad 6 ircuit nalyi Network heorem Boyletad

47 Superpoition heorem he uperpoition theorem can be ued to find the olution to network with two or more ource that are not in erie or parallel. he mot advantage of thi method i that it doe not require the ue of a mathematical technique uch a determinant to find the required voltage or current. he current through, or voltage acro, an element in a linear bilateral network i equal to the algebraic um of the current or voltage produced independently by each ource. Figure 9. review the variou ubtitution required when removing an ideal ource, and Figure 9. review the ubtitution with practical ource that have an internal reitance. FGU 9. emoving the effect of practical ource Number of network to be analyzed Number of independent ource FGU 9. emoving the effect of ideal ource 6 ircuit nalyi Network heorem Boyletad 4 6 ircuit nalyi Network heorem Boyletad 5 x. 9- Determine for the network of Fig. 9.. '' Ω ' + '' FGU 9. FGU 9.4 ' 0 6 ircuit nalyi Network heorem Boyletad 6 x. 9- Uing uperpoition, determine the current through the 4-Ω reitor of Fig Note that thi i a two-ource network of the type conidered in chapter 8. FGU 9.5 ' + // 4 Ω + Ω // 4 Ω 4 Ω + Ω 7 Ω 54 7 Ω FGU ( )( ) 5. Ω + 4 Ω 6 ircuit nalyi Network heorem Boyletad 7

48 x. 9- a. Uing uperpoition, find the current through the 6-Ω reitor of Fig b. Determine that uperpoition i not applicable to power level. FGU // 4 Ω + 4 Ω // Ω 4 Ω + 8 Ω Ω '' 48 4 Ω '' ' '' ( direction of ) FGU ircuit nalyi Network heorem Boyletad 8 FGU 9.9 FGU 9.0 a. conidering that the effect of the 6 ource ( Fig ): ' + 6 Ω + 6 Ω conidering that the effect of the 9 ource ( Fig. 9. ): '' ( Ω )( 9 ) 6 + Ω + 6 Ω 6 ircuit nalyi Network heorem FGU Boyletad 9. 9 he total current through the 6 Ω reitor ( Fig 9. ) i ' '' b. he power to the 6 Ω reitor i Power (8 ) ( 6 Ω ) 84 W FGU 9. x. 9-4 Uing the principle of uperpoition, find the current through the -kω reitor of Fig. 9.. ' ( 6 k Ω )( 6 m ) m + 6 k Ω + k Ω he calculated power to the 6 Ω reitor due to each ource, miu in g the principle of up erpoition, i ' P ( ) ( ) ( 6 Ω ) 4 W '' P ( ) ( 6 ) ( 6 Ω ) 6 W FGU 9. P + P 40 W 84 W becaue ( ) + ( 6 ) (8 ) 6 ircuit nalyi Network heorem 0 FGU ircuit nalyi Network heorem Boyletad

49 x. 9-5 Find the current through the -Ω reitor of the network of Fig he preence of three ource will reult in three different network to be analyzed. conidering that the effect of the 9 voltage ource ( Fig ): '' + FGU m 6 k k 05. Ω + Ω ' '' Since and have the ame direction through, the deired current i the um of the two : ' '' + m m. 5 m FGU 9.6 FGU ircuit nalyi Network heorem FGU 9.8 Boyletad FGU 9.9 conidering the effect of the ource (Fig. 97. ): ' + Ω + 4 Ω conidering that the effect of the 6 ource (Fig ): 6 '' + Ω + 4 Ω conidering the effect of the ource (Fig. 99. ): ''' (4Ω)( ) + Ω + 4Ω ny two-terminal, linear bilateral dc network can be replaced by an equivalent circuit coniting of a voltage ource and a erie reitor, a hown in Fig. 9.. hevenin heorem FGU 9. hevenin equivalent circuit he total current through the Ω reitor appear in Fig.. 90, and + + '' ''' ' 6 ircuit nalyi Method of nalyi FGU 9.0 Boyletad 4 FGU 9. he effect of applying hevenin theorem. 6 ircuit nalyi Network heorem Boyletad 5 4

50 x. 9-6 Find the hevenin equivalent circuit for the network in the haded area of the network of Fig hen find the current through for value of Ω, 0Ω, and 00Ω. FGU 9. Subtituting the hevenin equivalent circuit for a complex network.. emove that portion of the network acro which the hevenin equivalent circuit i to be found. n Fig. 9.(a), thi require that the road reitor be temporary removed from the network.. Make the terminal of the remaining two-terminal network.. alculate H by firt etting all ource to zero (voltage ource are replaced by hort circuit, and current ource by open circuit) and then finding the reultant reitance between the two marked terminal. 4. alculate H by firt returning all ource to their original poition and finding the open-circuit voltage between the marked terminal. 5. Draw the hevenin equivalent circuit with the portion of the circuit previouly 6 ircuit nalyi Method of nalyi Boyletad 6 removed replaced between the terminal of the equivalent circuit. FGU 9.4 FGU 9.5 dentifying the terminal of particular importance when applying hevenin theorem. // H ( Ω )( 6 Ω ) Ω + 6 Ω Ω FGU 9.6 Determining H for the network of Fig ircuit nalyi Network heorem Boyletad 7 x. 9-7 Find the hevenin equivalent circuit for the network in the haded area of the network of Fig FGU 9.7 FGU 9.9 Subtituting the hevenin equivalent circuit for the network external to in Fig. 9.. H + H + H FGU 9.8 ( 6 Ω )( 9 ) 6 6 Ω + Ω 6 Ω : 5. Ω + Ω 6 0 Ω : Ω + 0 Ω Ω : Ω + 00 Ω 8 FGU 9.0 FGU 9. FGU 9. H + 4 Ω + Ω 6 Ω 6 ircuit nalyi Network heorem Boyletad 9 5

