CIRCUITS 1 DEVELOP TOOLS FOR THE ANALYSIS AND DESIGN OF BASIC LINEAR ELECTRIC CIRCUITS

Transcription

1 CCUTS DEELOP TOOLS FO THE ANALYSS AND DESGN OF BASC LNEA ELECTC CCUTS

2 ESSTE CCUTS Here we introduce the basic concepts and laws that are fundamental to circuit analysis LEANNG GOALS OHM S LAW - DEFNES THE SMPLEST PASSE ELEMENT: THE ESSTO KCHHOFF S LAWS - THE FUNDAMENTAL CCUT CONSEATON LAWS- KCHHOFF CUENT (KCL) AND KCHHOFF OLTAGE (KL) LEAN TO ANALYZE THE SMPLEST CCUTS SNGLE LOOP - THE OLTAGE DDE SNGLE NODE-PA - THE CUENT DDE SEES/PAALLEL ESSTO COMBNATONS - A TECHNQUE TO EDUCE THE COMPLEXTY OF SOME CCUTS WYE - DELTA TANSFOMATON - A TECHNQUE TO EDUCE COMMON ESSTO CONNECTONS THAT AE NETHE SEES NO PAALLEL CCUTS WTH DEPENDENT SOUCES - (NOTHNG EY SPECAL)

3 ESSTOS i(t) v(t) A resistor is a passive element characterized by an algebraic relation between the voltage across its terminals and the current through it v( t) F( i( t)) General Model for a esistor A linear resistor obeys OHM s Law v( t) i( t) The constant,, is called the resistance of the component and is measured in units of Ohm () From a dimensional point of view Ohms is a derived unit of olt/amp Since the equation is algebraic the time dependence can be omitted Standard Multiples M k Mega Ohm(0 Kilo Ohm(0 of Ohm 6 ) ) olt A common occurrence is ma resulting in resistance in k Conductance f instead of expressing voltage as a function of current one expresses current in terms of voltage, OHM s law can be written i v We define G as Conductance of the component and write i Gv The unit of conductance is Siemens

4 Symbol Some practical resistors

5 v i Circuit epresent ation i Notice passive sign convention A touch of reality Two special resistor values i 0 v 0 Short Open Circuit Circuit 0 G G 0 Linear approximation Linear range Actual v- relationship v Ohm s Law is an approximation valid while voltages and currents remain in the Linear ange

6 v OHM S LAW POBLEM SOLNG TP i i Gv OHM's Law One equation and three variables. Given ANY two the third can be found Given current and resistance Find the voltage A 5 0[ ] Given Current and oltage Find esistance 0[ ] 4[ A] 5 Notice use of passive sign convention Given oltage and esistance Compute Current [ ] 4[ A] Determine direction of the current using passive sign convention Table Keeping Units Straight oltage Current esistance olts Amps Ohms olts ma k m A m m ma

7 GEN OLTAGE AND CONDUCTANCE EFEENCE DECTONS SATSFY PASSE SGN CONENTON UNTS? i( t) Gv( t) OHM S LAW CONDUCTANCE N SEMENS, OLTAGE N OLTS. HENCE CUENT N AMPEES i( t) 8[ A] OHM S LAW v( t) i( t) 4[ ] () i( t) i( t) [ A] 4 UNTS? v( t) i( t) OHM S LAW THE EXAMPLE COULD BE GEN LKE THS

8 ESSTOS AND ELECTC POWE esistors are passive components that can only absorb energy. Combining Ohm s law and the expressions for power we can derive several useful expressions P vi v i, or i Gv (Power) (Ohm's Law) Problem solving tip: There are four variables (P,v,i,) and two equations. Given any two variables one can find the other two. Given P, i P v, i v i Given i, v i, P vi i Given v, v v i, P vi Given P, P i, v i P f not given, the reference direction for voltage or current can be chosen and the other is given by the passive sign convention A MATTE OF UNTS Working with S units olt, Ampere Watt, Ohm, there is never a problem. One must be careful when using multiples or sub multiples. EXAMPLE : 40 k, i ma The basic strategy is to express all given variables in S units v (40*0 )*(*0 A) 80[ ] P i ( 40*0 )*(*0 A) 60*0 [ W ]

9 P? S S G S 0.50 [ A] 6 0 [ ] 500 [ S] P G [ A] P 0.50 [ W ] [ S] 5[ mw ]

10 P S 80[ mw ] S 5[ ] 4[ ma] P 800 [ W ] 40 A 5k

11 + - P 60W HALOGEN LAMP esistance of Lamp Current through Lamp Charge supplied by battery in min = / =.4 Ohms = P/ = 5A q current Q=5*60[C] SAMPLE POBLEM ecognizing the type of problem: Possibly useful relationships This is an application of Ohm s Law We are given Power and oltage. P We are asked for esistance, Current and Charge

12 ELECTC CCUT S AN NTECONNECTON OF ELECTCAL COMPONENTS TEMNALS COMPONENT a b characterized by the current through it and the voltage difference between terminals NODE NODE The concept of node is extremely important. We must learn to identify a node in any shape or form L TYPCAL LNEA CCUT vs + - v O C LOW DSTOTON POWE AMPLFE

13 CONENTON FO CUENTS T S ABSOLUTELY NECESSAY TO NDCATE THE DECTON OF MOEMENT OF CHAGED PATCLES. A POSTE ALUE FO THE CUENT NDCATES FLOW N THE DECTON OF THE AOW (THE EFEENCE DECTON) A NEGATE ALUE FO THE CUENT NDCATES FLOW N THE OPPOSTE DECTON THAN THE EFEENCE DECTON a a THE UNESALLY ACCEPTED CONENTON N ELECTCAL ENGNEENG S THAT CUENT S FLOW OF POSTE CHAGES. AND WE NDCATE THE DECTON OF FLOW FO POSTE CHAGES -THE EFEENCE DECTONa THE DOUBLE NDEX NOTATON F THE NTAL AND TEMNAL NODE AE LABELED ONE CAN NDCATE THEM AS SUBNDCES FO THE CUENT NAME 5A A b ab 5 A ab A A ba A b b POSTE CHAGES FLOW LEFT-GHT a a A ab A A ba A b b POSTE CHAGES FLOW GHT-LEFT ab ba

14 a A A b cb ab c 4A A This example illustrates the various ways in which the current notation can be used

15 THE + AND - SGNS DEFNE THE EFEENCE POLATY F THE NUMBE S POSTE PONT A HAS OLTS MOE THAN PONT B. F THE NUMBE S NEGATE PONT A HAS LESS THAN PONT B PONT A HAS MOE THAN PONT B PONT A HAS 5 LESS THAN PONT B

16 THE TWO-NDEX NOTATON FO OLTAGES NSTEAD OF SHOWNG THE EFEENCE POLATY WE AGEE THAT THE FST SUBNDEX DENOTES THE PONT WTH POSTE EFEENCE POLATY AB AB 5 BA 5 AB BA

17 ENEGY OLTAGE S A MEASUE OF ENEGY PE UNT CHAGE CHAGES MONG BETWEEN PONTS WTH DFFEENT OLTAGE ABSOB O ELEASE ENEGY THEY MAY TANSFE ENEGY FOM ONE PONT TO ANOTHE BASC FLASHLGHT Converts energy stored in battery to thermal energy in lamp filament which turns incandescent and glow EQUALENT CCUT The battery supplies energy to charges. Lamp absorbs energy from charges. The net effect is an energy transfer Charges gain energy here Charges supply Energy here

18 ENEGY OLTAGE S A MEASUE OF ENEGY PE UNT CHAGE CHAGES MONG BETWEEN PONTS WTH DFFEENT OLTAGE ABSOB O ELEASE ENEGY WHAT ENEGY S EQUED TO MOE 0[C] FOM PONT B TO PONT A N THE CCUT? THE CHAGES MOE TO A PONT WTH HGHE OLTAGE -THEY GANED (O ABSOBED) ENEGY THE CCUT SUPPLED ENEGY TO THE CHAGES AB W W Q 40J Q

