KIRCHHOFF CURRENT LAW

Transcription

1 KICHHOFF CUENT LAW ONE OF THE FUNDAMENTAL CONSEATION PINCIPLES IN ELECTICAL ENGINEEING CHAGE CANNOT BE CEATED NO DESTOYED

2 NODES, BANCHES, LOOPS A NODE CONNECTS SEEAL COMPONENTS. BUT IT DOES NOT HOLD ANY CHAGE. TOTAL CUENT FLOWING INTO THE NODE MUST BE EQUAL TO TOTAL CUENT OUT OF THE NODE (A CONSEATION OF CHAGE PINCIPLE) 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)

3 KICHHOFF CUENT LAW (KCL) SUM OF CUENTS FLOWING INTO A NODE IS EQUAL TO SUM OF CUENTS FLOWING OUT OF THE NODE 5A A current flowing is equivalent flowing out of to 5A into a node the negative the node ALGEBAIC SUM OF CUENT (FLOWING) OUT OF A NODE IS ZEO ALGEBAIC SUM OF CUENTS FLOWING INTO A NODE IS ZEO

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

5 A GENEALIZED NODE IS ANY PAT OF A CICUIT WHEE THEE IS NO ACCUMULATION OF CHAGE... O WE CAN MAKE SUPENODES BY AGGEGATING NODES Leaving : i Leaving 3: i Adding & 3:i i 6 i i 4 i 0 i i i5 i6 i7 INTEPETATION: SUM OF CUENTS LEAING NODES &3 IS ZEO ISUALIZATION: WE CAN ENCLOSE NODES &3 INSIDE A SUFACE THAT IS IEWED AS A GENEALIZED NODE (O SUPENODE) 0

6 POBLEM SOLING HINT: KCL CAN BE USED TO FIND A MISSING CUENT c 5A d 3A b I X a? SUM OF CUENTS INTO NODE IS ZEO 5A I ( 3A) 0 X I X A Which way are charges flowing on branch a-b?...and PACTICE NOTATION CONENTION AT THE SAME TIME... I I I I ab cb bd be A, 3A 4A? I be NODES: a,b,c,d,e BANCHES: a-b,c-b,d-b,e-b c a -3A A b 4A d I be =? 4A[ ( 3A)] ( A) 0 e

7 WITE ALL KCL EQUATIONS i ( t) i ( t) i ( t) 0 3 i ( t) i ( t) i ( t) i ( t) i ( t) i ( t) THE FIFTH EQUATION IS THE SUM OF THE FIST FOU... IT IS EDUNDANT!!!

8 FIND MISSING CUENTS KCL DEPENDS ONLY ON THE INTECONNECTION. THE TYPE OF COMPONENT IS IELEANT KCL DEPENDS ONLY ON THE TOPOLOGY OF THE CICUIT

9 WITE KCL EQUATIONS FO THIS CICUIT THE LAST EQUATION IS AGAIN LINEALY DEPENDENT OF THE PEIOUS THEE THE PESENCE OF A DEPENDENT SOUCE DOES NOT AFFECT APPLICATION OF KCL KCL DEPENDS ONLY ON THE TOPOLOGY

10 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 BIG node SUM OF CUENTS LEAING BIG NODE 0 I 40mA30mA 0mA 60mA 0 4 I4 70mA THE CUENT I5 BECOMES INTENAL TO THE NODE AND IT IS NOT NEEDED!!!

11 Find I Find I T I 50mA I T 0mA 40mA 0mA Find I I 3mA I Find 0 I and I I 4mA ma 0 0 ma 4 ma I 0

12 Find i x 0i i x x i x 4mA 44mA 0 i x 0i 0mAmA 0 x I 3 I I I 0 I 3 I 5 I4 I3 I 5 0 I = 4mA + - I I 4 I 5 = 5mA I = 6mA, I 3 = 8mA, I 4 = 4mA

13 + - I 3 DETEMINE THE CUENTS INDICATED I 4 I I I 4 ma 5mA + - I 5 I I 6 8mA I 5 5mA I ma, I 3mA, I 5mA 3 I I I 0 I 8mA 6 6 THE PLAN MAK ALL THE KNOWN CUENTS FIND NODES WHEE ALL BUT ONE CUENT AE KNOWN I 5 I I6 I 4 I3 I5 0 0