51 () 0 0 x. 9-8 Find the hevenin equivalent circuit for the network in the haded area of the network of Fig Note in thi example that there i no need for the ection of the network to be at preerved to be at the end of the configuration. FGU 9. H ( )( 4 Ω) 48 Ω FGU 9.5 FGU 9.6 FGU 9.4 Subtituting the hevenin equivalent circuit in the network external to the reitor of Fig // H ( 6 Ω )( 4 Ω ) 6 Ω + 4 Ω 4. Ω 6 ircuit nalyi Network heorem Boyletad 0 6 ircuit nalyi Method of nalyi FGU 9.7 Boyletad x. 9-9 Find the hevenin equivalent circuit for the network in the haded area of the network of Fig FGU 9.8 H FGU ( 6 Ω )(8 ) 6 Ω + 4 Ω 48. FGU 9.40 Subtituting the hevenin equivalent circuit in the network external to the reitor 4 of Fig FGU 9.4 FGU H // // 4 6 Ω // Ω + 4 Ω // Ω Ω + Ω 5 Ω 6 ircuit nalyi Network heorem Boyletad 6 ircuit nalyi Method FGU of nalyi 9.4 Boyletad 6

52 x. 9-0 (wo ource) Find the hevenin equivalent circuit for the network within the haded area of Fig FGU 9.44 FGU 9.46 Subtituting the hevenin equivalent circuit in the network external to the reitor of Fig ( 6 Ω )( 7 ) 48 6 Ω + Ω ( Ω )( 7 ) 54 Ω + 4 Ω H H FGU 9.45 FGU 9.48 FGU 9.47 H 4 + // // 4. k Ω k Ω // 4 k Ω // 6 k Ω 4. k Ω k Ω //. 4 k Ω 4. k Ω k Ω k Ω Boyletad 4 6 ircuit naly i FGU Method of 9.49 nalyi Boyletad 5 ( 0 ) 0 ' H ' // 4 k Ω // 6 k Ω. 4 k Ω xperimental Procedure FGU 9.50 ' ' + ( 4. k Ω )( 6 ) k Ω k Ω ' 45. H () 0 0 FGU 9.5 Subtituting the hevenin equivalent circuit in the network external to the reitor 6 ircuit nalyi Method of nalyi of Fig. ' Boyletad polarity of 6H 947 '' H ' // 08. k Ω // 6 k Ω k Ω FGU 9.5 ' ( k Ω )( 0 ) 5. ' k Ω + 4 k Ω '' ' '' H 5. H H H ( ) FGU 9.5 FGU 9.54 S H H H H S O where H H O S FGU ircuit nalyi Network heorem Boyletad 7 7

53 Norton heorem ny two-terminal, linear bilateral dc network can be replaced by an equivalent circuit coniting of a current ource and a parallel reitor, a hown in Fig emove that portion of the network acro which the hevenin equivalent circuit i found.. Make the terminal of the remaining two-terminal network.. alculate N by firt etting all ource to zero (voltage ource are replaced by hort circuit, and current ource by open circuit) and then finding the reultant reitance between the two marked terminal. 4. alculate N by firt returning all ource to their original poition and finding the hort-circuit current between the marked terminal. 5. Draw the Norton equivalent circuit with the portion of the circuit previouly removed replaced between the terminal of the equivalent circuit. FGU 9.56 Norton equivalent circuit 6 ircuit nalyi Network heorem Boyletad 8 FGU 9.57 Subtituting the Norton equivalent circuit for a complex network. 9 x. 9- Find the Norton equivalent circuit for the network in the haded area of Fig ( 06 ) Ω 0 N 9 Ω FGU 9.58 FGU 9.59 dentifying the terminal of particular interet for the network of Fig FGU 9.6 Determining N for the network of Fig FGU 9.60 Determining N for the network of Fig // N Ω // 6 Ω ( Ω )( 6 Ω ) Ω + 6 Ω Ω Boyletad 0 FGU 9.6 Subtituting the Norton equivalent circuit for the network external to the reitor of Fig FGU 9.6 onverting the Norton equivalent circuit of Fig. 9.6 to a hevenin equivalent circuit. 8

54 x. 9- Find the Norton equivalent circuit for the network external to the 9-Ω reitor in Fig FGU 9.67 Determining N for the network of Fig N + (5 Ω )( 0 Ω ) 5 Ω + 4 Ω FGU 9.64 FGU 9.65 N + 5 Ω + 4 Ω 9 Ω FGU 9.66 Boyletad FGU 9.68 Subtituting the Norton equivalent circuit for the network external to the reitor of Fig ircuit nalyi Network heorem Boyletad x. 9- (wo ource) Find the Norton equivalent circuit for the portion of the network external to the left of a-b in Fig FGU 9.7 FGU 9.7 FGU 9.70 FGU 9.69 // N 4 Ω // 6 Ω ( 4 Ω )( 6 Ω ) 4 Ω + 6 Ω 4. Ω FGU 9.7 Boyletad 4 FGU 9.74 Subtituting the Norton equivalent circuit for the network to the left of terminal a-b in Fig ' 7 N Ω '' N 8 '' ' N N N Boyletad 5 9

55 Maximum Power ranfer heorem load will receive maximum power from a linear bilateral dc network when it total reitive value i exactly equal to the hevenin reitance of the network a een by the load. HW 9-6 Uing uperpoition, find the voltage for the network in Fig ' 6.8kΩ (6 ) 6.8k Ω + kω.0 kω (9m) k Ω + 6.8kΩ 5.75 m Figure 9.4 Problem 6. H FGU 9.74 Defining the condition for maximum power to a load uing the hevenin equivalent circuit. N FGU 9.75 Defining the condition for maximum power to a load uing the Norton equivalent circuit. '' (5.75m)(6.8kΩ) 9.0 ' '' Homework 9:, 4, 6,, 4 6 ircuit nalyi Network heorem Boyletad 6 6 ircuit nalyi Network heorem Boyletad 7 0

