* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download KirchhoffsLaws7Ed
Survey
Document related concepts
Integrating ADC wikipedia , lookup
Nanofluidic circuitry wikipedia , lookup
Mathematics of radio engineering wikipedia , lookup
Schmitt trigger wikipedia , lookup
Power MOSFET wikipedia , lookup
Power electronics wikipedia , lookup
Operational amplifier wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Resistive opto-isolator wikipedia , lookup
Current source wikipedia , lookup
Wilson current mirror wikipedia , lookup
Topology (electrical circuits) wikipedia , lookup
Surge protector wikipedia , lookup
Opto-isolator wikipedia , lookup
Current mirror wikipedia , lookup
Transcript
KIRCHHOFF CURRENT LAW ONE OF THE FUNDAMENTAL CONSERVATION PRINCIPLES IN ELECTRICAL ENGINEERING “CHARGE CANNOT BE CREATED NOR DESTROYED” NODES, BRANCHES, LOOPS A NODE CONNECTS SEVERAL COMPONENTS. BUT IT DOES NOT HOLD ANY CHARGE. TOTAL CURRENT FLOWING INTO THE NODE MUST BE EQUAL TO TOTAL CURRENT OUT OF THE NODE (A CONSERVATION OF CHARGE PRINCIPLE) 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 BRANCH: Component connected between two nodes (e.g., component R4) NODE KIRCHHOFF CURRENT LAW (KCL) SUM OF CURRENTS FLOWING INTO A NODE IS EQUAL TO SUM OF CURRENTS FLOWING OUT OF THE NODE 5A 5A A GENERALIZED NODE IS ANY PART OF A CIRCUIT WHERE THERE IS NO ACCUMULATION OF CHARGE ... OR WE CAN MAKE SUPERNODES BY AGGREGATING NODES A current flowing into a node is equivalent to the negative flowing out of the node ALGEBRAIC SUM OF CURRENT (FLOWING) OUT OF A NODE IS ZERO ALGEBRAIC SUM OF CURRENTS FLOWING INTO A NODE IS ZERO Leaving 2 : i1 i6 i4 0 Leaving 3 : i2 i4 i5 i7 0 Adding 2 & 3 : i1 i2 i5 i6 i7 0 INTERPRETATION: SUM OF CURRENTS LEAVING NODES 2&3 IS ZERO VISUALIZATION: WE CAN ENCLOSE NODES 2&3 INSIDE A SURFACE THAT IS VIEWED AS A GENERALIZED NODE (OR SUPERNODE) PROBLEM SOLVING HINT: KCL CAN BE USED TO FIND A MISSING CURRENT SUM OF CURRENTS INTO NODE IS ZERO b IX ? c 5A WRITE ALL KCL EQUATIONS 5 A I X (3 A) 0 I X 2 A a Which way are charges flowing on branch a-b? 3A d ...AND PRACTICE NOTATION CONVENTION AT THE SAME TIME... I ab 2 A, I cb 3 A I bd 4 A I be ? NODES: a,b,c,d,e BRANCHES: a-b,c-b,d-b,e-b d c a -3A 2A 4A b Ibe = ? e Ibe 4 A [(3 A)] (2 A) 0 THE FIFTH EQUATION IS THE SUM OF THE FIRST FOUR... IT IS REDUNDANT!!! FIND MISSING CURRENTS KCL DEPENDS ONLY ON THE INTERCONNECTION. THE TYPE OF COMPONENT IS IRRELEVANT KCL DEPENDS ONLY ON THE TOPOLOGY OF THE CIRCUIT WRITE KCL EQUATIONS FOR THIS CIRCUIT •THE LAST EQUATION IS AGAIN LINEARLY DEPENDENT OF THE PREVIOUS THREE •THE PRESENCE OF A DEPENDENT SOURCE DOES NOT AFFECT APPLICATION 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 BIG node SUM OF CURRENTS LEAVING BIG NODE 0 I 4 40mA 30mA 20mA 60mA 0 I 4 70mA THE CURRENT I5 BECOMES INTERNAL TO THE NODE AND IT IS NOT NEEDED!!! Find I1 Find I T I1 50mA Find I1 10mA 4mA I1 0 IT 10mA 40mA 20mA Find I1 and