KIRCHHOFF CURRENT LAW

Transcription

1 KRCHHOFF CURRENT LAW ONE OF THE FUNDAMENTAL CONSERATON PRNCPLES N ELECTRCAL ENGNEERNG CHARGE CANNOT BE CREATED NOR DESTROYED

2 NODES, BRANCHES, LOOPS A NODE CONNECTS SEERAL COMPONENTS. BUT T DOES NOT HOLD ANY CHARGE. TOTAL CURRENT FLOWNG NTO THE NODE MUST BE EQUAL TO TOTAL CURRENT OUT OF THE NODE (A CONSERATON OF CHARGE PRNCPLE) NODE: point where two, or more, elements are joined (e.g., big node 1) LOOP: A closed path that never goes twice over a node (e.g., the blue line) The red path is NOT a loop NODE BRANCH: Component connected between two nodes (e.g., component R4)

3 KRCHHOFF CURRENT LAW (KCL) SUM OF CURRENTS FLOWNG NTO A NODE S EQUAL TO SUM OF CURRENTS FLOWNG OUT OF THE NODE 5A A current flowing is equivalent flowing out of 5A into a node to the negative the node ALGEBRAC SUM OF CURRENT (FLOWNG) OUT OF A NODE S ZERO ALGEBRAC SUM OF CURRENTS FLOWNG NTO A NODE S ZERO

4 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

5 A GENERALZED NODE S ANY PART OF A CRCUT WHERE THERE S NO ACCUMULATON OF CHARGE... OR WE CAN MAKE SUPERNODES BY AGGREGATNG NODES Leaving 2: i 1 Leaving 3: i Adding 2 & 3:i i 6 i i 4 i 0 i i2 i5 i6 i7 NTERPRETATON: SUM OF CURRENTS LEANG NODES 2&3 S ZERO SUALZATON: WE CAN ENCLOSE NODES 2&3 NSDE A SURFACE THAT S EWED AS A GENERALZED NODE (OR SUPERNODE) 0

6 PROBLEM SOLNG HNT: KCL CAN BE USED TO FND A MSSNG CURRENT c 5A d 3A b a? SUM OF CURRENTS NTO NODE S ZERO 5A ( 3A) 0 2A Which way are charges flowing on branch a-b?...and PRACTCE NOTATON CONENTON AT THE SAME TME... ab cb bd be 2A, 3A 4A? be NODES: a,b,c,d,e BRANCHES: a-b,c-b,d-b,e-b c a -3A 2A b 4A d be =? 4A[ ( 3A)] ( 2A) 0 e

7 WRTE ALL KCL EQUATONS i ( t) i ( t) i ( t) i ( t) i ( t) i ( t) i ( t) i ( t) i ( t) THE FFTH EQUATON S THE SUM OF THE FRST FOUR... T S REDUNDANT!!!

8 FND MSSNG CURRENTS KCL DEPENDS ONLY ON THE NTERCONNECTON. THE TYPE OF COMPONENT S RRELEANT KCL DEPENDS ONLY ON THE TOPOLOGY OF THE CRCUT

9 WRTE KCL EQUATONS FOR THS CRCUT THE LAST EQUATON S AGAN LNEARLY DEPENDENT OF THE PREOUS THREE THE PRESENCE OF A DEPENDENT SOURCE DOES NOT AFFECT APPLCATON 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 BG node SUM OF CURRENTS LEANG BG NODE 0 40mA 30mA 20mA 60mA mA THE CURRENT 5 BECOMES NTERNAL TO THE NODE AND T S NOT NEEDED!!!

