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RESISTIVE CIRCUITS Here we introduce the basic concepts and laws that are fundamental to circuit analysis LEARNING GOALS • OHM’S LAW - DEFINES THE SIMPLEST PASSIVE ELEMENT: THE RESISTOR • KIRCHHOFF’S LAWS - THE FUNDAMENTAL CIRCUIT CONSERVATION LAWS- KIRCHHOFF CURRENT (KCL) AND KIRCHHOFF VOLTAGE (KVL) • LEARN TO ANALYZE THE SIMPLEST CIRCUITS • SINGLE LOOP - THE VOLTAGE DIVIDER • SINGLE NODE-PAIR - THE CURRENT DIVIDER • SERIES/PARALLEL RESISTOR COMBINATIONS - A TECHNIQUE TO REDUCE THE COMPLEXITY OF SOME CIRCUITS • WYE - DELTA TRANSFORMATION - A TECHNIQUE TO REDUCE COMMON RESISTOR CONNECTIONS THAT ARE NEITHER SERIES NOR PARALLEL • CIRCUITS WITH DEPENDENT SOURCES - (NOTHING VERY SPECIAL) RESISTORS v(t ) i(t ) A resistor is a passive element characterized by an algebraic relation between the voltage across its terminals and the current through it v(t ) F (i (t )) General Model for a Resistor A linear resistor obeys OHM’s Law v(t ) Ri(t ) The constant, R, is called the resistance of the component and is measured in units of Ohm () From a dimensional point of view Ohms is a derived unit of Volt/Amp Since the equation is algebraic the time dependence can be omitted Standard Multiples of Ohm M Mega Ohm(106 ) k Kilo Ohm(103 ) A common occurrence is Volt mA resulting in resistance in k Conductance If instead a function current in law can be of expressing voltage as of current one expresses terms of voltage, OHM’s written i 1 v R 1 as Conductanc e R of the component and write i Gv We define G The unit of conductance is Siemens Some practical resistors Symbol i v Notice passive sign convention Two special resistor values R v0 Short Circuit Represent ation Circuit R0 G i “A touch of reality” i0 Open Circuit R G0 Linear approximation v Linear range Actual v-I relationship Ohm’s Law is an approximation valid while voltages and currents remain in the Linear Range GIVEN VOLTAGE AND CONDUCTANCE REFERENCE DIRECTIONS SATISFY PASSIVE SIGN CONVENTION i (t ) Gv (t ) OHM’S LAW UNITS? CONDUCTANCE IN SIEMENS, VOLTAGE IN VOLTS. HENCE CURRENT IN AMPERES i (t ) 8[ A] OHM’S LAW v (t ) Ri (t ) UNITS? 4[V ] (2)i (t ) i (t ) 2[ A] 4V v (t ) Ri (t ) OHM’S LAW THE EXAMPLE COULD BE GIVEN LIKE THIS DETERMINE CURRENT AND POWER ABSORBED BY RESISTOR 6mA V2 P VI I R R 2 P (12[V ])(6[mA]) 72[mW ] 0.6[mA ] V 6[V ] I R 10k VS2 P R VS 6[V ] VS2 (10 103 )(3.6 103W ) 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 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 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 A GENERALIZED NODE IS ANY PART OF