56 /6 ircuit nalyi apacitor OUNS ntroduction to apacitor he lectric Field apacitance lectrical and elecommunication ngineering echnology Department Profeor Jang Prepared by textbook baed on ntroduction to ircuit nalyi by obert Boyletad, Prentice Hall, 0 th edition. ranient in apacitive Network: harging Phae apacitor in Serie and Parallel Key Word: apacitor, lectric Field, apacitance, ranient 6 ircuit nalyi apacitor Boyletad ntroduction to apacitor hu far, the only paive device appearing in the cla ha been the reitor. We will now conider two additional paive device called the capacitor and the inductor, which are quite different from the reitor in purpoe, operation, and contruction. Unlike the reitor, both element diplay their total characteritic only when a change in voltage or current i made in the circuit in which they exit. n addition, if we conider the ideal ituation, they do not diipate energy a doe the reitor but tore it in a form that can be returned to the circuit whenever required by the circuit deign. he lectric Field he electric field i repreented by electric flux line, which are drawn to indicate the trength of the electric field at any point around any charged body; that i, the dener the line of flux, the tronger the electric field. n Fig. 0., the electric field trength i tronger at point a than at poition b becaue the flux line are dener at a than b. he flux per unit area (flux denity) i repreented by the capital letter D and i determined by D ψ ( flux / unit area) 6 ircuit nalyi apacitor Boyletad FGU 6 ircuit 0. nalyi Flux ditribution apacitor from an iolated poitive Boyletad charge. 4

57 Ψ Q ( coulom, ) F Q ( network / coulomb, N / ) kq r ( N / ) he lectric Field he electric field i repreented by electric flux line, which are drawn to indicate the trength of the electric field at any point around the charged body. he electric field trength at any point ditance r from a point charge of Q coulomb i directly proportional to the magnitude of the charge and inverely proportional to the ditance quared from the charge. lectric flux line alway extend from a poitively charged body to a negatively charged body, alway extend or terminate perpendicular to the charged urface, and never interect. FGU 0. lectric flux ditribution: (a) oppoite charge; (b) like charge. 6 ircuit nalyi apacitor Boyletad 5 Q apacitance capacitor i contructed imply of two parallel conducting plate eparated by inulating material (in thi cae, air). apacitance i a meaure of a capacitor ability to tore charge on it plate. capacitor ha a capacitance of farad if coulomb of charge i depoited on the plate by a potential difference of volt cro the plate. farad (F) Q coulomb () volt () FGU 0. lectric flux ditribution between the plate of a capacitor: (a) including fringing; (b) ideal. 6 ircuit nalyi apacitor Boyletad 6 D ε ( farad / meter, F / M ) ε εr ε 0 d ε εε r ε r ( F ) d d : lectric field (/m) D : Flux denity ε : Permittivity (F/m) : apacitance (F) Q : harge () : rea in quare meter d : Ditance in meter between the plate FGU 0.4 ffect of a dielectric on the field ditribution between the plate of a capacitor: (a) alignment of dipole in the dielectric; (b) electric field component between the plate of a capacitor 6 with ircuit a dielectric nalyi preent.. apacitor Boyletad 7 x. 0- Determine the capacitance of each capacitor on the right ide of Fig.0.5. a. (5 µ F ) 5 µ F b. ( 0. µ F ) µ F c.. 5 ( 0 µ F ) 50 µ F d. 4 (5) ( pf ) ( / ) ( 60 )( 000 pf ) 06. µ F FGU ircuit nalyi apacitor Boyletad 8

58 x. 0- For the capacitor of Fig. 0.6: a. Determine the capacitance. b. Determine the electric field trength between the plate if 450 are applied acro the plate. c. Find the reulting charge on each plate. a. b. c. o ε o d (.85 0 F / m)( 8 0.0m.5 0 m ε / m d.5 0 m Q Q ( )( 450 ) n ) F FGU ircuit nalyi apacitor Boyletad 9 FGU 0 7 Summary of capacitive element ranient in apacitive Network: harging Phae capacitor can be replaced by an open-circuit equivalent once the charging phae in a dc network ha paed. FG. 0.8 Baic charging network. FG. 0.9 i c during charging phae. FG 0.0 c during charging phae. Figure 0. he e -x function (x 0). FG. 0. Open-circuit equivalent for a FG. 0. Short-circuit equivalent for a capacitor 6 ircuit following nalyi the apacitor charging phae. capacitor Boyletad (witch cloed, t0). 6 ircuit nalyi apacitor Figure Boyletad 0.4 i c veru t during the charging phae. i e t/ t (ec ond, )

59 x. 0- a. Find the mathematical expreion for the tranient behavior of v, i, and v for the circuit of Fig. 0.8 when the witch i moved to poition. Plot the curve of v, i, and v. b. How much time mut pa before it can be aumed, for all practical purpoe, that i 0 and v volt? Figure 0.5 v c veru t during the charging phae. Figure 0.6 ffect of on the charging phae. t/ v ( e ) v e t/ Figure 0.7 v veru t during the charging phae. 6 ircuit nalyi apacitor Boyletad 6 a. τ (8 0 Ω )( 4 0 F ) FGU m t / τ t /( 0 ) v ( e ) 40 ( e ) i e t τ 80 e 8 k Ω / t /( 0 ) (5 0 ) e t /( 0 ) v e 40 e t / τ t /( 0 ) 6 ircuit nalyi apacitor Boyletad FGU b. 5 τ 5 ( m ) 60 m ranient in apacitive Network: Dicharging Phae v t/ e : dich arg ing i e t/ dich ing arg t/ v e dich arg ing x. 0-4 fter v in xample 0. ha reached it final value of 40, the witch i hown into poition, a hown in Fig. 0.. Find the mathematical expreion for the tranient behavior of v, i, and v after the cloing of the witch. Plot the curve for v, i, and v uing the defined direction and polaritie of Fig ume that t 0 when the witch i moved to poition. Figure 0.0 Demontrating the dicharge behavior of a capacitive network. Figure 0. he charging and dicharging cycle for the network of fig ircuit nalyi apacitor 5 τ m v t/ t / 0 e 40 ( e ) i 0 e t (5 ) e / t / 0 t v e 40 e FGU 0. / t / 0 FGU 0. 4