14 FIND I x I x 3mA I X I I X 0 I 4mAmA 0 I 3mA ma EIFICATION I b I ma X I X 4mA I ma b I b I x 4mA

15 This question tests KCL and convention to denote currents Use sum of currents leaving node = 0 A I X ( 5A) (3A) 0A 0 5A F I EF B I x 3A D I DE 0A E I EF I EG 4A C G I x -8A On BD current flows from B to D I EF 6A OnEF current flows from to E F 4A 0A 0 KCL

16 KICHHOFF OLTAGE LAW ONE OF THE FUNDAMENTAL CONSEATION LAWS IN ELECTICAL ENGINEING THIS IS A CONSEATION OF ENEGY PINCIPLE ENEGY CANNOT BE CEATE NO DESTOYED

17 KICHHOFF OLTAGE LAW (KL) KL IS A CONSEATION OF ENEGY PINCIPLE A POSITIE CHAGE GAINS ENEGY AS IT MOES TO A POINT WITH HIGHE OLTAGE AND ELEASES ENEGY IF IT MOES TO A POINT WITH LOWE OLTAGE W q( B A) B B A THOUGHT EXPEIMENT W q AB q A AB CA B B W q CA BC C W q BC q A IF THE CHAGE COMES BACK TO THE SAME INITIAL POINT THE NET ENEGY GAIN MUST BE ZEO (Conservative network) q q a c ab cd b d LOSES W q ab GAINS W q cd OTHEWISE THE CHAGE COULD END UP WITH INFINITE ENEGY, O SUPPLY AN INFINITE AMOUNT OF ENEGY q( ) 0 AB BC CD KL: THE ALGEBAIC SUM OF OLTAGE DOPS AOUND ANY LOOP MUST BE ZEO A B A OLTAGE A NEGATIE A ( ) B ISE IS DOP

18 POBLEM SOLING TIP: KL IS USEFUL TO DETEMINE A OLTAGE - FIND A LOOP INCLUDING THE UNKNOWN OLTAGE THE LOOP DOES NOT HAE TO BE PHYSICAL be S EXAMPLE :, be 3 DETEMINE THE OLTAGE 3 AE KNOWN 30[ ] 0 be LOOP abcdefa

19 BACKGOUND: WHEN DISCUSSING KCL WE SAW THAT NOT ALL POSSIBLE KCL EQUATIONS AE INDEPENDENT. WE SHALL SEE THAT THE SAME SITUATION AISES WHEN USING KL A SNEAK PEIEW ON THE NUMBE OF LINEALY INDEPENDENT EQUATIONS IN THE CICUIT DEFINE N B NUMBE OF NODES NUMBE OF BANCHES N B ( N ) LINEALY INDEPENDENT KCL EQUATIONS LINEALY INDEPENDENT KL EQUATIONS EXAMPLE: FO THE CICUIT SHOWN WE HAE N = 6, B = 7. HENCE THEE AE ONLY TWO INDEPENDENT KL EQUATIONS THE THID EQUATION IS THE SUM OF THE OTHE TWO!!

20 FIND THE OLTAGES ae, ec DEPENDENT SOUCES AE HANDLED WITH THE SAME EASE GIEN THE CHOICE USE THE SIMPLEST LOOP

21 ac ad ac bd 0 6 bd 4 0 bd MUST FIND FIST 0 0 eb DEPENDENT SOUCES AE NOT EALLY DIFFICULT TO ANALYZE EMINDE: IN A ESISTO THE OLTAGE AND CUENT DIECTIONS MUST SATISFY THE PASSIE SIGN CONENTION ad ad, eb

22 SAMPLE POBLEM k b, 4 x + - a DETEMINE P x ab k Power disipated the k resistor 4-8 on emember past topics We need to find a closed path where only one voltage is unknown FO X X X X ab 0 ab X

23 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 ]

24 SINGLE LOOP CICUITS BACKGOUND: USING KL AND KCL WE CAN WITE ENOUGH EQUATIONS TO ANALYZE ANY LINEA CICUIT. WE NOW STAT THE STUDY OF SYSTEMATIC, AND EFFICIENT, WAYS OF USING THE FUNDAMENTAL CICUIT LAWS a b 3 6 branches 6 nodes loop c 4 WITE 5 KCL EQS O DETEMINE THE ONLY CUENT FLOWING OLTAGE DIISION: THE SIMPLEST CASE KL ON THIS LOOP f 6 e 5 d ALL ELEMENTS IN SEIES ONLY ONE CUENT THE PLAN BEGIN WITH THE SIMPLEST ONE LOOP CICUIT EXTEND ESULTS TO MULTIPLE SOUCE AND MULTIPLE ESISTOS CICUITS IMPOTANT OLTAGE DIIDE EQUATIONS

