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Chapter 1 Basic Electric Circuit Concepts - 義守大學

1 Chapter 1 Basic Electric Circuit Concepts2 Basic CONCEPTSLEARNING GOALS System of Units: The SI standard system Systeme International unit (=International System of Units) ( ) Basic Quantities: Charge, current, voltage, power and energy Circuit Elements: Active and Passive3 at the foundation ofmodern science and technologyfrom the Physics Laboratory of NIST Detailed contents Values of the constants and related informationSearchable bibliography on the constants In-depth information on the SI, the modernmetric system Guidelines for the expressionof uncertainty in measurement About this reference. Feedback. Privacy Statement / Security Notice - NIST Disclaimer 6SI DERIVED Basic ELECTRICAL UNITS7 CURRENT AND VOLTAGE RANGES8 Strictly speaking current is a Basic quantity and charge is derived. However,physically the Electric current is created by a movement of charged particles.

Chapter 1 Basic Electric Circuit Concepts. 2 BASIC CONCEPTS LEARNING GOALS

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Transcription of Chapter 1 Basic Electric Circuit Concepts - 義守大學

1 1 Chapter 1 Basic Electric Circuit Concepts2 Basic CONCEPTSLEARNING GOALS System of Units: The SI standard system Systeme International unit (=International System of Units) ( ) Basic Quantities: Charge, current, voltage, power and energy Circuit Elements: Active and Passive3 at the foundation ofmodern science and technologyfrom the Physics Laboratory of NIST Detailed contents Values of the constants and related informationSearchable bibliography on the constants In-depth information on the SI, the modernmetric system Guidelines for the expressionof uncertainty in measurement About this reference. Feedback. Privacy Statement / Security Notice - NIST Disclaimer 6SI DERIVED Basic ELECTRICAL UNITS7 CURRENT AND VOLTAGE RANGES8 Strictly speaking current is a Basic quantity and charge is derived. However,physically the Electric current is created by a movement of charged particles.

2 +++)(tq+What is the meaning of a negative value for q(t)?PROBLEM SOLVING TIPPROBLEM SOLVING TIPIF THE CHARGE IS GIVEN DETERMINE THE CURRENT BYDIFFERENTIATIONIF THE CURRENT IS KNOWN DETERMINE THE CHARGE BYINTEGRATIONA PHYSICAL ANALOGY THAT HELPS VISUALIZE ELECTRICCURRENTS IS THAT OF WATER FLOW. CHARGES ARE VISUALIZED AS WATER PARTICLES9+++)(tq+EXAMPLE])[120sin(104)( 3 Cttq ==)(ti)120cos(1201043t ][A][)120cos( )(mAtti =EXAMPLE <= 000)(2tmAettitFIND THE CHARGE THAT PASSES DURING IN THE INTERVAL 0<t<1== 102dxeqx)21(212102102eeex = FIND THE CHARGE AS A FUNCTION OF TIME ==ttxdxedxxitq2)()(0)(0= tqt == >ttxedxetqt022)1(21)(0 And the units for the charge?..)1(212 =eqUnits?1012345610 102030 Charge(pC)Time(ms)Here we are given thecharge flow as functionof time.)/(1010010210101010931212sCsCm = =12345610 102030 Time(ms)) Current(nA4020 DETERMINE THE CURRENTTo determine current we must take ATTENTION TOUNITS11 CONVENTION FOR CURRENTSIT IS ABSOLUTELY NECESSARY TO INDICATETHE DIRECTION OF MOVEMENT OF CHARGED UNIVERSALLY ACCEPTED CONVENTION INELECTRICAL ENGINEERING IS THAT CURRENT ISFLOW OF POSITIVE WE INDICATE THE DIRECTION OF FLOWFOR POSITIVE CHARGES -THE REFERENCE DIRECTION-abaaabbbA3A3 A3A3 THE DOUBLE INDEX NOTATIONIF THE INITIAL AND TERMINAL NODE ARELABELED ONE CAN INDICATE THEM AS SUBINDICES FOR THE CURRENT NAMEabA5 AIab5=AIab3=AIba3 =AIab3 =AIba3=POSITIVE CHARGESFLOW LEFT-RIGHTPOSITIVE CHARGESFLOW RIGHT-LEFT baabII =A POSITIVE VALUE FORTHE CURRENT INDICATESFLOW IN THE DIRECTIONOF THE ARROW (THEREFERENCE DIRECTION)