19 THE OLTAGE DFFEENCE S 5 5 WHCH PONT HAS THE HGHE OLTAGE? AB 5

20 EXAMPLE A CAMCODE BATTEY PLATE CLAMS THAT THE UNT STOES 700mAHr AT 7.. WHAT S THE TOTAL CHAGE AND ENEGY STOED? CHAGE THE NOTATON 700mAHr NDCATES THAT THE UNT CAN DELE 700mA FO ONE FULL HOU Q C S 9.70 [ C] 600 s Hr Hr TOTAL ENEGY STOED THE CHAGES AE MOED THOUGH A 7. OLTAGE DFFEENTAL W Q[ C] J C 4 [ J] [ J] ENEGY AND POWE [C/s] PASS THOUGH THE ELEMENT EACH COULOMB OF CHAGE LOSES [J] O SUPPLES [J] OF ENEGY TO THE ELEMENT THE ELEMENT ECEES ENEGY AT A ATE OF 6[J/s] THE ELECTC POWE ECEED BY THE ELEMENT S 6[W] N GENEAL P w( t, t) HOW DO WE ECOGNZE F AN ELEMENT SUPPLES O ECEES POWE? t t p( x) dx

21 PASSE SGN CONENTON POWE ECEED S POSTE WHLE POWE SUPPLED S CONSDEED NEGATE a ab ab b P ab ab A CONSEQUENCE OF THS CONENTON S THAT THE EFEENCE DECTONS FO CUENT AND OLTAGE AE NOT NDEPENDENT -- F WE ASSUME PASSE ELEMENTS a ab F OLTAGE AND CUENT AE BOTH POSTE THE CHAGES MOE FOM HGH TO LOW OLTAGE AND THE COMPONENT ECEES ENEGY --T S A PASSE ELEMENT GEN THE EFEENCE POLATY EFEENCE DECTON FO CUENT b THS S THE EFEENCE FO POLATY a ab F THE EFEENCE DECTON FO CUENT S GEN EXAMPLE ab b a b ab ab 0 THE ELEMENT ECEES 0W OF POWE. WHAT S THE CUENT? SELECT EFEENCE DECTON BASED ON PASSE SGN CONENTON 0[ W ] ( 0 ) ab ab ab [ A] ab A

22 UNDESTANDNG PASSE SGN CONENTON We must examine the voltage across the component and the current through it S A A S P P S S AB A' B' AB A' B' B B oltage() Current A - A' S S positive positive supplies receives positive negative receives supplies negative positive receives supplies negative negative supplies receives ON S AB 0, 0 AB ON S A ' ' 0, ' ' B ON S A B A ' B' 0, A' B' 0 0

23 CHAGES ECEE ENEGY. THS BATTEY SUPPLES ENEGY CHAGES LOSE ENEGY. THS BATTEY ECEES THE ENEGY WHAT WOULD HAPPEN F THE CONNECTONS AE EESED N ONE OF THE BATTEES?

24 WHEN N DOUBT LABEL THE TEMNALS OF THE COMPONENT, 4A, A 4

25 COMPUTE POWE ABDOBED O SUPPLED BY EACH ELEMENT 4 A + - P (6 )(A) 6 A P (4 )( A) ( 4 )(A) 8 P = W P = 6W P = -48W P (8 )(A) MPOTANT: NOTCE THE POWE BALANCE N THE CCUT

26 CCUT ELEMENTS PASSE ELEMENTS OLTAGE DEPENDENT SOUCES UNTS FO, g, r,? NDEPENDENT SOUCES CUENT DEPENDENT SOUCES

27 EXECSES WTH DEPENDENT SOUCES FND O 40[ ] O FND O O 50mA

28 DETEMNE THE POWE SUPPLED BY THE DEPENDENT SOUCES 40 [ ] P ( 40[ ])( [ A]) 80[ W] P ( 0[ ])(4 4[ A]) 60[ W] TAKE OLTAGE POLATY EFEENCE TAKE CUENT EFEENCE DECTON

29 POWE ABSOBED O SUPPLED BY EACH ELEMENT P ( )(4A) 48[ W ] P (4 )(A) 48[ W ] P (8 )(A) 56[ W ] P DS P ( )( A) (4 )( A) 8[ W ] x 6 (6 )( 4A) 44[ W ] NOTCE THE POWE BALANCE

30 ESSTE CCUTS Here we introduce the basic concepts and laws that are fundamental to circuit analysis LEANNG GOALS OHM S LAW - DEFNES THE SMPLEST PASSE ELEMENT: THE ESSTO KCHHOFF S LAWS - THE FUNDAMENTAL CCUT CONSEATON LAWS- KCHHOFF CUENT (KCL) AND KCHHOFF OLTAGE (KL) LEAN TO ANALYZE THE SMPLEST CCUTS SNGLE LOOP - THE OLTAGE DDE SNGLE NODE-PA - THE CUENT DDE SEES/PAALLEL ESSTO COMBNATONS - A TECHNQUE TO EDUCE THE COMPLEXTY OF SOME CCUTS WYE - DELTA TANSFOMATON - A TECHNQUE TO EDUCE COMMON ESSTO CONNECTONS THAT AE NETHE SEES NO PAALLEL CCUTS WTH DEPENDENT SOUCES - (NOTHNG EY SPECAL)

31 ESSTOS i(t) v(t) A resistor is a passive element characterized by an algebraic relation between the voltage across its terminals and the current through it v( t) F( i( t)) General Model for a esistor A linear resistor obeys OHM s Law v( t) i( t) The constant,, is called the resistance of the component and is measured in units of Ohm () From a dimensional point of view Ohms is a derived unit of olt/amp Since the equation is algebraic the time dependence can be omitted Standard Multiples M k Mega Ohm(0 Kilo Ohm(0 of Ohm 6 ) ) olt A common occurrence is ma resulting in resistance in k Conductance f instead of expressing voltage as a function of current one expresses current in terms of voltage, OHM s law can be written i v We define G as Conductance of the component and write i Gv The unit of conductance is Siemens

32 Symbol Some practical resistors

33 v i Circuit epresent ation i Notice passive sign convention A touch of reality Two special resistor values i 0 v 0 Short Open Circuit Circuit 0 G G 0 Linear approximation Linear range Actual v- relationship v Ohm s Law is an approximation valid while voltages and currents remain in the Linear ange

34 v OHM S LAW POBLEM SOLNG TP i i Gv OHM's Law One equation and three variables. Given ANY two the third can be found Given current and resistance Find the voltage A 5 0[ ] Given Current and oltage Find esistance 0[ ] 4[ A] 5 Notice use of passive sign convention Given oltage and esistance Compute Current [ ] 4[ A] Determine direction of the current using passive sign convention Table Keeping Units Straight oltage Current esistance olts Amps Ohms olts ma k m A m m ma

35 GEN OLTAGE AND CONDUCTANCE EFEENCE DECTONS SATSFY PASSE SGN CONENTON UNTS? i( t) Gv( t) OHM S LAW CONDUCTANCE N SEMENS, OLTAGE N OLTS. HENCE CUENT N AMPEES i( t) 8[ A] OHM S LAW v( t) i( t) 4[ ] () i( t) i( t) [ A] 4 UNTS? v( t) i( t) OHM S LAW THE EXAMPLE COULD BE GEN LKE THS

36 ESSTOS AND ELECTC POWE esistors are passive components that can only absorb energy. Combining Ohm s law and the expressions for power we can derive several useful expressions P vi v i, or i Gv (Power) (Ohm's Law) Problem solving tip: There are four variables (P,v,i,) and two equations. Given any two variables one can find the other two. Given P, i P v, i v i Given i, v i, P vi i Given v, v v i, P vi Given P, P i, v i P f not given, the reference direction for voltage or current can be chosen and the other is given by the passive sign convention A MATTE OF UNTS Working with S units olt, Ampere Watt, Ohm, there is never a problem. One must be careful when using multiples or sub multiples. EXAMPLE : 40 k, i ma The basic strategy is to express all given variables in S units v (40*0 )*(*0 A) 80[ ] P i ( 40*0 )*(*0 A) 60*0 [ W ]