25 SUMMAY OF BASIC OLTAGE DIIDE v v( t) EXAMPLE: S, 90k, 30k 9 OLUME CONTOL? 5k

26 A PACTICAL POWE APPLICATION HOW CAN ONE EDUCE THE LOSSES?

27 THE CONCEPT OF EQUIALENT CICUIT THIS CONCEPT WILL OFTEN BE USED TO SIMPLFY THE ANALYSIS OF CICUITS. WE INTODUCE IT HEE WITH A EY SIMPLE OLTAGE DIIDE THE DIFFEENCE BETWEEN ELECTIC CONNECTION AND PHYSICAL LAYOUT SOMETIMES, FO PACTICAL CONSTUCTION EASONS, COMPONENTS THAT AE ELECTICALLY CONNECTED MAY BE PHYSICALLY QUITE APAT i i v S v S - i v S AS FA AS THE CUENT IS CONCENED BOTH CICUITS AE EQUIALENT. THE ONE ON THE IGHT HAS ONLY ONE ESISTO SEIES COMBINATION OF ESISTOS IN ALL CASES THE ESISTOS AE CONNECTED IN SEIES

28 CONNECTO SIDE ILLUSTATING THE DIFFEENCE BETWEEN PHYSICAL LAYOUT AND ELECTICAL CONNECTIONS PHYSICAL NODE PHYSICAL NODE SECTION OF 4.4 KB OICE/DATA MODEM COESPONDING POINTS COMPONENT SIDE

29 v v 5 v KL v FIST GENEALIZATION: MULTIPLE SOUCES i(t) v v 4 v v3 v v4 v5 v Collect all sources on one side v v v3 v4 v5 v v v eq v v - + v v 3 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 + -

30 SECOND GENEALIZATION: MULTIPLE ESISTOS FIND I,, P(30k) bd APPLY KL TO THIS LOOP APPLY KL TO THIS LOOP LOOP FO bd bd 0[ k ] I 0 ( KL) bd 0 POWE ON30k ESISTO 4 3 P I ( 0 A) (30*0 ) 30mW v i i i OLTAGE DIISION FO MULTIPLE ESISTOS

31 THE INESE OLTAGE DIIDE S OLTAGE DIIDE + - O S O "INESE" DIIDE S O COMPUTE S "INESE" DIIDE 0 0 S k 0

32 Find I and bd APPLY KL TO THIS LOOP 6 80kI 40kI 0 I 0. 05mA 40kI 0 0 bd bd

33 If ad = 3, find S 3 INESE DIIDE POBLEM S 3 9 0

34 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

35 EXAMPLE 9 A 0k B C + - KL FO DA I + - DETEMINE I USING KL 30k E I 0k DA CD DE 0.05mA D 30k * I. 5 KL : KL : - 0k *I 9 30k *I 0k *I 0 3 I 0. 05mA 60k DA 0k * I 0 DA. 5

36 EXAMPLE S Sometimes you may want to vary a bit b k S ab I 3 x + - APPLY KL TO THIS LOOP O KL HEE a KL HEE X P x ab ( 3x ) P ( 3 x ) is supplied the power absorbed or by the dependent source KL : KL : KL : P 4 3 X X 0 ab 3 0 ab ( 3 X ) 3 X 4 X I S X 0 4 OHMS' LAW: I ma 4k P [ ]*[ ma] mw ( 3 ) X X ab 0 (PASSIESIGN CONENTION)

37 SINGLE NODE-PAI CICUITS THESE CICUITS AE CHAACTEIZED BY ALL THE ELMENTS HAING THE SAME OLTAGE ACOSS THEM - THEY AE IN PAALLEL IN PACTICE NODES MAY ASSUME STANGE FOMS EXAMPLE OF SINGLE NODE-PAI LOW DISTOTION POWE AMPLIFIE THIS ELEMENT IS INACTE (SHOT-CICUITED)