I2 I 2 3mA I1 0 I1 4mA 12mA 0 Find ix 10i x i x 44mA 0 i x 10i x 120mA 12mA 0 i x 4mA I3 I 2 I1 0 I1 I3 I5 I 4 I3 0 I5 + - I2 I2 = 6mA, I3 = 8mA, I4 I4 = 4mA mA I1 = 14 _______ 5mA I5 = _______ DETERMINE THE CURRENTS INDICATED I3 I1 + - 5mA I 4 2mA + - I4 2I 2 I5 I6 I 2 8mA I1 2mA , I 2 3mA , I 3 5mA I6 I1 2 I 2 0 I6 8mA I5 I 2 I 6 0 I 4 I3 I5 0 I 5 5mA THE PLAN MARK ALL THE KNOWN CURRENTS FIND NODES WHERE ALL BUT ONE CURRENT ARE KNOWN FIND I x Ix 3mA I X I1 2 I X 0 I1 4mA 1mA 0 I1 3mA VERIFICATION I b 1mA I X 2mA 1mA 2 I X 4mA I b Ib 2I x 4mA This question tests KCL and convention to denote currents Use sum of currents leaving node = 0 A I X (5 A) (3 A) 10 A 0 5A F I EF B Ix D E I DE 10A I EF 4 A 10 A 0 I EG 4 A 3A C Ix G -8A On BD current flows fromB __ to D __ I EF 6A OnEF current flows from__ E toF__ KCL KIRCHHOFF VOLTAGE LAW ONE OF THE FUNDAMENTAL CONSERVATION LAWS IN ELECTRICAL ENGINERING THIS IS A CONSERVATION OF ENERGY PRINCIPLE “ENERGY CANNOT BE CREATE NOR DESTROYED” q C W q (VB VA ) W qVAB V B 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 B VB AB KVL IS A CONSERVATION OF ENERGY PRINCIPLE A “THOUGHT EXPERIMENT” V KIRCHHOFF VOLTAGE LAW (KVL) q W qVBC VA VCA W qVCA B VB VC IF THE CHARGE COMES BACK TO THE SAME INITIAL POINT THE NET ENERGY GAIN MUST BE ZERO (Conservative network) VA OTHERWISE THE CHARGE COULD END UP WITH INFINITE ENERGY, OR SUPPLY AN INFINITE AMOUNT OF ENERGY q Vab q q(VAB VBC VCD ) 0 a b Vcd c d LOSES W qVab KVL: THE ALGEBRAIC SUM OF VOLTAGE DROPS AROUND ANY LOOP MUST BE ZERO GAINS W qVcd V A B (V ) A B A VOLTAGE RISE IS A NEGATIVE DROP 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 Vbe VS VR VR VR 0 1 2 3 VR 12V 2 VR 18V 1 EXAMPLE : VR1 , VR3 ARE KNOWN DETERMINE THE VOLTAGE Vbe VR Vbe VR 30[V ] 0 1 LOOP abcdefa 3 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 IN THE CIRCUIT DEFINE N NUMBER OF NODES B NUMBER OF BRANCHES N 1 LINEARLY INDEPENDEN T KCL EQUATIONS B ( N 1) LINEARLY INDEPENDEN T KVL EQUATIONS 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!! FIND THE VOLTAGES Vae ,Vec GIVEN THE CHOICE USE THE SIMPLEST LOOP DEPENDENT SOURCES ARE HANDLED WITH THE SAME EASE Vad ______ 10V Vac ______ Vac 4 6 0 6V Vbd ______ 11V Vbd _______ Vbd 2 4 0 MUST FIND VR FIRST 12 VR 1 10VR 0 VR 1V 1 1 1 1 DEPENDENT SOURCES ARE NOT REALLY DIFFICULT TO ANALYZE Veb 4 6 12 0 REMINDER: IN A RESISTOR THE VOLTAGE AND CURRENT DIRECTIONS MUST SATISFY THE PASSIVE SIGN CONVENTION Vad 12 8 6 0 Vad _______, Veb ________ V V SAMPLE PROBLEM 4V V1 b Vx R 2k + - + - V1 12V , V2 4V a V2 DETERMINE Vx 4V Vab -8V Power disipated on the 2k resistor P2k Remember past topics We need to find a closed path where only one voltage is unknown FOR VX VX V2 V1 4 0 VX 4 12 4 0 VX V2 Vab 0 Vab V X V2 5k 10k 25V Vx + - V1 There are no loops with only one unknown!!! Vx/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!!! Vx 4 VX VX V1 0 4 2 VX VX 25[V ] VX 0 V 2 4 V1 X 5[V ] 4 VX 20[V ]