60 apacitor in Serie and Parallel Figure 0.4 Serie capacitor. Q Q Q Q + Q Q + Q + Q + + x. 0-5 For the circuit of Fig. 0.6: a. Find the total capacitance. b. Determine the charge on each plate. c. Find the voltage acro each capacitor. Figure 0.6 b. Q Q Q Q 6 (8 0 F )( 60 ) 480 µ a F 50 0 F 0 0 F µ F c Q F 6 Q F 6 Q F and Figure ircuit nalyi Parallel apacitor capacitor. Boyletad 7 6 ircuit nalyi apacitor Boyletad 8 x. 0-6 For the circuit of Fig. 0.7: a. Find the total capacitance. b. Determine the charge on each plate. c. Find the total charge. x. 0-7 Find the voltage acro and charge on each capacitor for the network of Fig ' + 4 µ F + µ F 6 µ F ' ( µ F )( 6 µ F ) µ F ' + µ F + 6 µ F 6 Q ( 0 F )( 0 ) 40 µ a. + + Figure µ F + 60 µ F + 00 µ F 060 µ F 6 b. Q (800 0 F )( 48 ) 8. 4 m 6 Q ( 60 0 F )( 48 ). 88 m 6 Q ( 00 0 F )( 48 ) m c. Q Q + Q + Q m 6 ircuit nalyi apacitor Boyletad 9 Figure 0.8 ' Q Q Q Q 40 µ 6 Q F ' Q 40 µ ' ' 6 Q ' F ' 6 Q ( 4 0 F )( 40 ) 60 µ Figure ircuit nalyi apacitor Boyletad 0 ' 6 Q ( 0 F )( 40 ) 80 µ 5

61 x. 0-8 Find the voltage acro and charge on capacitor of Fig. 0.0 after it ha charged up to it final value. x. 0-9 Find the voltage acro and charge on each capacitor for the network of Fig. 0.. after each ha charged up to it final value. Figure 0.0 Figure 6 0. ircuit nalyi apacitor (8 Ω )( 4 ) 6 4 Ω + 8 Ω Q 6 ( 0 0 F )( 6 ) 0 µ Boyletad (7 Ω)( 7 ) 76 7 Ω + Ω ( Ω)( 7 ) 6 7Ω + Ω Q 6 ( 0 F )( 6 ) µ Q ( 0 6 F)(56 ) 68 µ Figure 0. 6 ircuit nalyi apacitor Boyletad HW 0-54 Find the voltage acro and the charge on each capacitor for the circuit in Fig HW 0- epeat Problem for MΩ, and compare the reult. Figure 0.87 Problem. Figure 0.5 Problem 54. Homework 0:, 4, 4, 5,5, 54, 57 Homework 0:, 4, 7, 8 6

62 /6 ircuit nalyi Magnetim and nductor cknowledgement want to expre my gratitude to Prentice Hall giving me the permiion to ue intructor material for developing thi module. would like to thank the Department of lectrical and elecommunication ngineering echnology of NY for giving me upport to commence and complete thi module. hope thi module i helpful to enhance our tudent academic performance. lectrical and elecommunication ngineering echnology Department Profeor Jang Prepared by textbook baed on ntroduction to ircuit nalyi by obert Boyletad, Prentice Hall, th edition. OUNS ntroduction to nductor he Magnetic Field nductance - ranient: he Storage Phae hevenin quivalent Key Word: nductor, Magnetic Field, nductance, ranient, hevenin 6 ircuit nalyi Magnetim and nductor Boyletad ntroduction to nductor hree baic component appear in the majority of electrical/electronic ytem in ue today. hey include the reitor and the capacitor, which have already been introduced, and the inductor, to be examined in detail in thi module. ike the capacitor, the inductor exhibit it true characteritic only when a charge in voltage or current i made in the network. ecall from previou module that a capacitor can be replaced by an open-circuit equivalent under teady-tate condition. You will ee in thi module that an inductor can be replaced by a hort-circuit equivalent under teady-tate condition. Finally, you will learn that while reitor diipate the power delivered to them in form of heat, ideal capacitor tore the energy delivered to them in the form of an electric field. nductor are like capacitor in that they alo tore the energy delivered to them but in the form of a magnetic field. 6 ircuit nalyi Magnetim and nductor Boyletad