37 P? S S G S 0.50 [ A] 6 0 [ ] 500 [ S] P G [ A] P 0.50 [ W ] [ S] 5[ mw ]

38 P S 80[ mw ] S 5[ ] 4[ ma] P 800 [ W ] 40 A 5k

39 + - P 60W HALOGEN LAMP esistance of Lamp Current through Lamp Charge supplied by battery in min = / =.4 Ohms = P/ = 5A q current Q=5*60[C] SAMPLE POBLEM ecognizing the type of problem: Possibly useful relationships This is an application of Ohm s Law We are given Power and oltage. P We are asked for esistance, Current and Charge

40 KCHHOFF CUENT LAW ONE OF THE FUNDAMENTAL CONSEATON PNCPLES N ELECTCAL ENGNEENG CHAGE CANNOT BE CEATED NO DESTOYED

41 NODES, BANCHES, LOOPS A NODE CONNECTS SEEAL COMPONENTS. BUT T DOES NOT HOLD ANY CHAGE. TOTAL CUENT FLOWNG NTO THE NODE MUST BE EQUAL TO TOTAL CUENT OUT OF THE NODE (A CONSEATON OF CHAGE PNCPLE) NODE: point where two, or more, elements are joined (e.g., big node ) LOOP: A closed path that never goes twice over a node (e.g., the blue line) The red path is NOT a loop NODE BANCH: Component connected between two nodes (e.g., component 4)

42 KCHHOFF CUENT LAW (KCL) SUM OF CUENTS FLOWNG NTO A NODE S EQUAL TO SUM OF CUENTS FLOWNG OUT OF THE NODE 5A A current flowing is equivalent flowing out of 5A into a node to the negative the node ALGEBAC SUM OF CUENT (FLOWNG) OUT OF A NODE S ZEO ALGEBAC SUM OF CUENTS FLOWNG NTO A NODE S ZEO

43 A node is a point of connection of two or more circuit elements. t may be stretched out or compressed for visual purposes But it is still a node

44 A GENEALZED NODE S ANY PAT OF A CCUT WHEE THEE S NO ACCUMULATON OF CHAGE... O WE CAN MAKE SUPENODES BY AGGEGATNG NODES Leaving : i Leaving : i Adding & :i i 6 i i 4 i 0 i i i5 i6 i7 NTEPETATON: SUM OF CUENTS LEANG NODES & S ZEO SUALZATON: WE CAN ENCLOSE NODES & NSDE A SUFACE THAT S EWED AS A GENEALZED NODE (O SUPENODE) 0

45 POBLEM SOLNG HNT: KCL CAN BE USED TO FND A MSSNG CUENT c 5A d A b X a? SUM OF CUENTS NTO NODE S ZEO 5A ( A) 0 X X A Which way are charges flowing on branch a-b?...and PACTCE NOTATON CONENTON AT THE SAME TME... ab cb bd be A, A 4A? be NODES: a,b,c,d,e BANCHES: a-b,c-b,d-b,e-b c a -A A b 4A d be =? 4A[ ( A)] ( A) 0 e

46 Here we illustrate the use of a more general idea of node. The shaded surface encloses a section of the circuit and can be considered as a BG node SUM OF CUENTS LEANG BG NODE 0 40mA 0mA 0mA 60mA mA THE CUENT 5 BECOMES NTENAL TO THE NODE AND T S NOT NEEDED!!!

47 Find Find T 50mA T 0mA 40mA 0mA Find ma Find 0 and 4mA ma 0 0 ma 4 ma 0

48 Find i x 0i i x x i x 4mA 44mA 0 i x 0i 0mA ma 0 x = 4mA = 5mA = 6mA, = 8mA, 4 = 4mA

49 FND x x ma X X 0 4mA ma 0 ma ma EFCATON b ma X X 4mA ma b b x 4mA

50 KCHHOFF OLTAGE LAW ONE OF THE FUNDAMENTAL CONSEATON LAWS N ELECTCAL ENGNENG THS S A CONSEATON OF ENEGY PNCPLE ENEGY CANNOT BE CEATE NO DESTOYED

51 KCHHOFF OLTAGE LAW (KL) KL S A CONSEATON OF ENEGY PNCPLE A POSTE CHAGE GANS ENEGY AS T MOES TO A PONT WTH HGHE OLTAGE AND ELEASES ENEGY F T MOES TO A PONT WTH LOWE OLTAGE W q( ) B A B B A THOUGHT EXPEMENT W q AB q A AB CA B B W q CA BC C W q BC q A F THE CHAGE COMES BACK TO THE SAME NTAL PONT THE NET ENEGY GAN MUST BE ZEO (Conservative network) q q a c ab cd b d LOSES GANS W q ab W q cd OTHEWSE THE CHAGE COULD END UP WTH NFNTE ENEGY, O SUPPLY AN NFNTE AMOUNT OF ENEGY q( ) 0 AB BC CD KL: THE ALGEBAC SUM OF OLTAGE DOPS AOUND ANY LOOP MUST BE ZEO A B A OLTAGE A NEGATE A ( ) B SE S DOP

52 POBLEM SOLNG TP: KL S USEFUL TO DETEMNE A OLTAGE - FND A LOOP NCLUDNG THE UNKNOWN OLTAGE THE LOOP DOES NOT HAE TO BE PHYSCAL be S 0 8 EXAMPLE :, DETEMNE be THE OLTAGE AE KNOWN 0[ ] 0 be LOOP abcdefa

53 BACKGOUND: WHEN DSCUSSNG KCL WE SAW THAT NOT ALL POSSBLE KCL EQUATONS AE NDEPENDENT. WE SHALL SEE THAT THE SAME STUATON ASES WHEN USNG KL A SNEAK PEEW ON THE NUMBE OF LNEALY NDEPENDENT EQUATONS N THE CCUT DEFNE N B NUMBE OF NODES NUMBE OF BANCHES N B ( N ) LNEALY NDEPENDENT KCL EQUATONS LNEALY NDEPENDENT KL EQUATONS EXAMPLE: FO THE CCUT SHOWN WE HAE N = 6, B = 7. HENCE THEE AE ONLY TWO NDEPENDENT KL EQUATONS THE THD EQUATON S THE SUM OF THE OTHE TWO!!

54 0k 5k There are no loops with only one unknown!!! + - x x/ + The current through the 5k and 0k resistors is the same. Hence the voltage drop across the 5k is one half of the drop across the 0k!!! 5[ ] X X 0[ ] X 4 X x X 4 X 4 X 5[ 0 ]

55 SNGLE LOOP CCUTS BACKGOUND: USNG KL AND KCL WE CAN WTE ENOUGH EQUATONS TO ANALYZE ANY LNEA CCUT. WE NOW STAT THE STUDY OF SYSTEMATC, AND EFFCENT, WAYS OF USNG THE FUNDAMENTAL CCUT LAWS a b 6 branches 6 nodes loop c 4 WTE 5 KCL EQS O DETEMNE THE ONLY CUENT FLOWNG OLTAGE DSON: THE SMPLEST CASE KL ON THS LOOP f 6 e 5 d ALL ELEMENTS N SEES ONLY ONE CUENT THE PLAN BEGN WTH THE SMPLEST ONE LOOP CCUT EXTEND ESULTS TO MULTPLE SOUCE AND MULTPLE ESSTOS CCUTS MPOTANT OLTAGE DDE EQUATONS

