FINDTHEEQUIVALENTRESISTANCEUSINGTHERULES使用规则发现的等效电阻课件.pptx

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1、ELECTRIC CIRCUITAN ELECTRIC CIRCUIT IS A CONFIGURATION OF ELECTRONIC COMPONENTS THROUGH WHICH ELECTRICITY IS MADE TO FLOW.THE FIGURE BELOW SHOWS A TYPICAL EXAMPLE:OBJECTIVE:WE ARE GOING TO LEARN TO SOLVE SIMPLE ELECTRIC CIRCUITS USINGONLY:THE CONCEPT OF“EQUIVALENT RESISTANCE”FIRST OHMS LAWS1S2R1 R2R

2、3 R4 R5R6 V1V2 C1 C2WE ARE GOING TO READ FOUR SHORT TEXTS ON THE FOLLOWING TOPICS:ELECTRIC CURRENT READING 1ELECTROMOTIVE FORCES READING 2EQUIVALENT RESISTANCE READING 3FIRST OHMS LAW READING 4THEY ARE ESSENTIAL TO SOLVE ELECTRIC CIRCUIT(SIMPLEELECTRIC CIRCUIT)WRITE THE ITALIAN TRANSLATION OF THE FO

3、LLOWING WORDS ABOUT THE ELECTRIC CIRCUITS:ELECTRIC CURRENT _CHARGE _ELECTROMOTIVE FORCE _CONDUCTOR _VOLTAGE _WIRE _WIRES JOINED _BRANCH _ARM _CAPACITOR _AMMETER _RESISTANCE _DIRECT CURRENT(DC)_VOLTMETER _ ELECTRIC CURRENTAN ELECTRIC CURRENT IS A FLOW OF ELECTRIC CHARGE AND IT IS DEFINED AS THE AMOUN

4、T OF ELECTRIC CHARGE(MEASURED IN COULOMB)FLOWING THROUGH THE SURFACE IN THE TIME t :THE S.I.UNIT OF ELECTRIC CURRENT IS THE AMPERE(A),WHICH EQUALS A FLOW OF ONE COULOMB OF CHARGE PER SECOND.A DIRECT CURRENT(DC)IS A UNIDIRECTIONAL FLOW.CURRENT IS A SCALAR QUANTITY,BUT IN CIRCUIT ANALYSIS THE DIRECTIO

5、N OF CURRENT IS RELEVANT AN IS INDICATED BY ARROWS.CONVENTIONAL CURRENT:FOR HISTORICAL REASONS,ELECTRIC CURRENT IS SAID TO FLOW FROM THE POSITIVE PART OF A CIRCUIT TO THE MOST NEGATIVE PART(THIS WAS GUESSED AT BEFORE THE ELECTRONS WERE DISCOVERED).AN ELECTRIC CURRENT WILL ONLY FLOW WHEN:THERE IS A P

6、OTENTIAL DIFFERENCE BETWEEN TWO POINTS AND THE TWO POINTS ARE CONNECTED BY A CONDUCTOR.IN SOLID METALS,LIKE WIRES,THE POSITIVE CHARGE CARRIERS ARE MOTIONLESS,AND ONLY THE NEGATIVELY CHARGED ELECTRONS FLOW.AS THE ELECTRONS CARRY NEGATIVE CHARGE,THE ELECTRON CURRENT IS IN THE DIRECTION WHICH IS OPPOSI

7、TE TO THAT OF THE CONVENTIONAL(OR ELECTRIC)CURRENT.MOST FAMILIAR CONDUCTORS ARE METALLIC,HOWEVER,THERE ARE ALSO MANY NON-METALLIC CONDUCTORS,INCLUDING GRAPHITE,SOLUTIONS OF SALT,AND ALL PLASMAS.THE ISTRUMENT WHICH IS USED TO MEASURE THE CURRENT FLOWING IN A CONDUCTOR IS CALLED AN AMMETER.IN A CIRCUI