63 Magnetic Field Magnetim play an integral part in almot every electrical device ued today in indutry, reearch, or the home. he compa relie on a permanent magnet for indicating direction. Michael Faraday, Karl Friedrich Gau, and Jame lerk Maxwell continued to experiment in thi area and developed many of baic concept of electromagnetim magnetic effect induced by the flow of charge, or current. magnetic field exit in the region urrounding a permanent magnet, which can be repreented by magnetic flux line imilar to electric flux line. Magnetic flux line, however, do not have origin or terminating point a do electric flux line but exit in continuou loop, a hown in Fig... Figure. Flux ditribution for a permanent magnet. Figure 6. ircuit Flux nalyi ditribution apacitor for two adjacent, oppoite pole. Figure Boyletad. Flux ditribution for two adjacent, like pole. 4 Figure.6 Magnetic flux line around a current-carrying conductor. n the S ytem of unit, magnetic flux i meaured in weber (WB). he applied ymbol i Φ. he number of flux line per unit area, called the flux denity, i denoted by B and i meaured in tela (). Figure.7 Flux ditribution of a ingle-turn coil. Figure.8 Flux ditribution of a current carrying coil. Figure.9 lectromagnet. Figure.0 Determining the direction of flux for an electromagnet: (a) method; (b) notation. B Φ tela B Wb / m tela ( ) Φ weber ( Wb) m Wb/m 6 ircuit nalyi Magnetim and nductor Boyletad 5 6 ircuit nalyi Magnetim and nductor nductance Sending a current through a coil of wire etablihe a magnetic field through and urrounding the unit. hi component i called an inductor. t inductance level determine the trength of the magnetic field around the coil due to an applied current. inductor are deigned to et up a trong magnetic field linking the unit, wherea capacitor are deigned to et up a trong electric field between the plate. Figure.5 Some area of application of magnetic effect. Figure.6 Defining the B oyletad parameter for q. (.6). nductance nductor ontruction he level of inductance ha imilar contruction enitivitie in that i dependent on the area within the coil, the length of the unit, and the permeability of the core material. t i alo enitive to the number of turn of wire in the coil a depicted by q. (.6) and defined in Fig..6 for two of the mot popular hape: µ N l μ : Permeability (Wb/ m) N : Number of turn (t) : m (.6) l : m : henrie (H) µ r µ o N 7 µ r where µ µ r µ o or 4 π 0 l µ N N l ( henrie, H ) he inductance of an inductor with a o µ r µ r o l ferromagnetic core i μ r time the inductance obtained with an air core. 6 ircuit nalyi Magnetim and nductor Boyletad 7

64 x. - For the air-core coil in Fig..8: a. Find the inductance. b. Find the inductance if a metallic core with μ r 000 i inerted in the coil. m a. d in. 6.5 mm in. π d π(6.5mm).7 µm 4 4 m l in. 5.4 mm 9.7 in. 7 µ r N 7 ()(00t) (.7µm 4π 0 4π 0 l 5.4 mm b. µ (000)(5.68 µh).6 mh r 0 Figure.8 ir-core coil for example.. ) 5.68 µh 6 ircuit nalyi Magnetim and nductor Boyletad 8 x. - n Fig..9, if each inductor in the left column i changed to the type appearing in the right column, find the new inductance level. For each change, aume that the other factor remain the ame. a. he only change wa the number of turn, but it i a quared factor, reulting in () o (4)(0 μh) 80 μh b. n thi cae, the area i three time the original ize, and the number of turn i ½. Since the area i in the numerator, it increae the inductance by a factor of three. he drop in the number of turn reduce the inductance by a factor of (½) ¼. herefore, () (/4) (/4)(6 μh) μh Figure.9 nductor for example.. o c. Both μ and the number of turn have increaed, although the increae in the number of turn i quared. he increaed length reduce the inductance. herefore, () (00) (4. 0 )(0µH) 4.mH.5 o 6 ircuit nalyi apacitor Boyletad 9 Figure.0 nductor coil ymbol. nductance ype of nductor nductor, like capacitor and reitor, can be categorized under the general heading fixed or variable. he ymbol for a fixed air-core inductor i provided in Fig..0(a), for an inductor with a ferromagnetic core in Fig..0(b), for a tapped coil in Fig..0(c), and for a variable inductor in Fig..0(d). Practical quivalent nductor nductor, like capacitor, are not ideal. ociated with every inductor i a reitance determined by the reitance of the turn of wire and by core loe. Both element are included in the equivalent circuit in Fig..4. For mot application in thi text, the capacitance can be ignored, reulting in the equivalent model in Fig..5. nduced oltage v Faraday law of electromagnetic induction i one of the mot important in thi field becaue it enable u to etablih ac and dc voltage with generator. f we move a conductor through a magnetic field o that it cut magnetic line of flux a hown in Fig..8. f we go a tep further and move a coil of N turn through the magnetic field a hown in Fig..9, a voltage will be induced acro the coil a determined by Faraday law: dφ e N ( volt, ) dt Figure.5 Practical equivalent model for an inductor. Figure.4 omplete equivalent model for an inductor. Figure.8 Generating an induced voltage by Figure.9 Demontrating Faraday law. moving a conductor through a magnetic field. 6 ircuit nalyi Magnetim and nductor Boyletad