38 LOW OLTAGE POWE SUPPLY FO CT - PATIAL IEW SAMPLE PHYSICAL NODES COMPONENT SIDE CONNECTION SIDE

39 BASIC CUENT DIIDE APPLY KCL p THE CUENT DIISION THE CUENT i(t) ENTES THE NODE AND SPLITS - IT IS DIIDED BETWEEN THE CUENTS i(t) AND i(t) USE OHM S LAW TO EPLACE CUENTS DEFINE PAALLEL ESISTANCE COMBINATION i( t) v( t) p v( t) i( t ) I 4 (5) mai I I (5 4 5 )

40 FIND I,, I O WHEN IN DOUBT EDAW THE CICUIT TO HIGHLIGHT ELECTICAL CONNECTIONS!! IS EASIE TO SEE THE DIIDE 80k * I 4

41 CA STEEO AND CICUIT MODEL 5mA 5mA POWE PE SPEAKE LEANING EXTENSION - CUENT DIIDE THEE IS MOE THAN ONE OPTION TO COMPUTE I I 0 (6) 0 40 KCL : I 6 I 0 POWE : I USING ESISTANCEIN k, CUENT DIIDE 40 I (6) 4mA 0 40 P 44*40mW 5. 76W I CUENT IN ma YIELD POWE IN mw ma

42 FIST GENEALIZATION: MULTIPLE SOUCES APPLY KCL TO THIS NODE EQUIALENTSOUCE DEFINE PAALLEL ESISTANCE COMBINATION i O ( t) v( t) v( t) i ( t ) O p

43 FIND O AND THE POWE SUPPLIED BY THE SOUCES 6k 0mA 3k 5mA O 5mA p O 6k *3k p k 6k 3k P P O 6mA 0 5mA 50mW O O (5mA) ( 0mA) 00mW

44 APPLY KCL TO THIS NODE SECOND GENEALIZATION: MULTIPLE ESISTOS ) ( ) ( ) ( ) ( ) ( ) ( t i t i t v t i t i t v O k p K k k O P General current divider Ohm s Law at every resistor

45 vt () v v v 4k 6k k Notice use of passive sign convention Once v(t) is known all other variables can be determined; e.g., v v v KCL :6mA 4mA 0 4k 6k k k 7 3v v 48 v 0 4 6v 0 v 4 P 6k v 6 6k 6k.667mW

46 FIND i AND THE POWE SUPPLIED BY THE SOUCE 0k 5k 8mA i 4k 0k 5k 5 4 p 4k 0k 5k 0k k k i (8) 4mA 4k v 4k * i 6 P v( 8mA) 8mW v( t) PiO ( t) v( t) i ik ( t) k K ( t) p k i O ( t) p k AN ALTENATIE APPOACH 8mA i 4k 4k General current divider

47 FIND THE CUENT I L COMBINE THE SOUCES COMBINE ESISTOS ma STATEGY: CONET THE POBLEM INTO A BASIC CUENT DIIDE BY COMBINING SOUCES AND ESISTOS. THE NEXT SECTION EXPLOES IN MOE DETAIL THE IDEA OF COMBINING ESISTOS NOTICE THE MINUS SIGN

48 B I 9mA 6k 3k C 6k 3k I I 3 ma ma 9 9 [ ] 3 I I A I 6k C B 3k 6k I B 3k A 9mA I 6k 3k C 3k A I 9mA 6k DIFFEENT LOOKS FO THE SAME ELECTIC CICUIT

49 B I 6k C I 6k 9mA 3k 3k A B EDAWING A CICUIT MAY, SOMETIMES, HELP TO ISUALIZE BETTE THE ELECTICAL CONNECTIONS I A 9mA I 6k 3k 6k 3k C

50 + k 4k 3k _ 0mA Determine power delivered by source P p *(0mA) p p k 3 k 4k 3k 6 3 k 4 P P 3 *0 *(0* W 3 3 ) [ A]

51 SEIES PAALLEL ESISTO COMBINATIONS UP TO NOW WE HAE STUDIED CICUITS THAT CAN BE ANALYZED WITH ONE APPLICATION OF KL(SINGLE LOOP) O KCL(SINGLE NODE-PAI) WE HAE ALSO SEEN THAT IN SOME SITUATIONS IT IS ADANTAGEOUS TO COMBINE ESISTOS TO SIMPLIFY THE ANALYSIS OF A CICUIT NOW WE EXAMINE SOME MOE COMPLEX CICUITS WHEE WE CAN SIMPLIFY THE ANALYSIS USING THE TECHNIQUE OF COMBINING ESISTOS PLUS THE USE OF OHM S LAW SEIES COMBINATIONS PAALLEL COMBINATION G G G... p G N