56 SUMMAY OF BASC OLTAGE DDE v v( t) EXAMPLE : S, 90k, 0k 9 OLUME CONTOL? 5k

57 A PACTCAL POWE APPLCATON HOW CAN ONE EDUCE THE LOSSES?

58 THE CONCEPT OF EQUALENT CCUT THS CONCEPT WLL OFTEN BE USED TO SMPLFY THE ANALYSS OF CCUTS. WE NTODUCE T HEE WTH A EY SMPLE OLTAGE DDE THE DFFEENCE BETWEEN ELECTC CONNECTON AND PHYSCAL LAYOUT SOMETMES, FO PACTCAL CONSTUCTON EASONS, COMPONENTS THAT AE ELECTCALLY CONNECTED MAY BE PHYSCALLY QUTE APAT i i v S v S - i v S AS FA AS THE CUENT S CONCENED BOTH CCUTS AE EQUALENT. THE ONE ON THE GHT HAS ONLY ONE ESSTO SEES COMBNATON OF ESSTOS N ALL CASES THE ESSTOS AE CONNECTED N SEES

59 v v 5 v KL v FST GENEALZATON: MULTPLE SOUCES i(t) v v 4 v v v v4 v5 v Collect all sources on one side v v v v4 v5 v v v eq v v - + v v oltage sources in series can be algebraically added to form an equivalent source. We select the reference direction to move along the path. oltage drops are subtracted from rises 0 v eq + -

60 SECOND GENEALZATON: MULTPLE ESSTOS FND,, P(0k) bd APPLY KL TO THS LOOP APPLY KL TO THS LOOP LOOP FO bd bd 0[ k ] 0 ( KL) bd 0 POWE ON 0k ESSTO 4 P ( 0 A) (0*0 ) 0mW v i i i OLTAGE DSON FO MULTPLE ESSTOS

61 THE NESE OLTAGE DDE S OLTAGE DDE + - O S O "NESE" DDE S O COMPUTE S "NESE" DDE 0 0 S k 0

62 f ad =, find S NESE DDE POBLEM S 9 0

63 Notice use of passive sign convention 80 k * i( t) KL : 6 80 k * i( t) 40 k * i( t) 0 it () 40 k * i( t) 6 i( t) 0.05mA 0k Knowing the current one can compute ALL the remaining voltages and powers

64 EXAMPLE S Sometimes you may want to vary a bit b k S ab x + - APPLY KL TO THS LOOP O KL HEE a KL HEE X P x ab ( x) P ( x ) is supplied the power absorbed or by the dependent source KL : KL: KL: P 4 X X 0 ab 0 ab ( X ) X 4 X S OHMS' LAW : P X 0 X ab 0 (PASSE SGN CONENTON) 4 ma 4k ) [ ]*[ ma] mw ( X

65 SEES PAALLEL ESSTO COMBNATONS UP TO NOW WE HAE STUDED CCUTS THAT CAN BE ANALYZED WTH ONE APPLCATON OF KL(SNGLE LOOP) O KCL(SNGLE NODE-PA) WE HAE ALSO SEEN THAT N SOME STUATONS T S ADANTAGEOUS TO COMBNE ESSTOS TO SMPLFY THE ANALYSS OF A CCUT NOW WE EXAMNE SOME MOE COMPLEX CCUTS WHEE WE CAN SMPLFY THE ANALYSS USNG THE TECHNQUE OF COMBNNG ESSTOS PLUS THE USE OF OHM S LAW SEES COMBNATONS PAALLEL COMBNATON G G G... p G N

66 FST WE PACTCE COMBNNG ESSTOS 6k k k SEES (0K,K)SEES 6k k 4k 5k k k

67 EXAMPLES COMBNATON SEES-PAALLEL 9k f the drawing gets confusing edraw the reduced circuit and start again 8k 9k 6k ESSTOS AE N SEES F THEY CAY EXACTLY THE SAME CUENT 6k 6k 0k ESSTOS AE N PAALLEL F THEY AE CONNECTED EXACTLY BETWEEN THE SAME TWO NODES

68 EFFECT OF ESSTO TOLEANCE NOMNAL ESSTO ESSTO ALUE : TOLEANCE : 0%.7k ANGES FO CUENT AND POWE? NOMNAL CUENT : ma.7 NOMNAL POWE : _ 0 P 7. 04mW.7 MNMUM CUENT : MAXMUM CUENT : min max 0.67 ma mA MNMUM POWE( min ):.67mW MAXMUM POWE : 4.5mW THE ANGES FO CUENT AND POWE AE DETEMNED BY THE TOLEANCE BUT THE PECENTAGE OF CHANGE MAY BE DFFEENT FOM THE PECENTAGE OF TOLEANCE. THE ANGES MAY NOT EEN BE SYMMETC

69 CCUT WTH SEES-PAALLEL ESSTO COMBNATONS THE COMBNATON OF COMPONENTS CAN EDUCE THE COMPLEXTY OF A CCUT AND ENDE T SUTABLE FO ANALYSS USNG THE BASC TOOLS DEELOPED SO FA. COMBNNG ESSTOS N SEES ELMNATES ONE NODE FOM THE CCUT. COMBNNG ESSTOS N PAALLEL ELMNATES ONE LOOP FOM THE CCUT GENEAL STATEGY: EDUCE COMPLEXTY UNTL THE CCUT BECOMES SMPLE ENOUGH TO ANALYZE. USE DATA FOM SMPLFED CCUT TO COMPUTE DESED AABLES N OGNAL CCUT - HENCE ONE MUST KEEP TACK OF ANY ELATONSHP BETWEEN AABLES

70 k k k LEANNG BY DONG OLTAGE DDE : k O ( ) k k k k k CUENT DDE : k O (A) A k k

71 AN EXAMPLE OF BACKTACKNG ma xz 6.5mA ma.5ma O 6 0.5mA A STATEGY. ALWAYS ASK: WHAT ELSE CAN COMPUTE? b 6k * b k 4 4 a k * xz a xz b 4k 5 6k * 4k 5 O * xz

72 FND O STATEGY: FND USE OLTAGE DDE + - OLTAGE O 0k 0k DDE 0k 0k 40k 6 60k 0k 60k 0k FND 9 0k () 6 0k 0k S 0.5mA 6 60 k *0. ma 0.05mA THS S AN NESE POBLEM WHAT CAN BE COMPUTED? S 0k *0.5mA 6 6 0k SEES PAALLEL

73 Y TANSFOMA TONS THS CCUT HAS NO ESSTO N SEES O PAALLEL F NSTEAD OF THS WE COULD HAE THS THEN THE CCUT WOULD BECOME LKE THS AND BE AMENABLE TO SEES PAALLEL TANSFOMATONS

74 Y b a ab ) ( ab ) ( b a ) ( c b ) ( a c SUBTACT THE FST TWO THEN ADD TO THE THD TO GET a Y c b a a b b a c b c b EPLACE N THE THD AND SOLE FO Y a a c c b b a c a c c b b a b a c c b b a Y

75 LEANNG EXAMPLE: APPLCATON OF WYE-DELTA TANSFOMATON COMPUTE S c DELTA CONNECTON c k 6k k 6k 8k a b a b c Y a k 9k (k 6k) k EQ 6k 0 S. ma k b ONE COULD ALSO USE A WYE - DELTA TANSFOMATON...

