KRCHHOFF CURRENT LAW ONE OF THE FUNDAMENTAL CONSERATON PRNCPLES N ELECTRCAL ENGNEERNG CHARGE CANNOT BE CREATED NOR DESTROYED
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)
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
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
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 0 2 4 5 7 1 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
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
WRTE ALL KCL EQUATONS i ( t) i ( t) i ( t) 0 1 2 3 i ( t) i ( t) i ( t) 0 1 4 6 i ( t) i ( t) i ( t) 0 3 5 8 THE FFTH EQUATON S THE SUM OF THE FRST FOUR... T S REDUNDANT!!!
FND MSSNG CURRENTS KCL DEPENDS ONLY ON THE NTERCONNECTON. THE TYPE OF COMPONENT S RRELEANT KCL DEPENDS ONLY ON THE TOPOLOGY OF THE CRCUT
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
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 0 4 4 70mA THE CURRENT 5 BECOMES NTERNAL TO THE NODE AND T S NOT NEEDED!!!
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
Find i x 10i i x x i x 4mA 44mA 0 i x 10i 120mA 12mA 0 x 3 2 1 1 0 3 5 4 3 5 0 1 = 14mA + - 2 4 5 = 5mA 2 = 6mA, 3 = 8mA, 4 = 4mA
+ - 3 DETERMNE THE CURRENTS NDCATED 4 1 2 2 4 2mA 5mA + - 5 2 6 8mA 5 5mA ma, 3mA, 5mA 1 2 2 3 0 8mA 6 1 2 2 6 THE PLAN MARK ALL THE KNOWN CURRENTS FND NODES WHERE ALL BUT ONE CURRENT ARE KNOWN 5 2 6 4 3 5 0 0
FND x x 3mA 1 2 0 1 4mA 1mA 0 1 3mA 1mA ERFCATON b 2 1mA 4mA 2mA b b 2 x 4mA
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
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
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
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
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!!
FND THE OLTAGES ae, ec DEPENDENT SOURCES ARE HANDLED WTH THE SAME EASE GEN THE CHOCE USE THE SMPLEST LOOP
ac 4 6 0 ad ac bd 10 6 bd 2 4 0 bd MUST 11 FND R 1 FRST 12 R 110 R 0 R 1 1 1 1 eb 4 6 12 0 DEPENDENT SOURCES ARE NOT REALLY DFFCULT TO ANALYZE REMNDER: N A RESSTOR THE OLTAGE AND CURRENT DRECTONS MUST SATSFY THE PASSE SGN CONENTON ad 12 8 6 0 ad, eb
SAMPLE PROBLEM 1 + - 4 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 2 4 12 1 4 0 4 0 ab 0 2 ab 2
10k 5k There are no loops with only one unknown!!! + - x 25 4 1 - 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[ ] 2 4 0 + - x 1 1 4 4 2 5[ 0 ]