8、T IT MUST BE PLACED IN SERIES WITH THE PARTS THROUGH WHICH THE CURRENT TO BE MEASURED IS PASSING.tQi ELECTROMOTIVE FORCE(EMF)THE ELECTROMOTIVE FORCE(OFTEN ABBREVIATE“EMF”AND DENOTED )IS AN ARCHAIC TERM THAT INDICATES AN ELECTRIC POTENTIAL.WHEN WE COSIDER AN“IDEAL BATTERY”(THE INTERNAL RESISTANCE IS

9、ZERO)THE POTENTIAL DIFFERENCE ACROSS THE BATTERY EQUALS THE ELECTROMOTIVE FORCE(=).THE ELECTROMOTIVE FORCE,WHICH TRIES TO MOVE A POSITIVE CHARGE FROM A HIGHER TO A LOWEL POTENTIAL,THERE MUST BE ANOTHER FORCE TO MOVE CHARGE FROM A POTENTIAL TO A HIGHER INSIDE THE BATTERY.THIS SO-CALLED FORCE IS CALLE

10、D THE ELECTROMOTIVE FORCE,OR EMF.THE INSTRUMENT WHICH IS USED TO MEASURE THE POTENTIAL DIFFERENCE BETWEEN TWO POINTS IN A CIRCUIT IS THE VOLTMETER.IT IS CONNECTED IN PARALLEL WITH THE TWO POINTS WHOSE POTENTIAL IS BEING MEASURED.OHMS LAWIN 1826 A GERMAN SCIENTIST GEORG SIMON OHM,WHO EXPERIMENTED WIT

11、H CIRCUITS,FOUND OUT THE RELATIONSHIPS BETWEEN CURRENT AND VOLTAGE:THE POTENTIAL DIFFERENCE BETWEEN THE ENDS OF A METALLIC CONDUCTOR IS DIRECTLY PROPORTIONAL TO THE CURRENT FLOWING.IT IS CALLED OHMS LAW AND CAN BE FORMALLY DEFINED AS FOLLOWS(TEMPERATURE IS CONSTANT):THE VALUE OF GIVES AN INDICATION

12、OF HOW A CURRENT CAN REALLY FLOW IN A PARTICULAR CONDUCTOR AND IT IS CALLED RESISTANCE.HENCE CAN BE WRITTEN THIS FORMULA IS OFTEN EXPRESSED MATHEMATICALLY AS WHERE:V IS THE APPLIED VOLTAGE,I IS THE CURRENT,R IS THE RESISTANCEWITH OHMS LAW IS POSSIBLE TO CALCULATE THE CURRENT IN AN (IDEAL)RESISTOR(OR

13、 OTHER OHMIC DEVICE)DIVIDING VOLTAGE BY RESISTANCE:IVCONSTANTIVRIViRVRVI EQUIVALENT RESISTANCEIF TWO(OR MORE)RESISTORS R1 AND R2(OR R1,RN)ARE CONNECTED IN SERIES OR IN PARALLEL,WE CAN REPLACE THEM WITH THEIR EQUIVALENT RESISTANCE AND REDRAW THE CIRCUIT,IN THIS CASE WE GET A SEMLIFIED VERSION CIRCUIT

14、 AND WE CALL THIS EQUIVALENT RESISTANCE R12 IF R1 AND R2 ARE CONNECTED IN PARALLELIF R1 AND R2 ARE CONNECTED IN SERIESTHE VALUE OF THE EQUIVALENT RESISTANCE R12 IS GIVEN BY:R1 R2 R12R1R2R122112111RRR2112RRRNnRRR1.111.1NNRRR.1.1THE TOTAL CURRENT I THROUGH THE COMBINATION EQUALS THE SUM OF CURRENTS IN

15、 EACH BRANCH OF THE CIRCUIT21iiiSERIES CONNECTIONi i1 i2PARALLEL CONNECTIONMATCH EACH CIRCUIT SYMBOL WITH ITS COMPONENTWIRERESISTORVOLTMETERLAMP(LIGHTING)WIRES JOINED (JUNCTION,NODE)AMMETERBATTERYSWITCHBRANC(ARM)CAPACITOR SWITCH A V COMPLETE THE FOLLOWING SENTENCES:IN THE S.I.:THE AMPERE IS THE UNIT