52 FIST WE PACTICE COMBINING ESISTOS 6k 3k 3k SEIES (0K,K)SEIES 6k k 4k 5k 3k k

53 3k 6k k k If things get confusing 6k (4k k) k k 6k

54 EXAMPLES COMBINATION SEIES-PAALLEL 9k If the drawing gets confusing edraw the reduced circuit and start again 8k 9k 6k ESISTOS AE IN SEIES IF THEY CAY EXACTLY THE SAME CUENT 6k 6k 0k ESISTOS AE IN PAALLEL IF THEY AE CONNECTED EXACTLY BETWEEN THE SAME TWO NODES

55 AN INESE SEIES PAALLEL COMBINATION Given the final value Find a proper combination SIMPLE CASE ONLY MUST BE 600m WHEN I 3A 0. ESISTOS AE AAILABLE.6 EQUIED A ONLY NOT SO SIMPLE CASE MUST BE 600m WHEN I 9A 0. ESISTOS AE AAILABLE.6 EQUIED A

56 EFFECT OF ESISTO TOLEANCE NOMINAL ESISTO ESISTO ALUE: TOLEANCE: 0%.7k ANGES FO CUENT AND POWE? NOMINAL CUENT : 0 I mA.7 NOMINAL POWE : _ 0 P mW.7 MINIMUM CUENT : MAXIMUM CUENT : I I min max mA mA MINIMUMPOWE(I MAXIMUMPOWE : min ): 33.67mW 4.5mW THE ANGES FO CUENT AND POWE AE DETEMINED BY THE TOLEANCE BUT THE PECENTAGE OF CHANGE MAY BE DIFFEENT FOM THE PECENTAGE OF TOLEANCE. THE ANGES MAY NOT EEN BE SYMMETIC

57 CICUIT WITH SEIES-PAALLEL ESISTO COMBINATIONS THE COMBINATION OF COMPONENTS CAN EDUCE THE COMPLEXITY OF A CICUIT AND ENDE IT SUITABLE FO ANALYSIS USING THE BASIC TOOLS DEELOPED SO FA. COMBINING ESISTOS IN SEIES ELIMINATES ONE NODE FOM THE CICUIT. COMBINING ESISTOS IN PAALLEL ELIMINATES ONE LOOP FOM THE CICUIT GENEAL STATEGY: EDUCE COMPLEXITY UNTIL THE CICUIT BECOMES SIMPLE ENOUGH TO ANALYZE. USE DATA FOM SIMPLIFIED CICUIT TO COMPUTE DESIED AIABLES IN OIGINAL CICUIT - HENCE ONE MUST KEEP TACK OF ANY ELATIONSHIP BETWEEN AIABLES

58 4k k k FIST EDUCE IT TO A SINGLE LOOP CICUIT SECOND: BACKTACK USING KL, KCL OHM S 6k 6k 6k I 3 OHM' S: a I KCL : I I I3 0 6k OHM'S: b 3k * I3 I4 I3 4 4k * I KCL : I 5 OHM'S: I 4 C I 3 0 3k * I OTHE OPTIONS... 5 b 4 I k a 3 3 () 9

59 k k k LEANING BY DOING OLTAGE DIIDE: k O (3 ) k k k k k CUENT DIIDE: k I O (3A) A k k

60 AN EXAMPLE OF BACKTACKING I 3mA 3 xz 6.5mA ma.5ma O mA A STATEGY. ALWAYS ASK: WHAT ELSE CAN I COMPUTE? b 6k * I I I 3 b 3k 4 I I 3 4 a k * I I I xz a xz b 4k I 5 6k * I 4k I 5 I O * xz

61 FIND O STATEGY: FIND USE OLTAGE DIIDE + - OLTAGE O 0k 0k DIIDE 0k 0k 40k 6 60k 30k 60k 0k FIND 9 0k () 6 0k 0k S 0.5mA 6 60 k *0. ma 0.05mA THIS IS AN INESE POBLEM WHAT CAN BE COMPUTED? S 0k *0.5mA 6 I 6 0k SEIES PAALLEL