76 LEANNG EXAMPLE SHOULD KEEP THESE TWO NODES! CONET THS Y NTO A DELTA? F WE CONET TO Y NTO A DELTA THEE AE SEES PAALLEL EDUCTONS! Y a a c c b b a c a c c b b a b a c c b b a * * 6 k k k k 4mA 6k 6k 6k k k O THE ESULTNG CCUT S A CUENT DDE

77 6 k k 9k CCUT AFTE PAALLEL ESSTO EDUCTON 4mA 6k O 9k O 6k 8 4mA ma O 6k 8k 8 9k 9k ma 4 NOTCE THAT BY KEEPNG O O THE FACTON WE PESEE FULL NUMECAL ACCUACY WYE DELTA

78 NODE ANALYSS One of the systematic ways to determine every voltage and current in a circuit The variables used to describe the circuit will be Node oltages -- The voltages of each node with respect to a pre-selected reference node

79 T S NSTUCTE TO STAT THE PESENTATON WTH A ECAP OF A POBLEM SOLED BEFOE USNG SEES/ PAALLEL ESSTO COMBNATONS COMPUTE ALL THE OLTAGES AND CUENTS N THS CCUT

80 4k k k SECOND: BACKTACK USNG KL, KCL OHM S 6k 6k 6k FST EDUCE TO A SNGLE LOOP CCUT a OHM' S: KCL: 0 6k OHM' S: b k * 4 4 4k * KCL: 5 OHM'S: 4 C 0 k * OTHE OPTONS... 5 b 4 k a () 9

81 THE NODE ANALYSS PESPECTE S S S 0 a KL a a a KL a b 0 b b 5 KL c 5 b 0 c EFEENCE WHAT S THE PATTEN??? THEE AE FE NODES. F ONE NODE S SELECTED AS EFEENCE THEN THEE AE FOU OLTAGES WTH ESPECT TO THE EFEENCE NODE 5 b c ONCE THE OLTAGES AE KNOWN THE CUENTS CAN BE COMPUTED USNG OHM S LAW THEOEM: F ALL NODE OLTAGES WTH ESPECT TO A COMMON EFEENCE NODE AE KNOWN THEN ONE CAN DETEMNE ANY OTHE ELECTCAL AABLE FO THE CCUT v v m v N v DLL QUESTON ca A GENEAL EW

82 THE EFEENCE DECTON FO CUENTS S ELEANT v ' v i' USNG THE LEFT-GHT EFEENCE DECTON THE OLTAGE DOP ACOSS THE ESSTO MUST HAE THE POLATY SHOWN vm vn ' i i OHM'S LAW i PASSE SGN CONENTON ULES! F THE CUENT EFEENCE DECTON S EESED... THE PASSE SGN CONENTON WLL ASSGN THE EESE EFEENCE POLATY TO THE OLTAGE ACOSS THE ESSTO ' vn vm OHM'S LAW i

83 DEFNNG THE EFEENCE NODE S TAL 4 THE STATEMENT 4 UNTL THE EFEENCE PONT S DEFNED BY CONENTON THE GOUND SYMBOL SPECFES THE EFEENCE PONT. S MEANNGLES S ALL NODE OLTAGES AE MEASUED WTH ESPECT TO THAT EFEENCE PONT?

84 THE STATEGY FO NODE ANALYSS S a b c. DENTFY ALL NODES AND SELECT A EFEENCE NODE. DENTFY KNOWN NODE a : a s a a b 9k 6k b : 4 5 b a b b c k 4k c : 6 c b c 0 9k k EFEENCE. AT EACH NODE WTH UNKNOWN OLTAGE WTE A KCL EQUATON (e.g.,sum OF CUENT LEANG =0) 4. EPLACE CUENTS N TEMS OF NODE OLTAGES AND GET ALGEBAC EQUATONS N THE NODE OLTAGES... SHOTCUT: SKP WTNG THESE EQUATONS... AND PACTCE WTNG THESE DECTLY

85 WHEN WTNG A NODE EQUATON... AT EACH NODE ONE CAN CHOSE ABTAY DECTONS FO THE CUENTS b a b c a c a b a b c ' d d ' ' c d d AND SELECT ANY FOM OF KCL. WHEN THE CUENTS AE EPLACED N TEMS OF THE NODE OLTAGES THE NODE EQUATONS THAT ESULT AE THE SAME O EQUALENT CUENTS LEANG 0 a b b d b c 0 0 CUENTS NTO NODE 0 a b b d b c 0 0 CUENTS LEANG 0 ' ' ' b a b d c b 0 0 CUENTS NTO ' ' ' b a b d c b 0 0 NODE 0 WHEN WTNG THE NODE EQUATONS WTE THE EQUATON DECTLY N TEMS OF THE NODE OLTAGES. BY DEFAULT USE KCL N THE FOM SUM-OF-CUENTS-LEANG = 0 THE EFEENCE DECTON FO THE CUENTS DOES NOT AFFECT THE NODE EQUATON

86 CCUTS WTH ONLY NDEPENDENT SOUCES HNT: THE FOMAL MANPULATON OF EQUATONS MAY BE SMPLE F ONE USES CONDUCTANCES NSTEAD OF NODE USNG ESSTANCE S i A v v v 0 WTH CONDUCTANCES EODENG TEMS i A Gv G( v v ) NODE EODENG TEMS THE MODEL FO THE CCUT S A SYSTEM OF ALGEBAC EQUATONS THE MANPULATON OF SYSTEMS OF ALGEBAC EQUATONS CAN BE EFFCENTLY DONE USNG MATX ANALYSS

87 EXAMPLE WTE THE KCL NODE WE SUALZE THE CUENTS LEANG AND WTE THE KCL EQUATON EPEAT THE POCESS AT NODE i v v v v 4 0 O SUALZE CUENTS GONG NTO NODE

88 ANOTHE EXAMPLE OF WTNG NODE EQUATONS B B 5mA A MAK THE NODES (TO NSUE THAT NONE S MSSNG) A 8k k 8k k C SELECT AS EFEENCE k 8k 8k B k WTE KCL AT EACH NODE N TEMS OF A A NODE A 5mA B B 5mA 0

89 A MODEL S SOLED BY MANPULATON OF EQUATONS AND USNG MATX ANALYSS NUMECAL MODEL LEANNG EXAMPLE USE GAUSSAN ELMNATON THE NODE EQUATONS i A ma, i B k, 4mA 6k THE MODEL EPLACE ALUES AND SWTCH NOTATON TO UPPE CASE ALTENATE MANPULATON ADD EQS * /k * / 6k GHT HAND SDE S OLTS. COEFFS AE NUMBES 4 */ (and add equations) [ ] 4 60[ ]

90 SOLUTON USNG MATX ALGEBA PLACE N MATX FOM AND DO THE MATX ALGEBA... USE MATX ANALYSS TO SHOW SOLUTON PEFOM THE MATX MANPULATONS A Adj ( A) A FO THE ADJONT EPLACE EACH ELEMENT BY TS COFACTO SAMPLE 8k 0 k 40 6k

91 AN EXAMPLE OF NODE ANALYSS v COULD WTE EQUATONS BY v CONDUCTANCES CONNECTED TO NODE CONDUCTANCES BETWEEN & CONDUCTANCES BETWEEN & CONDUCTANCES BETWEEN &

92 WTNG EQUATONS BY NSPECTON FO CCUTS WTH ONLY NDEPENDENT SOUCES THE MATX S ALWAYS SYMMETC THE DAGONAL ELEMENTS AE POSTE THE OFF-DAGONAL ELEMENTS AE NEGATE Conductances connected to node Conductances between and Conductances between and Conductances between and ALD ONLY FO CCUTS WTHOUT DEPENDENT SOUCES

93 k k ma 6 4 USNG KCL 0 6 k k ma BY NSPECTON ma k k k 4 6 ma k k k 6 LEANNG EXTENSON

94 LEANNG EXTENSON 6mA Node analysis : ma 6mA 0 : 6mA 0 6k N MOST CASES THEE AE SEEAL DFFEENT WAYS OF SOLNG A POBLEM 6 8mA k (6 ma) ma k 6k 6k (6 ma) 4mA k 6k NODE EQS. BY NSPECTON 0 6mA k 0 ma 6k k 6 Once node voltages are known k 6k k CUENTS COULD BE COMPUTED DECTLY USNG KCL AND CUENT DDE!!