11 Find 1 Find T 1 50mA T 10mA 40mA 20mA Find 1 2 3mA 1 Find 0 1 and 2 1 4mA 12mA 0 10 ma 4 ma 1 0

12 Find i x 10i i x x i x 4mA 44mA 0 i x 10i 120mA 12mA 0 x = 14mA = 5mA 2 = 6mA, 3 = 8mA, 4 = 4mA

13 + - 3 DETERMNE THE CURRENTS NDCATED mA 5mA mA 5 5mA ma, 3mA, 5mA mA THE PLAN MARK ALL THE KNOWN CURRENTS FND NODES WHERE ALL BUT ONE CURRENT ARE KNOWN

14 FND x x 3mA mA 1mA 0 1 3mA 1mA ERFCATON b 2 1mA 4mA 2mA b b 2 x 4mA

15 This question tests KCL and convention to denote currents Use sum of currents leaving node = 0 A ( 5A) (3A) 10A 0 5A F EF B x 3A D DE 10A E EF EG 4A C G x -8A On BD current flows from B to D EF 6A OnEF current flows from to E F 4A10A 0 KCL

16 KRCHHOFF OLTAGE LAW ONE OF THE FUNDAMENTAL CONSERATON LAWS N ELECTRCAL ENGNERNG THS S A CONSERATON OF ENERGY PRNCPLE ENERGY CANNOT BE CREATE NOR DESTROYED

17 KRCHHOFF OLTAGE LAW (KL) KL S A CONSERATON OF ENERGY PRNCPLE A POSTE CHARGE GANS ENERGY AS T MOES TO A PONT WTH HGHER OLTAGE AND RELEASES ENERGY F T MOES TO A PONT WTH LOWER OLTAGE W q( ) B A B B A THOUGHT EPERMENT W q AB q A AB CA B B W q CA BC C W q BC q A F THE CHARGE COMES BACK TO THE SAME NTAL PONT THE NET ENERGY GAN MUST BE ZERO (Conservative network) q q a c ab cd b d LOSES GANS W q ab W q cd OTHERWSE THE CHARGE COULD END UP WTH NFNTE ENERGY, OR SUPPLY AN NFNTE AMOUNT OF ENERGY q( ) 0 AB BC CD KL: THE ALGEBRAC SUM OF OLTAGE DROPS AROUND ANY LOOP MUST BE ZERO A B A OLTAGE A NEGATE A ( ) B RSE S DROP

18 PROBLEM SOLNG TP: KL S USEFUL TO DETERMNE A OLTAGE - FND A LOOP NCLUDNG THE UNKNOWN OLTAGE THE LOOP DOES NOT HAE TO BE PHYSCAL be S R 1 R 2 R 3 0 R 18 1 R 12 2 EAMPLE :, DETERMNE R 1 be R1 R R3 THE OLTAGE 3 ARE KNOWN 30[ ] 0 be LOOP abcdefa

19 BACKGROUND: WHEN DSCUSSNG KCL WE SAW THAT NOT ALL POSSBLE KCL EQUATONS ARE NDEPENDENT. WE SHALL SEE THAT THE SAME STUATON ARSES WHEN USNG KL A SNEAK PREEW ON THE NUMBER OF LNEARLY NDEPENDENT EQUATONS N THE CRCUT DEFNE N B NUMBER OF NODES NUMBER OF BRANCHES N 1 B ( N 1) LNEARLY NDEPENDENT KCL EQUATONS LNEARLY NDEPENDENT KL EQUATONS EAMPLE: FOR THE CRCUT SHOWN WE HAE N = 6, B = 7. HENCE THERE ARE ONLY TWO NDEPENDENT KL EQUATONS THE THRD EQUATON S THE SUM OF THE OTHER TWO!!

20 FND THE OLTAGES ae, ec DEPENDENT SOURCES ARE HANDLED WTH THE SAME EASE GEN THE CHOCE USE THE SMPLEST LOOP

21 ac ad ac bd 10 6 bd bd MUST 11 FND R 1 FRST 12 R 110 R 0 R eb DEPENDENT SOURCES ARE NOT REALLY DFFCULT TO ANALYZE REMNDER: N A RESSTOR THE OLTAGE AND CURRENT DRECTONS MUST SATSFY THE PASSE SGN CONENTON ad ad, eb

22 SAMPLE PROBLEM R 2k b 1 12, 2 4 x + - a 2 DETERMNE P x ab 2k Power disipated the 2k resistor 4-8 on Remember past topics We need to find a closed path where only one voltage is unknown FOR ab 0 2 ab 2

23 10k 5k There are no loops with only one unknown!!! + - x x/2 + The current through the 5k and 10k resistors is the same. Hence the voltage drop across the 5k is one half of the drop across the 10k!!! 25[ ] 20[ ] x [ 0 ]