62 Y TANSFOMATIONS THIS CICUIT HAS NO ESISTO IN SEIES O PAALLEL IF INSTEAD OF THIS WE COULD HAE THIS THEN THE CICUIT WOULD BECOME LIKE THIS AND BE AMENABLE TO SEIES PAALLEL TANSFOMATIONS

63 Y b a ab ) ( 3 ab 3 3 ) ( b a 3 3 ) ( c b 3 3 ) ( a c SUBTACT THE FIST TWO THEN ADD TO THE THID TO GET a Y c b a a b b a 3 3 c b c b EPLACE IN THE THID 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 3 Y

64 LEANING EXAMPLE: APPLICATION OF WYE-DELTA TANSFOMATION COMPUTE I S c DELTA CONNECTION c 3 k 6k k 6k 8k a b a b c Y a 3k 9k (k 6k) k EQ 6k 0 I S. ma k b ONE COULD ALSO USE A WYE - DELTA TANSFOMATION...

65 LEANING EXAMPLE SHOULD KEEP THESE TWO NODES! CONET THIS Y INTO A DELTA? IF WE CONET TO Y INTO A DELTA THEE AE SEIES PAALLEL EDUCTIONS! Y a a c c b b a c a c c b b a b a c c b b a 3 3* * 36 k k k k 4mA 36k 36k 36k k k O THE ESULTING CICUIT IS A CUENT DIIDE

66 36 k k 9k CICUIT AFTE PAALLEL ESISTO EDUCTION 4mA 36k I O 9k O 36k 8 I 4mA ma O 36k 8k 3 8 9k I 9k ma 4 NOTICE THAT BY KEEPING O O 3 THE FACTION WE PESEE FULL NUMEICAL ACCUACY WYE DELTA

67 CICUITS WITH DEPENDENT SOUCES A CONENTION ABOUT DEPENDENT SOUCES. UNLESS OTHEWISE SPECIFIED THE CUENT AND OLTAGE AIABLES AE ASSUMED IN SI UNITS OF Amps AND olts DEPENDENT AIABLE D I X CONTOLLING AIABLE FO THIS EXAMPLE THE MULTIPLIE MUST HAE UNITS OF OHM OTHE I D D X X DEPENDENT ( scalar) ( Siemens) SOUCES I I ( scalar) D X AN ALTENATIE DESCIPTION I, D X ma ASSUMES CUENT IN UNITS AE EXPLICIT ma FIND GENEAL STATEGY TEAT DEPENDENT SOUCES AS EGULA SOUCES AND ADD ONE MOE EQUATION FO THE CONTOLLING AIABLE O A k * I KL A A PLAN: SINGLE LOOP CICUIT. USE KL TO DETEMINE CUENT KL: 3k * I 5k * I A ONE EQUATION, TWO UNKNOWNS. CONTOLLING AIABLE POIDES EXTA EQUATION EPLACE AND SOLE FO THE CUENT USE OHM S LAW I ma O 5k * I 0 0

68 FIND O KCL TO THIS NODE. THE DEPENDENT SOUCE IS JUST ANOTHE SOUCE A PLAN: IF _s IS KNOWN _0 CAN BE DETEMINED USING OLTAGE DIIDE. TO FIND _s WE HAE A SINGLE NODE-PAI CICUIT THE EQUATION FO THE CONTOLLING AIABLE POIDES THE ADDITIONAL EQUATION ALGEBAICALLY, THEE AE TWO UNKNOWNS AND JUST ONE EQUATION SUBSTITUTION OF I_0 YIELDS */6k 5 60 S OLTAGE DIIDE 4k 4k k ( 3 O S ) NOTICE THE CLEE WAY OF WITING ma TO HAE OLTS IN ALL NUMEATOS AND THE SAME UNITS IN DENOMINATO

69 FIND O KL TO THIS LOOP A PLAN: ONE LOOP POBLEM. FIND THE CUENT THEN USE OHM S LAW. THE DEPENDENT SOUCE IS ONE MOE OLTAGE SOUCE THE EQUATION FO THE CONTOLLING AIABLE POIDES THE ADDITIONAL EQUATION EPLACE AND SOLE FO CUENT I AND FINALLY

70 FIND G v v O i ( t) ( t) KCL A PLAN: ONE LOOP ON THE LEFT - KL ONE NODE-PAI ON IGHT - KCL KL KL KCL g m v g vo( t) ( t) 0 L ALSO A OLTAGE DIIDE