95 CCUTS WTH DEPENDENT SOUCES CANNOT BE MODELED BY NSPECTON. THE SYMMETY S LOST. A POCEDUE FO MODELNG WTE THE NODE EQUATONS USNG DEPENDENT SOUCES AS EGULA SOUCES. FO EACH DEPENDENT SOUCE WE ADD ONE EQUATON EXPESSNG THE CONTOLLNG AABLE N TEMS OF THE NODE OLTAGES 0 v v v i o 0 v v v i A MODEL FO CONTOLLNG AABLE v i o NUMECAL EXAMPLE ma v k k v k v k k v k k v v EPLACE AND EAANGE i A v v 4k * / * / 6k ] [ 0 ADDNG THE EQUATONS ] [ CCUTS WTH DEPENDENT SOUCES LEANNG EXAMPLE

96 LEANNG EXAMPLE: CCUT WTH OLTAGE-CONTOLLED CUENT EPLACE AND EAANGE WTE NODE EQUATONS. TEAT DEPENDENT SOUCE AS EGULA SOUCE CONTNUE WTH GAUSSAN ELMNATON... O USE MATX ALGEBA EXPESS CONTOLLNG AABLE N TEMS OF NODE OLTAGES FOU EQUATONS N OU UNKNOWNS. SOLE USNG FAOTE TECHNQUE

97 USNG MATLAB TO SOLE THE NODE EQUATONS 4 k, 4k, i [ A/ ] A ma, i k, B 4mA, DEFNE THE COMPONENTS OF THE CCUT DEFNE THE MATX G Entries in a row are separated by commas (or plain spaces). ows are separated by semi colon» =000;=000;=000; 4=4000; %resistances in Ohm» ia=0.00;ib=0.004; %sources in Amps» alpha=; %gain of dependent source» G=[(/+/), -/, 0; %first row of the matrix -/, (/+alpha+/), -(alpha+/); %second row 0, -/, (/+/4)], %third row. End in comma to have the echo G = DEFNE GHT HAND SDE ECTO » =[ia;-ia;ib]; %end in ";" to skip echo SOLE LNEA EQUATON» =G\ % end with carriage return and get the echo =

98 LEANNG EXTENSON: FND NODE OLTAGES EAANGE AND MULTPLY BY 0k 40[ ] */ and add eqs NODE : 4mA 0k : O 0k 0k CONTOLLNG AABLE (N TEMS ON NODE OLTAGES) O 0 k EPLACE 4mA 0k 0k 0k 0k 0k 0 0

99 FND THE OLTAGE O LEANNG EXTENSON NODE EQUATONS x x ma 0 k 6k * / 6k x O O 0 * /k 6k k k x O [ ] x x 0 NOTCE EPLACEMENT OF DEPENDENT SOUCE N TEMS OF NODE OLTAGE 4[ ] O 4[ ]

100 CCUTS WTH NDEPENDENT OLTAGE SOUCES nodes plus the reference. n principle one needs equations... but two nodes are connected to the reference through voltage sources. Hence those node voltages are known!!! Hint: Each voltage source connected to the reference node saves one node equation One more example. Only one KCL is necessary 6k k [ ] k 4 6[ ] SOLNG ( ) 6[ ] ( 0 THESE AE THE EMANNG TWO NODE EQUATONS THE EQUATONS ).5[ ] 0

101 Problem.67 (6th Ed) Find _0 S 4 4 O + - S S S DENTFY AND LABEL ALL NODES WTE THE NODE EQUATONS NOW WE LOOK WHAT S BENG ASKED TO DECDE THE SOLUTON STATEGY. 0 ONLY AE NEEDED FO, O = k; = k, = k, 4 = k s =ma, s = 4mA, s = 4mA, s = NODE OLTAGE : S [ : S [ : S 4[ 4 : S k k S [ ma] 4[ ma] k k k k

102 [ ma] k [ ma] k 0 k k k 4 4 [ ma] 4[ ma] k TO SOLE BY HAND ELMNATE DENOMNATOS 4 0 */k 0 */k */k 4[ ] 4 5 [ ] 4 4[ ] Add + 4 6[ ] () () () ALTENATE: USE LNEA ALGEBA [ ] 4 40[ ] 0[ ] 4 56[ ] 4[ ] FNALLY!! 0 4[ ] * / and add So. What happens when sources are connected between two non reference nodes?

103 THE SUPENODE TECHNQUE We will use this example to introduce the concept of a SUPENODE SUPENODE S Conventional node analysis requires all currents at a node 6mA 4 ma 0 S S k eqs, unknowns...panic!! The current through the source is not related to the voltage of the source Math solution: add one equation 6[ ] Efficient solution: enclose the source, and all elements in parallel, inside a surface. Apply KCL to the surface!!! 4mA 0 6k k 6 ma The source current is interior to the surface and is not required We STLL need one more equation 6[ ] Only eqs in two unknowns!!!

104 ALGEBAC DETALS The Equations () () Solution 6mA 4mA 6k k 6[ ] 0 * /k. Eliminate denominators in Eq(). Multiply by... 4[ ] 6[ ]. Add equations toeliminate 0[ ] 0[ ]. Use Eq() to compute 6[ ] 4[ ]

105 s S FND THE NODE OLTAGES AND THE POWE SUPPLED BY THE OLTAGE SOUCE s S 0k, 0[ ], s 4k 0[ ma], s 6[ ma] 0 0[ ] 0mA 0 0k 0k */0k 00[ ] adding: 60[ ] 00 40[ ] TO COMPUTE THE POWE SUPPLED BY OLTAGE SOUCE WE MUST KNOW THE CUENT THOUGH T 6mA 8mA 0k 0 k P 0[ ] 8[ ma] 60mW BASED ON PASSE SGN CONENTON THE POWE S ECEED BY THE SOUCE!!

106 LEANNG EXAMPLE WTE THE SUPENODE CONSTANT : v v v A KCL (leaving supernode) : THEE EQUATONS N THEE UNKNOWNS

107 LEANNG EXAMPLE FND O SUPENODE, 6 4 SUPENODE CONSTANT KNOWN NODE OLTAGES SUPENODE

108 LEANNG EXTENSON SUPENODE SOUCES CONNECTED TO THE EFEENCE CONSTANT EQUATON SUPENODE k S NOT NEEDED FO O 6 ( 4) 0 * / k k k k * / and add 5 8 OHM'S LAW O. 8mA k

109 WTE THE NODE EQUATONS supernode dentify all nodes, select a reference and label nodes Nodes connected to reference through a voltage source oltage sources in between nodes and possible supernodes EQUATON BOOKKEEPNG: KCL@ _, KCL@ supernode, constraints equations and one known node Supernodes can be more complex 4 KCL@_ (Careful not to omit any current) CONSTANTS DUE TO OLTAGE SOUCES S 5 S 5 4 S 5 EQUATONS N FE UNKNOWNS. 4 0

110 CCUTS WTH DEPENDENT SOUCES PESENT NO SGNFCANT ADDTONAL COMPLEXTY. THE DEPENDENT SOUCES AE TEATED AS EGULA SOUCES WE MUST ADD ONE EQUATON FO EACH CONTOLLNG AABLE

111 LEANNG EXAMPLE FND O OLTAGE SOUCE CONNECTED TO EFEENCE KCL@ 0 6 x k k : CONTOLLNG AABLE N TEMS OF NODE OLTAGES x 6 k 0 k 6k 6k * / 6k 0 6 O ma k EPLACE

112 SUPE NODE WTH DEPENDENT SOUCE OLTAGE SOUCE CONNECTED TO EFEENCE 6 SUPENODE CONSTANT KCL AT SUPENODE CONTOLLNG AABLE N TEMS OF NODES x x * /k ( 6)

113 CUENT CONTOLLED OLTAGE SOUCE CONSTANT DUE TO SOUCE k x CONTOLLNG AABLE N TEMS OF NODES k x x k 4mA ma 0 k k 4[ ]* / and add 0 8[ ] KCL AT SUPENODE 4 O ma k

114 An example with dependent sources k k X S k x 000 a x 6k a has units of [olt/amp] DENTFY AND LABEL NODES nodes are connected to the reference through voltage sources EXPESS CONTOLLNG AABLE N TEMS OF NODE OLTAGES X S 000a x X k S X X k X k X v k 0 ( X EPLACE x N (8 a) X S k * a k ) What happens when a=8? S X X EPLACE N KCL ( a X X a X / ) 0