16、 _ ITS SYMBOL IS _THE VOLT IS THE UNIT _ ITS SYMBOL IS _THE COULOMB IS THE UNIT _ ITS SYMBOL IS _THE ELECTRIC CURRENT IS GIVEN BY _THE UNIT OF THE ELECTRICAL CURRENT IS _ DEFINED AS _THE DIRECTION OF CURRENT IS RELEVANT AND IT IS INDICATED BY _WHAT IS THE ELECTRIC CURRENT?_CHARGE,CURRENT AND TIME AR

17、E RELATED BY _IF THE CHARGE ON 1 ELECTRON IS_,FIND HOW MANY ELECTRONS ARE INVOLVED IF A CURRENT FLOW RESULT IN THE MOVEMENT OF 3.60 105 OF CHARGE:_HOW MANY ELECTRONS FLOW THROUGH A BATTERY THAT DELIVERS A CURRENT OF 1.5 A FOR 10 s?_THE POTENTIAL DIFFERENCE(VOLTAGE)ACROSS AN IDEAL CONDUCTOR IS _TO TH

18、E CURRENT THROUGH IT.USE OHMS LAW TO FIND THE POYENTIAL DIFFERENCE BETWEEN TWO POINTS INCLUDING A RESISTANCE R=8 WHEN THIS IS RUN THROUGH BY A CURRENT OF 0.25A?FIND THE EQUIVALENT RESISTANCE USING THE RULES OF RESISTORS IN SERIES OR IN PRARALLEL R1=20 ;R2=40 ;R3=35 ;R4=15 .COMPLETE:R1R2R3 R4 R1 R1 R

19、23 R4 R234 R1234 CONNECTION EQUIVALENT RESISTANCE R2 R3 in seriesR23=R2+R3=.EQUIVALENT RESISTANCE=CONSIDER THE CIRCUIT IN THE FOLLOWING FIGURE:HOW MANY NODES ARE THERE?_HOW MANY BRANCHES ARE THERE?_DRAW AN ARROW FOR EACH BRANCH TO INDICATE THE DIRECTION OF CONVENTIONAL CURRENTDRAW AGAIN THE CIRCUIT

20、AND INSERT AN AMMETER AND A VOLTMETER IN THE CORRECT PLACE TO MEASURE THE CURRENT AND THE POTENTIAL DIFFERENCE ACROSS THE RESISTOR R5R1 R2R3R4+R5 HOW CAN WE USE THE CONCEPTS EQUIVALENT RESISTANCEOHMS LAWTO SOLVE THE ELECTRIC CIRCUITS?EQUIVALENT RESISTANCEREPLACING EACH GROUP OF RESISTORS WITH THEIR

21、EQUIVALENT RESISTANCE AND REDRAWING THE CIRCUIT,WE GET A NEW SEMPLIFIED VERSION OF CIRCUIT.WE CONTINUE TO SIMPLIFY THE NEW VERSION OF THE CIRCUIT,AS FOR AS WE GET A SOURCE OF ELECTRICAL ENERGY THAT WILL PRODUCE A POTENTIAL DIFFERENCE BETWEEN TWO POINTS WHICH IS CONNECTED WITH ONLY ONE RESISTANCEFIRS

22、T OHMS LAWTO DETERMINE:THE CURRENT IN A RESISTOR(R)WHICH IS GIVEN BY VOLTAGE(V)DIVIDED BY RESISTANCE:THE POTENTIAL DIFFERENCE BETWEEN TWO POINTS WHICH INCLUDE A RESISTANCE(R)IS GIVEN BY THE PRODUCT OF THE RESISTANCE AND THE CURRENT FLOWING THROUGH THE RESISTANCE:RVi iRVCONSIDER THE ELECTRIC CIRCUIT