A CIRCUIT WHERE THERE IS NO ACCUMULATION OF CHARGE ... OR WE CAN MAKE SUPERNODES BY AGGREGATING NODES 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) WRITE ALL KCL EQUATIONS i1 ( t ) i2 ( t ) i3 ( t ) 0 i1 ( t ) i4 ( t ) i6 ( t ) 0 i3 ( t ) i5 ( t ) i8 ( t ) 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 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 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” V A B (V ) A B A VOLTAGE RISE IS A NEGATIVE DROP KIRCHHOFF VOLTAGE LAW (KVL) KVL IS A CONSERVATION OF ENERGY PRINCIPLE KVL: THE ALGEBRAIC SUM OF VOLTAGE DROPS AROUND ANY LOOP MUST BE ZERO 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 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 ] SINGLE LOOP CIRCUITS BACKGROUND: USING KVL AND KCL WE CAN WRITE ENOUGH EQUATIONS TO ANALYZE ANY LINEAR CIRCUIT. WE NOW START THE STUDY OF SYSTEMATIC, AND EFFICIENT, WAYS OF USING THE FUNDAMENTAL CIRCUIT LAWS a 1 2 b 6 branches 6 nodes 1 loop 3 c 4 WRITE 5 KCL EQS OR DETERMINE THE ONLY CURRENT FLOWING f 6 e 5 d ALL ELEMENTS IN SERIES ONLY ONE CURRENT THE PLAN • BEGIN WITH THE SIMPLEST ONE LOOP CIRCUIT • EXTEND RESULTS TO MULTIPLE SOURCE • AND MULTIPLE RESISTORS CIRCUITS IMPORTANT VOLTAGE DIVIDER EQUATIONS VOLTAGE DIVISION: THE SIMPLEST CASE KVL ON THIS LOOP SUMMARY OF BASIC VOLTAGE DIVIDER v R1 R1 v (t ) R1 R2 EXAMPLE : VS 9V , R1 90k, R2 30k VOLUME CONTROL? R1 15k A “PRACTICAL” POWER APPLICATION HOW CAN ONE REDUCE THE LOSSES? THE CONCEPT OF EQUIVALENT CIRCUIT THIS CONCEPT WILL OFTEN BE USED TO SIMPLFY THE ANALYSIS OF CIRCUITS. WE INTRODUCE IT HERE WITH A VERY SIMPLE VOLTAGE DIVIDER i vS R1 i vS + - R2 i R1 R2 + - vS R1 R2 AS FAR AS THE CURRENT IS CONCERNED BOTH CIRCUITS ARE EQUIVALENT. THE ONE ON THE RIGHT HAS ONLY ONE RESISTOR SERIES COMBINATION OF RESISTORS R1 R2 R1 R2 IN ALL CASES THE RESISTORS ARE CONNECTED IN SERIES FIRST GENERALIZATION: MULTIPLE SOURCES v2 v R1 Voltage sources in series can be algebraically added to form an equivalent source. v3 + + - v5 i(t) R2 + - R1 + - v1 We select the reference direction to move along the path. v Voltage drops are subtracted from rises R2 + - KVL v4 vR1 v2 v3 vR 2 v4 v5 v1 0 R1 Collect all sources on one side v1 v2 v3 v4 v5 vR1 vR 2 v v eq R1 vR 2 veq + - R2 SECOND GENERALIZATION: MULTIPLE RESISTORS FIND I ,Vbd , P (30k ) APPLY KVL TO THIS LOOP APPLY KVL TO THIS LOOP LOOP FOR Vbd Vbd 12 20[k ] I 0 (KVL) Vbd 10V POWER ON 30k RESISTOR P I 2 R (104 A) 2 (30 *103 ) 30mW v R Ri i i VOLTAGE DIVISION FOR MULTIPLE RESISTORS SINGLE NODE-PAIR CIRCUITS THESE CIRCUITS ARE CHARACTERIZED BY ALL THE ELMENTS HAVING THE SAME