FIND THE EQUIVALENT RESISTANCE USING THE RULES 使用规则发现的等效电阻.ppt

SOLUTION OF ELECTRIC CIRCUIT,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 LAW,S1,S2,R1,R2,R3,R4,R5,R6,V1,V2,C1,C2,WE 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 FOLLOWING WORDS ABOUT THE ELECTRIC CIRCUITS:ELECTRIC CURRENT _CHARGE _ELECTROMOTIVE FORCE _CONDUCTOR _VOLTAGE _WIRE _WIRES JOINED _BRANCH _ARM _CAPACITOR _AMMETER _RESISTANCE _DIRECT CURRENT(DC)_VOLTMETER _,READING 1 ELECTRIC CURRENT,AN ELECTRIC CURRENT IS A FLOW OF ELECTRIC CHARGE AND IT IS DEFINED AS THE AMOUNT 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 DIRECTION 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 POTENTIAL 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 OPPOSITE 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 CIRCUIT IT MUST BE PLACED IN SERIES WITH THE PARTS THROUGH WHICH THE CURRENT TO BE MEASURED IS PASSING.,READING 2 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 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 CALLED 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.,READING 3 OHMS LAW,IN 1826 A GERMAN SCIENTIST GEORG SIMON OHM,WHO EXPERIMENTED WITH 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 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 OTHER OHMIC DEVICE)DIVIDING VOLTAGE BY RESISTANCE:,READING 4 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 AND WE CALL THIS EQUIVALENT RESISTANCE R12,IF R1 AND R2 ARE CONNECTED IN PARALLEL,IF R1 AND R2 ARE CONNECTED IN SERIES,R1,R2,R12,R1,R2,R12,i,i1,i2,PARALLEL CONNECTION,TEST 1:MATCH EACH CIRCUIT SYMBOL WITH ITS COMPONENT,A,V,TEST 2:COMPLETE THE FOLLOWING SENTENCES:IN THE S.I.:THE AMPERE IS THE UNIT _ 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 ARE 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 THE 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?,EXERCISE 1:FIND THE EQUIVALENT RESISTANCE USING THE RULES OF RESISTORS IN SERIES OR IN PRARALLEL R1=20;R2=40;R3=35;R4=15.COMPLETE:,R1,R2,R3,R4,R1,R1,R23,R4,R234,R1234,TEST 3: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 AND INSERT AN AMMETER AND A VOLTMETER IN THE CORRECT PLACE TO MEASURE THE CURRENT AND THE POTENTIAL DIFFERENCE ACROSS THE RESISTOR R5,R1,R2,R3,R4,+,R5,HOW CAN WE USE THE CONCEPTS,EQUIVALENT RESISTANCEOHMS LAWTO SOLVE THE ELECTRIC CIRCUITS?EQUIVALENT RESISTANCEREPLACING EACH GROUP OF RESISTORS WITH THEIR 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 RESISTANCEFIRST 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:,EXERCISE 2:CONSIDER THE ELECTRIC CIRCUIT SHOWN IN THE DIAGRAM BELOW(THE INTERNAL RESISTANCE OF THE BATTERY CAN BE IGNORED),A1,A2,S1,S2,R1,R2,R3,R4,EMF,+,THERE ARE FOUR POSSIBLE CIRCUIT DIAGRAMS:,(a),R1,R2,R3,R123,(b),R1,R2,R4,(c),R1,R2,R4,(d),R1,R2,R3,R4,WHAT DOES EACH VOLTMETER IN THE CIRCUIT BELOW INDICATE?,V0,V2,V4,V4,V2,V0,+,S1,S2,R1,R2,R3,R4,EXERCISE 3:THREE IDENTICAL LAMPS(EACH BULB HAS A RESISTANCE 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),+,S1,S2,L1,L2,L3,A,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 WHICH 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,