115 LOOP ANALYSS The second systematic technique to determine all currents and voltages in a circuit T S DUAL TO NODE ANALYSS - T FST DETEMNES ALL CUENTS N A CCUT AND THEN T USES OHM S LAW TO COMPUTE NECESSAY OLTAGES THEE AE STUATON WHEE NODE ANALYSS S NOT AN EFFCENT TECHNQUE AND WHEE THE NUMBE OF EQUATONS EQUED BY THS NEW METHOD S SGNFCANTLY SMALLE

116 Apply node analysis to this circuit There are 4 non reference nodes There is one super node There is one node connected to the reference through a voltage source We need three equations to compute all node voltages BUT THEE S ONLY ONE CUENT FLOWNG THOUGH ALL COMPONENTS AND F THAT CUENT S DETEMNED ALL OLTAGES CAN BE COMPUTED WTH OHM S LAW STATEGY:. Apply KL (sum of voltage drops =0). Use Ohm s Law to express voltages in terms of the loop current. [ ] 8[ ] [ ] 8[ ] ESULT S ONE EQUATON N THE LOOP CUENT!!! SHOTCUT 4 Skip this equation 0 Write this one directly 0

117 LOOPS, MESHES AND LOOP CUENTS a b 7 c f 6 e 5 d A BASC CCUT EACH COMPONENT S CHAACTEZED BY TS OLTAGE ACOSS AND TS CUENT THOUGH A LOOP S A CLOSED PATH THAT DOES NOT GO TWCE OE ANY NODE. THS CCUT HAS THEE LOOPS fabef ebcde 4 fabcdef A MESH S A LOOP THAT DOES NOT ENCLOSE ANY OTHE LOOP. fabef, ebcde AE MESHES A LOOP CUENT S A (FCTCOUS) CUENT THAT S ASSUMED TO FLOW AOUND A LOOP,, AE LOOP CUENTS A MESH CUENT S A LOOP CUENT ASSOCATED TO A MESH., AE MESH CUENTS CLAM: N A CCUT, THE CUENT THOUGH ANY COMPONENT CAN BE EXPESSED N TEMS OF THE LOOP CUENTS EXAMPLES FACT: NOT EEY LOOP CUENT S EQUED TO COMPUTE ALL THE CUENTS THOUGH COMPONENTS a a f be bc b 7 c f 6 e 5 d A BASC CCUT THE DECTON OF THE LOOP CUENTS S SGNFCANT 4 USNG LOOP a f b e b c TWO CUENTS FO EEY CCUT THEE S A MNMUM NUMBE OF LOOP CUENTS THAT AE NECESSAY TO COMPUTE EEY CUENT N THE CCUT. SUCH A COLLECTON S CALLED A MNMAL SET (OF LOOP CUENTS).

118 FO A GEN CCUT LET B NUMBE OF BANCHES N NUMBE OF NODES THE MNMUM EQUED NUMBE OF LOOP CUENTS S L B ( N ) MESH CUENTS AE ALWAYS NDEPENDENT AN EXAMPLE DETEMNATON OF LOOP CUENTS KL ON LEFT MESH KL ON GHT MESH v v v v 0 S 4 5 USNG OHM S LAW v i, v i, v ( i i ) v i, v i EPLACNG AND EAANGNG B 7 N 6 L 7 (6 ) TWO LOOP CUENTS AE EQUED. THE CUENTS SHOWN AE MESH CUENTS. HENCE THEY AE NDEPENDENT AND FOM A MNMAL SET N MATX FOM i v S i v 4 5 S THESE AE LOOP EQUATONS FO THE CCUT

119 WTE THE MESH EQUATONS v BOOKKEEPNG BANCHES = 8 NODES = 7 LOOP CUENTS NEEDED = i AND WE AE TOLD TO USE MESH CUENTS! THS DEFNES THE LOOP CUENTS TO BE USED DENTFY ALL OLTAGE DOPS v i v ( i i v 4 ) v 5 i 4 i 5 WTE KL ON EACH MESH TOP MESH: v BOTTOM: v USE OHM S LAW S v vs v v5 v4 vs v 0 0

120 DEELOPNG A SHOTCUT WTE THE MESH EQUATONS WHENEE AN ELEMENT HAS MOE THAN ONE LOOP CUENT FLOWNG THOUGH T WE COMPUTE NET CUENT N THE DECTON OF TAEL 5 4 DAW THE MESH CUENTS. OENTATON CAN BE ABTAY. BUT BY CONENTON THEY AE DEFNED CLOCKWSE NOW WTE KL FO EACH MESH AND APPLY OHM S LAW TO EEY ESSTO. AT EACH LOOP FOLLOW THE PASSE SGN CONENTON USNG LOOP CUENT EFEENCE DECTON ( ) ( ) 0

121 LEANNG EXAMPLE: FND o USNG LOOP ANALYS AN ALTENATE SELECTON OF LOOP CUENTS SHOTCUT: POLATES AE NOT NEEDED. APPLY OHM S LAW TO EACH ELEMENT AS KL S BENG WTTEN EAANGE k 6k 6k 9k */ k mA k 6k and add 5 4 ma EXPESS AABLE OF NTEEST AS FUNCTON OF LOOP CUENTS O NOW EAANGE O THS SELECTON S MOE EFFCENT k 6k */ 6k 9k 9 */ and substract 4k 8 ma 4

122 F THE CCUT CONTANS ONLY NDEPENDENT SOUCE THE MESH EQUATONS CAN BE WTTEN BY NSPECTON A PACTCE EXAMPLE MUST HAE ALL MESH CUENTS WTH THE SAME OENTATON N LOOP K THE COEFFCENT OF k S THE SUM OF ESSTANCES AOUND THE LOOP. LOOP coefficient of coefficien t of coefficient of 4k 6k 0 6k HS 6[ ] THE GHT HAND SDE S THE ALGEBAC SUM OF OLTAGE SOUCES AOUND THE LOOP (OLTAGE SES - OLTAGE DOPS) THE COEFFCENT OF j S THE SUM OF ESSTANCES COMMON TO BOTH k AND j AND WTH A NEGATE SGN. LOOP k 6k LOOP k 9k 6 Loop LOOP coefficient of 0 coefficient of 9k k coefficient of k HS 6[ ] (6 k) ( k) (k 6k k) 0

123 LEANNG EXTENSON. DAW THE MESH CUENTS. WTE MESH EQUATONS MESH k 4k k) k [ ] ( k (k 6k) (6 MESH ) DDE BY k. GET NUMBES FO COEFFCENTS ON THE LEFT AND ma ON THE HS. SOLE EQUATONS 8 [ ma] 8 9[ ma] * / 4 0 [ ma] O and add 6k [ ] 5

124 WTE THE MESH EQUATONS k 4k 4k 4 k. DAW MESH CUENTS 6k 9 BOOKKEEPNG: B = 7, N = 4. WTE MESH EQUATONS. USE KL MESH : k 6k( ) 0 MESH : 4k( 4) 4k( ) MESH : 9 6k( ) 4k( ) MESH 4: 4k( ) k CHOOSE YOU FAOTE TECHNQUE TO SOLE THE SYSTEM OF EQUATONS EQUATONS BY NSPECTON 8k 6k 8k 4k 4k4 6k 4k 0k 9 k 6k 9 4 4

125 CCUTS WTH NDEPENDENT CUENT SOUCES KL THEE S NO ELATONSHP BETWEEN AND THE SOUCE CUENT! HOWEE... MESH CUENT S CONSTANED MESH EQUATON MESH ma BY NSPECTON k 8k k (ma) 9 ma O 6k [ 8k 4 ] CUENT SOUCES THAT AE NOT SHAED BY OTHE MESHES (O LOOPS) SEE TO DEFNE A MESH (LOOP) CUENT AND EDUCE THE NUMBE OF EQUED EQUATONS TO OBTAN APPLY KL TO ANY CLOSED PATH THAT NCLUDES

126 LEANNG EXAMPLE COMPUTE O USNG MESH ANALYSS KL FO o TWO MESH CUENTS AE DEFNED BY CUENT SOUCES ma ma MESH 4 USE KL TO COMPUTE o BY NSPECTON k(4ma) 4k( ma) k k 4k k 4 ma