VOLTAGE ACROSS THEM - THEY ARE IN PARALLEL V EXAMPLE OF SINGLE NODE-PAIR V THIS ELEMENT IS INACTVE (SHORT-CIRCUITED) BASIC CURRENT DIVIDER Rp THE CURRENT DIVISION APPLY KCL THE CURRENT i(t) ENTERS THE NODE AND SPLITS - IT IS DIVIDED BETWEEN THE CURRENTS i1(t) AND i2(t) USE OHM’S LAW TO REPLACE CURRENTS DEFINE “PARALLEL RESISTANCE COMBINATION” i (t ) 1 v (t ) Rp v (t ) R1 R2 i (t ) R1 R2 4 1 I I I (5) I1 (5) 1mA 2 1 1 5 1 4 FIND I1 , I2 , VO WHEN IN DOUBT… REDRAW THE CIRCUIT TO HIGHLIGHT ELECTRICAL CONNECTIONS!! IS EASIER TO SEE THE DIVIDER 80k * I 2 24V FIRST GENERALIZATION: MULTIPLE SOURCES APPLY KCL TO THIS NODE EQUIVALENT SOURCE DEFINE “PARALLEL RESISTANCE COMBINATION” iO ( t ) 1 v (t ) Rp v (t ) R1 R2 iO ( t ) R1 R2 FIND VO AND THE POWER SUPPLIED BY THE SOURCES 6k 10mA 3k Rp 5mA VO 15mA VO 10V VO P15mA VO (15mA ) 6k * 3k Rp 2 k 6k 3k 150mW P6 mA VO (10mA ) 100mW SECOND GENERALIZATION: MULTIPLE RESISTORS APPLY KCL TO THIS NODE Ohm’s Law at every resistor v ( t ) RP i O ( t ) R v (t ) i K (t ) p iO (t ) ik (t ) Rk Rk General current divider SERIES PARALLEL RESISTOR COMBINATIONS UP TO NOW WE HAVE STUDIED CIRCUITS THAT CAN BE ANALYZED WITH ONE APPLICATION OF KVL(SINGLE LOOP) OR KCL(SINGLE NODE-PAIR) WE HAVE ALSO SEEN THAT IN SOME SITUATIONS IT IS ADVANTAGEOUS TO COMBINE RESISTORS TO SIMPLIFY THE ANALYSIS OF A CIRCUIT NOW WE EXAMINE SOME MORE COMPLEX CIRCUITS WHERE WE CAN SIMPLIFY THE ANALYSIS USING THE TECHNIQUE OF COMBINING RESISTORS… … PLUS THE USE OF OHM’S LAW SERIES COMBINATIONS PARALLEL COMBINATION G p G1 G2 ... GN FIRST WE PRACTICE COMBINING RESISTORS 3k SERIES 6k||3k (10K,2K)SERIES 6k || 12k 4k 5k 3k 12k 4k || 12k 12k FIRST REDUCE IT TO A SINGLE LOOP CIRCUIT SECOND: “BACKTRACK” USING KVL, KCL OHM’S 6k I3 KCL : I1 I 2 I 3 0 Va OHM' S : I 2 6k …OTHER OPTIONS... OHM' S : Vb 3k * I 3 12 I4 I3 4 12 Vb 4k * I 4 6k || 6k KCL : I 5 I 4 I 3 0 OHM' S : VC 3k * I 5 I1 12V 12k Va 3 (12) 39 2k || 2k 1k VOLTAGE DIVIDER : VO LEARNING BY DOING 1k (3V ) 1V 1k 2k 1k 1k 2k CURRENT DIVIDER : I O 1k (3 A) 1A 1k 2k AN EXAMPLE OF “BACKTRACKING” 1.5mA I1 3mA V xz 6V 3V 1.5mA 1mA VO 36V 3V 0.5mA A STRATEGY. ALWAYS ASK: “WHAT ELSE CAN I COMPUTE?” Vb 6k * I 4 I3 Vb 3k I2 I3 I4 Va 2k * I 2 V xz Va Vb V I 5 xz 4k I1 I 2 I 5 VO 6k * I1 V xz 4k * I1 FIND VO 60k V1 6V FIND VS 2V 30k || 60k 20k STRATEGY : FIND V1 USE VOLTAGE DIVIDER 9V 12V VOLTAGE DIVIDER 20k VO V1 20k 40k 0.05mA I1 VS 20k * 0.15mA 6V 20k 0.15mA 6V THIS IS AN INVERSE PROBLEM WHAT CAN BE COMPUTED? 