65 he polarity of the induced voltage acro the coil i uch that it oppoe the increaing level of current in the coil a hown in Fig..0. n other word, the changing current through the coil induce a voltage acro the coil that i oppoing the applied voltage that etablihe the increae in current to the firt place. he quicker the change in current through the coil, the greater the oppoing induced voltage to quelch the attempt of the current to increae in magnitude. hi effect i a reult of an important law referred to a enz law, which tate that dφ an induced effect i alway uch a N ( henrie, H ) di to oppoe the caue that produced it. f the inductance level i very mall, there will be almot no change in flux linking the coil, and the induced voltage acro the coil will be very mall. hat i dφ dφ di e N N dt di dt and dφ N ( henrie, H ) di di e dt ( volt, ) n network analyi, the voltage induced acro an inductor will alway have a polarity that oppoe the applied voltage. herefore, the following notation i ued for the induced voltage acro an inductor: the larger the inductance and/or the more rapid the change in current through a coil, the larger will the induced voltage acro the coil Figure.0 Demontrating the effect of enz law. di dt ( volt, ) be 6 ircuit nalyi apacitor Boyletad - ranient he Storage e Phae great number of imilaritie exit between the analye of inductive and capacitive network. he torage waveform have the ame hape, and time contant are defined for each configuration. he circuit in Fig.. i ued to decribe the charging phae of capacitor, with a imple replacement of the capacitor by an ideal inductor. t i important to remember that the energy i tored in the form of an electric field between the plate of a capacitor. For inductor, on the other hand, energy i tored in the form of magnetic field linking the coil. t the intant the witch i cloed, the choking action of the coil prevent an intantaneou change in current through the coil, reulting in i 0 a hown in Fig..(a). he abence of a current through the acro the reitor a determined by v i i (0 ) 0, a hown in Fig..(c). pplying K around the cloed loop reult in volt acro the coil at the intant the witch i cloed, a hown in Fig..(b). Figure. Baic - tranient network. Figure 6. ircuit i nalyi apacitor Boyletad, v, and v for the circuit in Fig.. following the cloing of the witch. he equation for the tranient repone of the current through an inductor if the following: t / τ i ( e ) ( ampere, ) with the time contant now defined by τ (econd, ) f we keep contant and increae a hown in Fig.. for increaing level of. he change in tranient repone i expected becaue the higher the inductance level, the greater the choking action on the changing current level, and longer it will take to reach-teady tate condition. he equation for the voltage acro the coil i the following: t / τ v e ( volt, ) and for the voltage acro reitor: t / τ v ( e ) ( volt, ) Figure. ffect of on the hape of the i torage waveform. Since the waveform are imilar to thoe obtained for capacitive network, we will aume that the torage phae paed and teady-tate condition have been etablihed once a period of time equal to five time contant ha occurred. 6 ircuit nalyi Magnetim and nductor Boyletad 4 n addition, ince τ / will alway have ome numerical value, even though it may be very mall at time, the tranient period of 5τ will alway have ome numerical value. herefore, the current cannot change intantaneouly in an inductive network. f we examine the condition that exit at the intant the witch i cloed, we find that the voltage acro the coil i volt, although the current i zero ampere a hown in Fig..4. n eence, therefore, the inductor take on the characteritic of an open circuit at the intant the witch i cloed. However, if we conider the condition that exit when teady-tate condition have been etablihed, we find that the voltage acro the coil i zero volt and the current i a maximum value of / ampere a hown in Fig..5. n eence, therefore, the inductor take on the characteritic of a hort circuit when teady-tate condition have been etablihed. 6 ircuit nalyi Magnetim and nductor Figure.4 ircuit in Figure. the intant the witch i cloed. Figure.5 Boyletad ircuit in Figure. under teady-tate condition. 4

66 x. - Find the mathematical expreion for the tranient behavior of i and v for the circuit in Fig..6 if the witch i cloed at t 0. Sketch the reulting curve. Firt, the time contant i determined : hen the maximum or teady tate τ Subtituting 4 H kω 50 kω i 5 m( e m m 5 0 into q.(.) : t/m ) current i 5 m Figure.7 i and v for the network in Fig..6. Uing q.(.5) : v 50e he reulting t/m Figure.6 - circuit for xample.. waveform appear in Fig..7. nitial ondition Since the current through a coil cannot change intantaneouly, the current through a coil begin the tranient phae at the initial value etablihed by the network (note Fig..8) before the witch wa cloed. t then pae through the tranient phae until it reache the teady-tate level after about five time contant. he teadytate level of the inductor current can be found by ubtituting it hort-circuit equivalent and finding the reulting current through the element. Uing the tranient equation developed in the previou dicuion, an equation for the current i can be written for the entire time interval in Fig..8; that i t / τ i + ( ) e f i f with ( f i ) repreenting the total change during the tranient phae. However, by multiplying through and rearranging term: t / τ t / τ i + e + e t /τ 6 ircuit nalyi Magnetim and nductor Boyletad 6 6 ircuit nalyi apacitor i f i Boyletad 7 we find i e f f f t / τ f i + ( + e i t / τ i )( e i ) Figure.8 Defining the three phae of a tranient waveform. - ranient he eleae Phae n the analyi of - circuit, we found that the capacitor could hold it charge and tore energy in the form of an electric field for a period of time determined by the leakage factor. n - circuit, the energy i tored in the form of a magnetic field etablihed by the current through the coil. Unlike the capacitor, however, an iolated inductor cannot continue to tore energy, becaue the abence of a cloed path caue the current to drop to zero, releaing the energy tored in the form of a magnetic field. f the erie - circuit in Fig..4 reache teady-tate condition and the witch i quickly opened, a park will occur acro the contact due to the rapid change in current di/dt of the equation v (di/dt) etablihe a high voltage v acro the coil that, in conjunction with the applied voltage, appear acro the point of the witch. hi i the ame mechanim ued in the ignition ytem of a car to ignite the fuel in the cylinder. Figure.4 Demontrating the effect of opening a witch in erie with an inductor with a teady-tate current. hevenin quivalent τ / h x. -6 For the network in Fig..46: a. Find the mathematical expreion for the tranient behavior of the current i and the voltage v after the cloing of the witch ( i 0 m). b. Draw the reultant waveform for each. a. pplying thevenin ' theorem to the 80 mh inductor ( Fig..47) yield 0 k Ω h 0 kω N Figure.47 Demontrating h for the network in Fig..46. Figure.46 xample.6. 6 ircuit nalyi Magnetim and nductor Boyletad 8 6 ir cuit nalyi apacitor Boyletad 9 5