127 LEANNG EXTENSONS WE ACTUALLY NEED THE CUENT ON THE GHT MESH. HENCE, USE MESH ANALYSS MESH : 4mA MESH : [ ] 4k( ) 6k mA 44mA ma O 6k [ ] 5 MESH : 4mA MESH : k k 0 4 O 6 4 ma 6k 8[ ]

128 Problem.46 (6th Ed). Write loop equations. S S + - Determine O k 4k k S = ma, S = 6. Select loop currents. 6k SELECTNG THE SOLUTON METHOD + O non-reference nodes. meshes One current source, one super node BOTH APPOACHES SEEM COMPAABLE. CHOOSE LOOP ANALYSS n this case we use meshes. We note that the current source could define one mesh. _ Loop S Loop S 4k( ) k( ) 0 Loop k( ) 6k k( ) 0 4 Since we need to compute o it is efficient to solve for only. HNT: Divide the loop equations by k. Coefficients become numbers and voltage source becomes ma. Loop Loop We use the fact that = s S 4 (6 4)[ ma] * / k */ and add eqs 6 4 S 4mA O 6k 7 ma

129 CUENT SOUCES SHAED BY LOOPS - THE SUPEMESH APPOACH. WTE CONSTANT EQUATON DUE TO MESH CUENTS SHANG CUENT SOUCES 4mA. WTE EQUATONS FO THE OTHE MESHES ma 4. DEFNE A SUPEMESH BY (MENTALLY) EMONG THE SHAED CUENT SOUCE. SELECT MESH CUENTS 5. WTE KL FO THE SUPEMESH 6 k k k( ) k ( ) 0 SUPEMESH NOW WE HAE THEE EQUATONS N THEE UNKNOWNS. THE MODEL S COMPLETE

130 CUENT SOUCES SHAED BY MESHES - THE GENEAL LOOP APPOACH THE STATEGY S TO DEFNE LOOP CUENTS THAT DO NOT SHAE CUENT SOUCES - EEN F T MEANS ABANDONNG MESHES FO CONENENCE STAT USNG MESH CUENTS UNTL EACHNG A SHAED SOUCE. AT THAT PONT DEFNE A NEW LOOP. N ODE TO GUAANTEE THAT F GES AN NDEPENDENT EQUATON ONE MUST MAKE SUE THAT THE LOOP NCLUDES COMPONENTS THAT AE NOT PAT OF PEOUSLY DEFNED LOOPS A POSSBLE STATEGY S TO CEATE A LOOP BY OPENNG THE CUENT SOUCE THE LOOP EQUATONS FO THE LOOPS WTH CUENT SOUCES AE ma 4mA THE LOOP EQUATON FO THE THD LOOP S 6 [ ] k k( ) k( ) k ( ) 0 THE MESH CUENTS OBTANED WTH THS METHOD AE DFFEENT FOM THE ONES OBTANED WTH A SUPEMESH. EEN FO THOSE DEFNED USNG MESHES.

131 S ( FND OLTAGES 4 S S 4 ACOSS S ESSTOS S Now we need a loop current that does not go over any current source and passes through all unused components. HNT: F ALL CUENT SOUCES AE EMOED THEE S ONLY ONE LOOP LEFT MESH EQUATONS FO LOOPS WTH CUENT SOUCES s S S KL OF EMANNG LOOP For loop analysis we notice... Three independent current sources. Four meshes. One current source shared by two meshes. Careful choice of loop currents should make only one loop equation necessary. Three loop currents can be chosen using meshes and not sharing any source. SOLE FO THE CUENT 4. USE OHM S LAW TO C0MPUTE EQUED OLTAGES ( ) ( 4 ) 4 4 ( 4 ) ( 4 ) ) ) ( ) (

132 - + 4 S S S S A COMMENT ON METHOD SELECTON The same problem can be solved by node analysis but it requires equations S S S S S

133 CCUTS WTH DEPENDENT SOUCES Treat the dependent source as though it were independent. Add one equation for the controlling variable k ma X 4 BY SOUCES DETEMNED CUENTS MESH 0 ) ( ) ( 4 k k k x MESH : 0 ) ( ) ( 4 4 k k MESH 4: ) ( 4 k x x CONTOLLN G AABLES COMBNE EQUATONS. DDE BY k

134 SOLE USNG MATLAB PUT N MATX FOM Since we divided by k the HS is ma and all the coefficients are numbers >> is the MATLAB prompt. What follows is the command entered DEFNE THE MATX» =[,0,0,0; %FST OW,, -, 0; %SECOND OW 0,,,-; %THD OW 0,-,-,] %FOUTH OW = DEFNE THE GHT HAND SDE ECTO» =[4;0;8;] = SOLE AND GET THE ANSWE» =\ The answers are in ma =

135 LEANNG EXTENSON: Dependent Sources Find o USNG MESH CUENTS USNG LOOP CUENTS We treat the dependent source as one more voltage source x LOOP k( ) 4k 0 MESH k 4k( ) 0 6 4k( ) MESH 0 x k LOOP k( ) 6k 0 x NOW WE EXPESS THE CONTOLLNG AABLE N TEMS OF THE LOOP CUENTS x k 4k 4k( ) and solve... x 4k 4k 0k ma,. 5mA 0 EPLACE AND EAANGE 6k 6k O SOLUTONS 6k 9[ ] 6k 8k 0.5mA,. 5mA NOTCE THE DFFEENCE BETWEEN MESH CUENT AND LOOP CUENT EEN THOUGH THEY AE ASSOCATED TO THE SAME PATH The selection of loop currents simplifies expression for x and computation of o.

136 DEPENDENT CUENT SOUCE. CUENT SOUCES NOT SHAED BY MESHES WE AE ASKED FO o. WE ONLY NEED TO SOLE FO EPLACE AND EAANGE x x k 4k( ) 8k k 4mA ma 8 We treat the dependent source as a conventional source Equations for meshes with current sources O 6k [ ] 4 Then KL on the remaining loop(s) And express the controlling variable, x, in terms of loop currents

137 DAW MESH CUENTS WTE MESH EQUATONS. MESH MESH : k 4k( ) CONTOLLNG AABLE N TEMS OF LOOP CUENTS x : k x k 4k( ) 0 0 EPLACE AND EAANGE 6k 6k 0 4k 6k SOLE FO k 6mA O k [ ]

138 n the following we shall solve using loop analysis two circuits that had previously been solved using node analysis This is one circuit. we recap first the node analysis approach and then we solve using loop analysis

139 LEANNG EXAMPLE FND THE OLTAGE o ECAP OF NODE : k AT SUPE NODE X 4 ma 0 k k k k k k : ma 0 CONTOLLNG AABLE X SOLE EQUATONS NOW X 6 X X AABLE OF NTEEST O

140 DETEMNE o USNG LOOP ANALYSS Write loop equations Loop : ma Loop : ma Loop : k k( ) 0 X Loop 4: 4 k( ) k 0 4 X 4 Controlling variable: k( ) X 4 k k 6 4 ma, ma 4 4k ariable of nterest O k STAT SELECTON USNG MESHES SELECT A GENEAL LOOP TO AOD SHANG A CUENT SOUCE

141 LEANNG EXAMPLE Find the current o ECAP OF NODE : node: 6 (constraint eq.) X k k k k k 5 4 : 0 5 X k k CONTOLLNG AABLES X X 4 k 7 eqs in 7 variables AABLE OF NTEEST O 5 k

142 Find the current o using mesh analysis Write loop/mesh equations Select mesh currents Loop: k k( ) k( ) 0 4 Loop : k( ) 6 k( ) 0 5 Loop : X Loop 4: k( ) 0 4 X Loop 5: k( ) k( ) X Loop 6: k( ) k k( ) Controlling variables X X k 8 eqs in 8 unknowns 5 6 ariable of interest: O 6