20k + - V1 60k * 0.1mA V1 20k (12) 6V 20k 20k SERIES PARALLEL 6V 120k http://www.wiley.com/college/irwin/0470128690/animations/swf/D2Y.swf Y TRANSFORMATIONS THIS CIRCUIT HAS NO RESISTOR IN SERIES OR PARALLEL IF INSTEAD OF THIS WE COULD HAVE THIS THEN THE CIRCUIT WOULD BECOME LIKE THIS AND BE AMENABLE TO SERIES PARALLEL TRANSFORMATIONS Rab R2 || ( R1 R3 ) Y Rab Ra Rb Y Ra R1 Rb R1 R2 ( R1 R3 ) R R 1 2 R Ra Rb 3 Ra R R Ra b 3 R1 R2 R3 R1 R2 R3 Rb R2 RR R2 b 1 Rc R1 Rc REPLACE IN THE THIRD AND SOLVE FOR R1 R2 R3 R ( R R2 ) Rb Ra Rb Rb Rc Rc Ra Rb Rc 3 1 R1 R2 R3 R 1 R1 R2 R3 Rb R3 R1 Rc R R Rb Rc Rc Ra R1 R2 R3 R2 a b R1 ( R2 R3 ) Rc Rc Ra Y R1 R2 R3 R R Rb Rc Rc Ra R3 a b Ra SUBTRACT THE FIRST TWO THEN ADD TO THE THIRD TO GET Ra Y LEARNING EXAMPLE: APPLICATION OF WYE-DELTA TRANSFORMATION c R1 R3 R2 12k 6k 12k 6k 18k R1 R2 Ra R1 R2 R3 Rb R2 R3 R1 R2 R3 Rc R3 R1 R1 R2 R3 Y a b a c DELTA CONNECTION b COMPUTE IS REQ 6k 3k 9k || (2k 6k ) 10k IS 12V 1.2mA 12k ONE COULD ALSO USE A WYE - DELTA TRANSFORMATION ... CIRCUITS WITH DEPENDENT SOURCES A CONVENTION ABOUT DEPENDENT SOURCES. UNLESS OTHERWISE SPECIFIED THE CURRENT AND VOLTAGE VARIABLES ARE ASSUMED IN SI UNITS OF Amps AND Volts DEPENDENT VARIABLE GENERAL STRATEGY TREAT DEPENDENT SOURCES AS REGULAR SOURCES AND ADD ONE MORE EQUATION FOR THE CONTROLLING VARIABLE VA FIND VO VD I X CONTROLLING VARIABLE FOR THIS EXAMPLE THE MULTIPLIER MUST HAVE UNITS OF OHM OTHER DEPENDENT SOURCES VD VX ( scalar) I D VX ( Siemens) I D I X ( scalar) A PLAN: SINGLE LOOP CIRCUIT. USE KVL TO DETERMINE CURRENT KVL : 12 3k * I1 VA 5k * I1 0 ONE EQUATION, TWO UNKNOWNS. CONTROLLING VARIABLE PROVIDES EXTRA EQUATION AN ALTERNATIVE DESCRIPTIO N V UNITS VD I X , 2 mA ASSUMES CURRENT IN mA KVL ARE EXPLICIT VA 2k * I1 REPLACE AND SOLVE FOR THE CURRENT I1 2mA USE OHM’S LAW VO 5k * I1 10V KCL TO THIS NODE. THE DEPENDENT SOURCE IS JUST ANOTHER SOURCE FIND VO A PLAN: IF V_s IS KNOWN V_0 CAN BE DETERMINED USING VOLTAGE DIVIDER. TO FIND V_s WE HAVE A SINGLE NODE-PAIR CIRCUIT THE EQUATION FOR THE CONTROLLING VARIABLE PROVIDES THE ADDITIONAL EQUATION ALGEBRAICALLY, THERE ARE TWO UNKNOWNS AND JUST ONE EQUATION SUBSTITUTION OF I_0 YIELDS VOLTAGE DIVIDER * / 6k 5VS 60 VO NOTICE THE CLEVER WAY OF WRITING mA TO HAVE VOLTS IN ALL NUMERATORS AND THE SAME UNITS IN DENOMINATOR 4k 2 VS (12)V 4k 2k 3 FIND VO KVL TO THIS LOOP A PLAN: ONE LOOP PROBLEM. FIND THE CURRENT THEN USE OHM’S LAW. THE DEPENDENT SOURCE IS ONE MORE VOLTAGE SOURCE THE EQUATION FOR THE CONTROLLING VARIABLE PROVIDES THE ADDITIONAL EQUATION REPLACE AND SOLVE FOR CURRENT I … AND FINALLY FIND G vO ( t ) vi (t ) KCL A PLAN: ONE LOOP ON THE LEFT - KVL ONE NODE-PAIR ON RIGHT - KCL KVL KVL KCL gm v g ( t ) ALSO A VOLTAGE DIVIDER vO ( t ) 0 RL