KIRCHHOFF CURRENT LAW

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1 KRCHHOFF CURRENT LAW One of the fundamental conservation principles n electrical engineering 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 節點 node: 兩個以上的電路元件連接的地方本質節點 essential node: 三個以上電路元件連接的一種節點路徑 path: 由基本元件連成, 但每個元件只能通過一次的路線分支 branch: 連接二個節點的路徑本質分支 essential branch: 連接二個本質節點但不通過別的本質節點的路徑迴路 loop: 終點跟起點是同一個節點的路徑網目 mesh: 不再含其他迴路的一種迴路, 也就是最小的迴路

4 KRCHHOFF CURRENT LAW (KCL) Sum of currents flowing into a node is Equal to sum of currents flowing out of The node 5A A current flowing is equivalent flowing out of 5A into a node to the negative the node Algebraic sum of current (flowing) out of A node is zero Algebraic sum of currents flowing into a Node is zero

5 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

6 A generalized node is any part of a circuit where there is no accumulation 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 i i2 i5 i6 i7 nterpretation: sum of currents leaving nodes 2&3 is zero isualization: we can enclose nodes 2&3 inside a surface that is viewed as a generalized node (or supernode)

7 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 5 A ( 3A) 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? 4 A [ ( 3A)] ( 2A) e

8 WRTE ALL KCL EQUATONS i1() t i2() t i3() 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!!!

9 FND MSSNG CURRENTS KCL depends only on the interconnection. The type of component is irrelevant KCL depends only on the topology of the circuit

10 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

11 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 4mA 3mA 2mA 6mA 4 4 7mA THE CURRENT 5 BECOMES NTERNAL TO THE NODE AND T S NOT NEEDED!!!

Find 1 Find T 1 5mA ma ma ma T 1 4 2 Find 1

12 Find 1 Find T 1 5mA ma ma ma T Find 1 2 3mA 1 Find 1 and 2 1 4mA 12mA 1 ma 4 ma 1

13 Find i x 1i i x x i x 4mA 44mA i x 1 i 12mA 12mA x mA mA 5 2 6mA, 3 8mA, 4 4mA

14 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

15 FND x x 3mA mA 1mA 1 3mA 1mA ERFCATON b 2 1mA 4mA 2mA b b 2 x 4mA

16 A This question tests KCL and convention to denote currents Use sum of currents leaving node ( 5A) (3A) 1A 5A EF F B x 3A D DE 1A E EG 4A C G x -8A On BD current flows from B to D EF 6A OnEF current flows from to E EF F 4 A 1A KCL

17 KRCHHOFF OLTAGE LAW One of the fundamental conservation laws in electrical enginering THS S A CONSERATON OF ENERGY PRNCPLE ENERGY CANNOT BE CREATE NOR DESTROYED

18 KRCHHOFF OLTAGE LAW (KL) KL S A CONSERATON OF ENERGY PRNCPLE A positive charge gains energy as it moves to a point with higher voltage and releases energy if it moves to a point with lower voltage q q a c ΔW q( ) B A q ab cd A b d LOSES B B GANS ΔW q ab ΔW q cd A THOUGHT EPERMENT ΔW q AB q A AB CA B B ΔW q CA BC C ΔW q BC if the charge comes back to the same initial point the net energy gain must be zero (conservative network) Otherwise the charge could end up with infinite energy, or supply an infinite amount of energy q( ) AB BC CA KL: the algebraic sum of voltage drops around any loop must be zero A B A OLTAGE A NEGATE A ( ) B RSE S DROP

19 Problem solving tip: kvl is useful to determine a voltage - find a loop including the unknown voltage The loop does not have to be physical be S R 1 R 2 R 3 R 18 1 R 12 2 EAMPLE :, DETERMNE R 1 3[ ] be R1 R3 THE OLTAGE R 3 ARE KNOWN be LOOP abcdefa

20 background: when discussing kcl we saw that not all possible kcl equations are independent. we shall see that the same situation arises when using kvl A sneak preview on the number of linearly independent equations N THE CRCUT DEFNE N B NUMBER NUMBER OF NODES OF BRANCHES N 1 B ( N 1) LNEARLY NDEPENDENT KCL EQUATONS LNEARLY NDEPENDENT KL EQUATONS Example: For the circuit shown we have N 6, B 7. Hence there are only two independent Kvl equations The third equation is the sum of the other two!!

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

22 ac 4 6 ad ac bd 1 6 bd 2 4 bd MUST 11 FND R 1 FRST 12 R 1 1 R R DEPENDENT SOURCES ARE NOT REALLY DFFCULT TO ANALYZE 1 eb REMNDER: N A RESSTOR THE OLTAGE AND CURRENT DRECTONS MUST SATSFY THE PASSE SGN CONENTON ad ad, eb

23 SAMPLE PROBLEM 1-4 R 2kΩ b x 1 12, 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 2 ab 2

24 1kΩ 5kΩ There are no loops with only one unknown!!! - x x/2 The current through the 5k and 1k resistors is the same. Hence the voltage drop across the 5k is one half of the drop across the 1k!!! 25[ ] 2[ ] x [ ]

KIRCHHOFF CURRENT LAW

KIRCHHOFF CURRENT LAW 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