23、SHOWN IN THE DIAGRAM BELOW(THE INTERNAL RESISTANCE OF THE BATTERY CAN BE IGNORED)IN THIS EXERCISE WE ARE GOING TO LEARN HOW TO CALCOLATE THE CURRENT FLOWING IN EACH ARM OF THE CIRCUIT.EXAMINE THE CIRCUIT DIAGRAM AND LIST ITS COMPONENTS:ONE BATTERY _ _ _ _ LETS REDRAW THE CIRCUIT DIAGRAM WITHOUT DRAW

24、ING:THE INSTRUMENT USED TO MEASURE IT THE BRANCHES WHERE THERE IS AN OPEN SWITCH(SINCE THEY ARENT RUN THROUGH BY ELECTRIC CURRENT)A1 A2 S1 S2R1 R2 R3 R4 EMF+EFM=10 VR1=2 R2=3 R3=4 R4=6 THERE ARE FOUR POSSIBLE CIRCUIT DIAGRAMS:(a)(b)(c)(d)SWITCHESCIRC.DIAGRCONNECTIONS1 AND S2 OPENS1 OPEN AND S2 CLOSE

25、DS1 CLOSED AND S2 OPEN (a)R1-R2-R3 IN SERIESS1 AND S2 CLOSEDCOMPLETE:(a)USE “EQUIVALENT RESISTANCE”V1+V2+V3=.V1=.=V2=.=V3=R3 i=R123=USE FIRST OHMS LAWR1 R2 R3R123.i(b)USE “EQUIVALENT RESISTANCE”V1+V2+V3=.V1=.=V2=.=V4=.=.USE FIRST OHMS LAWR1.iR2 R4(c)USE “EQUIVALENT RESISTANCE”USE FIRST OHMS LAWR1 R2

26、 R4(d)USE “EQUIVALENT RESISTANCE”i12=i4=.i3=V34=V12=i=USE OHMS LAWR1 R2R3 R4 WHAT DOES EACH VOLTMETER IN THE CIRCUIT BELOW INDICATE?V0V2 V4 SWITCHES S1 AND S2 OPEN S1 OPEN AND S2 CLOSED S1 CLOSED AND S2 OPEN S1 AND S2 CLOSEDCOMPLETE:V4 V2 V0+S1 S2 R1 R2 R3R4 THREE IDENTICAL LAMPS(EACH BULB HAS A RES

27、ISTANCE R)ARE CONNECTED AS SHOWN IN THE CIRCUIT DIAGRAM BELOW.THE POTENTIAL DIFFERENCE ACROSS THE BATTERY IS 5.7 V(THE INTERNAL RESISTANCE OF THE BATTERY CAN BE IGNORED)CONSIDER THE POSITION(OPEN/CLOSED)OF THE SWITCHES S1 AND S2 AND PUT(V)FOR EACH LIGHTED LAMP(L1,L2,L3)POSITION SWITCHES L1 L2 L3 S1

28、AND S2 CLOSED S1 CLOSED AND S2 OPEN S1 OPEN AND S2 CLOSED S1 AND S2 OPEN+S1S2L1 L2 L3A LETS CONSIDER NOW THE SWITCH S1 CLOSED;DESCRIBE WHAT HAPPENS TO THE GROUP OF PARALLEL LAMPS L2 AND L3 WHEN WE CLOSE THE SWITCH S2.WHY?LABORATORY EXPERIMENTCHECK YOUR ANSWERS THERE ARE MORE COMPLICATED CIRCUITS WHI

29、CH CANNOT BE SIMPLY REDUCED TO A PARALLEL OR SERIES CIRCUIT USING EQUIVALENT RESISTANCES.THESE ONES NEED TO BE SOLVED USING TWO LAWS:KIRCHHOFFS CURRENT LAW;KIRCHHOFFS VOLTAGE LAW.OFTEN,WHEN WE USE KIRCHHOFFS LAWS,WE GET A LOT OF EQUATIONS WHICH ARE COMPLICATED TO SOLVE.THE ANALISYS OF THIS CIRCUIT IS MORE SIMPLE IF WE USE THE FOLLOWING LAWS:THEVENINNORTON

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