67 pplying the voltage divider rule (Fig..48), we obtain Figure.48 Determining th for the network in Fig..46. h in Fig..49. ( + ) (4k Ω + 6kΩ)( ) k Ω + 4kΩ + 6kΩ (0k Ω )( ) 6 40 k Ω he hevenin equivalent circuit i hown Uing q. (.) t / τ ( e ) h i 80 0 H τ Ω m and h h h Ω i 0.6 m( e Uing q.(.5) : v e t/ 8 µ 8 µ ) 0.6 m b. See Fig..50. t/τ h Figure.49 he reulting hevenin equivalent o that t/ 8 µ v 6 e Figure.50 he reulting waveform for i and v for the network in Fig..46. circuit for the network in Fig ircuit nalyi Magnetim and nductor Boyletad 0 6 ircuit nalyi Magnetim and nductor Boyletad HW - For Fig..94: a. Determine the mathematical expreion for i and v following the cloing of the witch. b. Determine i and v after one time contant. Figure.94 Problem. Homework :, 4, 8, 0,, 4, 6 ircuit nalyi Magnetim and nductor Boyletad 6

68 /6 ircuit nalyi Magnetic ircuit OUNS ntroduction to Magnetic Field eluctance Ohm aw for the Magnetic ircuit Magnetizing Force and Flux Φ lectrical and elecommunication ngineering echnology Department Profeor Jang Prepared by textbook baed on ntroduction to ircuit nalyi by obert Boyletad, Prentice Hall, th edition. Hyterei mpere ircuital aw Serie Magnetic ircuit Key Word: Magnetic field, eluctance, Flux, Hyterei, mpere ircuital aw 6 ircuit nalyi Magnetic ircuit Boyletad ntroduction to Magnetic Field he magnetic field ditribution around a permanent magnet or electromagnet wa covered in previou module. he flux denity i defined by q... Φ B Wb/m tela(f) B Φ weber (Wb) m he preure on the ytem to etablih magnetic line of force i determined by the applied magnetomotive force which i directly related to the number of turn and current of the magnetizing coil a appearing in q... F ampere turn ( t) F N N turn ( t) ampere ( ) he level of magnetic flux etablihed in a ferromagnetic core i a direction function of the permeability of the material. Ferromagnetic material have a very high level of permeability while non-magnetic material uch a air and wood have very low level. he radio of the permeability of the material to that of air i called the relative permeability and i defined by q... µ 7 µ r µ o 4 π 0 Wb/ m µ 6 ircuit nalyi Magnetic ircuit Boyletad o eluctance he reitance of a material to the flow of charge i determined for electric circuit by the equation l ρ ( ohm, Ω) he reluctance of a material to the etting up of magnetic flux line in the material i determined by the following equation: l µ (rel, or t/wb) Where i the reluctance, l i the length of the magnetic path, and i the croectioned area. he t in the unit t/wb i the number of turn of the applied winding. Note that the reitance and reluctance are inverely proportional to the area, indicating that an increae in area reult in a reduction in each and an increae in the deired reult: current and flux. For an increae in length, the oppoite i true, and the deired effect i reduced. he reluctance, however, i inverely proportional to the permeability, while the reitance i directly proportional to the reitivity. 6 ircuit nalyi Magnetic ircuit Boyletad 4

69 Ohm aw for the Magnetic ircuit Figure.6 Magnetic flux line around a current-carrying conductor. n the S ytem of unit, magnetic flux i meaured in weber (WB). he applied ymbol i Φ. he number of flux line per unit area, called the flux denity, i denoted by B and i meaured in tela (). Figure.7 Flux ditribution of a ingle-turn coil. Figure.8 Flux ditribution of a current carrying coil. Figure.9 lectromagnet. Figure.0 Determining the direction of flux for an electromagnet: (a) method; (b) notation. B Φ tela B Wb / m tela ( ) Φ weber ( Wb) m Wb/m 6 ircuit nalyi Magnetic ircuit Boyletad 5 For the magnetic circuit, the effect deired i the flux Φ. he caue i the magnetomotive force (mmf) F, which i the external force (or preure ) required to et up the magnetic flux line within the magnetic material. he oppoition to the etting of the flux Φ i the reluctance. Subtituting, we have F Φ Since F N, above equation clearly reveal that an increae in the number of turn or the current through the wire in Fig.. reult in an increaed preure on the ytem to etablih the flux line through the core. Figure. Defining the component of a magnetomotive force. 6 ircuit nalyi Magnetic ircuit Boyletad 6 Subtituting for the magnetomotive force reult in 6 ircuit nalyi Magnetic ircuit Magnetizing Force he magnetomotive force per unit length i called the magnetizing force (H). n equation form, For the magnetic circuit in Fig.., if N 40 t and l 0. m, then N 40 t H 00 t / m l 0. m Note in Fig.. that the direction of the flux Φ can be determined by placing the finger of your right hand in the direction of the thumb. t i intereting to realize that the magnetizing force i independent of the type of core material it i determined olely by the number of turn, the current, and the length of the core. Boyletad F H l N H l ( t / m) ( t / m) Figure. Defining the magnetizing force of a magnetic circuit. he flux denity and the magnetizing force are related by the following equation : B µ H Hyterei curve of the flux denity B veru the magnetizing force H of a material i of particular important to the engineer. typical B-H curve force a ferromagnetic material uch a can be derived uing the etup in Fig..4. he core i initially unmagnetized, and the current 0. f the current i increaed to ome value above zero, the magnetizing force H increae to a value determined by N H l he flux Φ and the flux denity B (B Φ/) alo Figure.4 Serie magnetic circuit ued to increae with the current (or H). f the material ha define the hyterei curve. no reidual magnetim, and the magnetizing force H i increaed from zero to ome value H a, the B-H curve follow the path hown in Fig..5 between 0 to a. f the magnetizing force H i increaed until aturation (H ) occur, the curve continue a hown in the figure to point b. When aturation occur, the flux denity ha, for all practical purpoe, reached it maximum value. 7 6 ircuit nalyi Magnetic ircuit Boyletad Figure.5 Hyterei curve.

70 f the magnetizing force i reduced to zero by letting decreae to zero, the curve follow the path of the curve follow the path of the curve between b and c. he flux denity B, which remain when the magnetizing force i zero, i called the reidual flux denity. t i thi reidual flux denity that make it poible to create permanent magnet. f the current i reerved, developing a magnetizing force, H, the flux denity when H d i reached. he magnetizing force H d required to coerce the flux denity to reduce it level to zero i called the coercive force. the force H i increaed until aturation again occur and i then reerved and brought back to zero, the path def reult. f the magnetizing force i increaed in the poitive direction (+H), the curve trace the path hown from f to b. the entire curve repreented by bcdefb i called the hyterei curve. f the entire cycle i repeated, the curve obtained for the ame core will be determined by the maximum H applied. hree hyterei loop for the ame material for maximum value of H le than the aturation value are hown in Fig..6. n addition, the aturation curve i repeated for comparion purpoe. Figure.4 Defining the normal magnetization curve. 6 ircuit nalyi Magnetic ircuit Boyletad 9 mpere ircuital aw he imilarity between the analye of electric and magnetic circuit ha been demontrated to ome extent for the quantitie in able.. f we apply the caue analogy to K ( able. 0), we obtain the following: lectric Magnetic F 0 (for magnetic ircuit) ircuit ircuit aue F which, in word, tate that the algebraic um of the rie and drop of the mmf around a cloed ffect Φ loop of a magnetic circuit i equal to zero. Oppoition bove equation i referred a mpere circuital law. When it i applied to magnetic circuit, ource of mmf are expreed by the eulting in the following for magnetic circuit F Φ (t) equation F N (t) he equation for the mmf drop acro a portion of a magnetic circuit can be found by applying the relationhip lited in able.; that i, for electric circuit, Where Φ i the flux paing through a ection of the magnetic circuit and i the reluctance of that ection. more practical equation for the mmf drop i, F H l (t) where H i the magnetizing force on a ection of a magnetic circuit and l i the length of the ection. Flux Φ f we continue to apply the relationhip decribed in the previou module to K, we find that the um of the fluxe entering a junction i equal to the um of the fluxe leaving a junction; that i, for the circuit in Fig.., or Φ Φ both of which are equivalent. a b Φ + Φ b c + Φ Φ c ( at junction a) ( at junction b) Figure. Flux ditribution of a erie-parallel magnetic network. a Serie Magnetic ircuit Determining N n one type, Φ i given, and the impreed mmf N mut be computed. hi i the type of problem encountered in the deign of motor, generator, and tranformer. n the other type, N i given, and the flux Φ of the magnetic circuit mut be found. hi type of problem i encountered primarily in the deign of magnetic amplifier. hi ection conider only erie magnetic circuit in which the flux Φ i the ame throughout. 6 ircuit nalyi Magnetic ircuit Boyletad x. - For the erie magnetic circuit in Fig..: a. Find the value of required to develop a magnetic flux of Φ Wb. b. Determine µ and µ r for the material under thee condition. he magnetic circuit can be repreented by the ytem hown in Fig..(a). he electric circuit analogy i hown in Fig..(b). nalogie of thi type can be very helpful in the olution of magnetic circuit. able. i for part (a) of thi problem. a. he flux denity B i φ Wb B m Uing the B-H curve in Fig..8, we can determine the magnetizing force H: H (cat teel) 70 t/m pplying mpere circuital law yield N Hl Hl (70t / m)(0.6m) and 68m N 400t Figure. xample.. 6 ircuit nalyi Magnetic ircuit Boyletad

71 b. he permeabili ty of the material can be found B 0. µ.76 0 Wb / m H 70 t / m and the relative permeabili ty i µ.76 0 µ µ 4 π 0 Section o One continuou ection Figure. (a) magnetic circuit equivalent and (b) electric circuit analogy. Φ (Wb) able. (m ) B() 0 - H(t/m) l(m) 0.6 Hl(t) 6 ircuit nalyi Magnetic ircuit Boyletad ir Gap et conider the effect that an air gap ha on a magnetic circuit. Note the preence of air gap in the magnetic circuit of the motor and meter in Fig..5. he preading of the flux line outide the common area of the core for the air gap in Fig..8(a) i known a fringing. For our purpoe, we hall ignore thi effect and aume the flux ditribution to be a in Fig..8(b). Φ g B g g he flux denity of the air gap in Fig..8(b) i given by where, for our purpoe, Φ g Φ core and g core Figure.8 ir gap: (a) with fringing; (b) ideal. For the mot practical application, the permeability of air i taken to be equal to that of free pace. he magnetizing force of the air gap i then determined by Bg H g µ and the mmf drop acro the air gap i equal to H g g. n equation for H g i a follow: g g H g 7 µ o 4 π 0 5 and H ( ) B ( t / m) o g g 6 ircuit nalyi Magnetic ircuit Boyletad 4 B B x. -4 Find the value of required to etablih a magnetic flux of Φ Wb in the erie magnetic circuit in Fig..9. n equivalent magnetic circuit and it electric circuit analogy are hown in Fig..0. he flux denity for each ection i 4 Φ Wb B m From the B-H curve in Fig..8, H ( cat teel) 80 t / m Figure.9 elay for xample.4. pplying quation, H g ( ) Bg ( )(0.5 ).98 0 t / m he mmf drop are H (80t / m)(00 0 m) 8 t H core g g core 5 (.98 0 t / m)( 0 pplying mpere circuital law, N H + H core core (00) 84 t g and g m) 796 t Figure 6.0 ircuit (a) Magnetic nalyi Magnetim circuit equivalent and nductor and (b) electric circuit analogy Boyletad for the relay in Fig t t 4. HW -6 epeat Problem 5 for Φ 7,000 maxwell and an impreed mmf of 0 gilbert. 0 gilbert.67 0 rel ( GS) Φ 7,000 max well Homework :, 4, 6, ircuit nalyi Magnetic ircuit Boyletad 6 4