3 A NEGATIVE VALUE FORTHE CURRENT INDICATESFLOW IN THE OPPOSITEDIRECTION THAN THE REFERENCE DIRECTION12 This example illustrates the various waysin which the current notation can be usedbaIA3== =abcbIAIAI42A2c13 CONVENTIONS FOR VOLTAGES ONE DEFINITION FOR VOLTTWO POINTS HAVE A VOLTAGE DIFFERENTIAL OFONE VOLT IF ONE COULOMB OF CHARGE GAINS (OR LOSES) ONE JOULE OF ENERGY WHEN IT MOVES FROM ONE POINT TO THE OTHER+abC1IF THE CHARGE GAINSENERGY MOVING FROMa TO b THEN b HAS HIGHERVOLTAGE THAN IT LOSES ENERGY THENb HAS LOWER VOLTAGETHAN aDIMENSIONALLY VOLT IS A DERIVED UNITsAmN ==COULOMBJOULEVOLTVOLTAGE IS ALWAYS MEASURED IN A RELATIVE FORM AS THE VOLTAGE DIFFERENCEBETWEEN TWO POINTSIT IS ESSENTIAL THAT OUR NOTATION ALLOWS US TO DETERMINE WHICH POINT HAS THE HIGHER VOLTAGE14 THE + AND - SIGNS DEFINE THE REFERENCEPOLARITYVIF THE NUMBER V IS POSITIVE POINT A HAS VVOLTS MORETHAN POINT THE NUMBER V IS NEGATIVE POINT A HAS|V| LESSTHAN POINT BPOINT A HAS 2V MORETHAN POINT BPOINT A HAS 5V LESSTHAN POINT B15 THE TWO-INDEX NOTATION FOR VOLTAGESINSTEAD OF SHOWING THE REFERENCE POLARITYWE AGREE THAT THE FIRST SUBINDEX DENOTESTHE POINT WITH POSITIVE REFERENCE POLARITYVVAB2=VVAB5 =VVBA5=BAABVV =16 ENERGYVOLTAGE IS A MEASURE

4 OF ENERGY PER UNIT MOVING BETWEEN POINTS WITH DIFFERENT VOLTAGE ABSORB ORRELEASE ENERGY THEY MAY TRANSFER ENERGY FROM ONE POINT TO ANOTHERBASIC FLASHLIGHTC onverts energy stored in batteryto thermal energy in lamp filamentwhich turns incandescent and glowsThe battery supplies energy to absorbs energy from net effect is an energy transferEQUIVALENT CIRCUITC harges gainenergy hereCharges supplyEnergy here17 WHAT ENERGY IS REQUIRED TO MOVE 120[C] FROMPOINT B TO POINT A IN THE Circuit ?VVAB2=JVQWQWV240== =THE CHARGES MOVE TO A POINT WITH HIGHERVOLTAGE -THEY GAINED (OR ABSORBED) ENERGYTHE Circuit SUPPLIED ENERGY TO THE CHARGESENERGYVOLTAGE IS A MEASURE OF ENERGY PER UNIT MOVING BETWEEN POINTS WITH DIFFERENT VOLTAGE ABSORB ORRELEASE ENERGY18 EXAMPLEA CAMCODER BATTERY PLATE CLAIMS THATTHE UNIT STORES 2700mAHr AT IS THE TOTAL CHARGE AND ENERGYSTORED?CHARGETHE NOTATION 2700mAHr INDICATES THATTHE UNIT CAN DELIVER 2700mA FOR ONE FULL HOUR][ = = TOTAL ENERGY STOREDTHE CHARGES ARE MOVED THROUGH A DIFFERENTIAL][ ][ ][43 JJCJVCQW = = =ENERGY AND POWER2[C/s] PASSTHROUGH THE ELEMENTEACH COULOMB OF CHARGE LOSES 3[J]OR SUPPLIES 3[J] OF ENERGY TO THE ELEMENTTHE ELEMENT RECEIVES ENERGY AT A RATE OF 6[J/s]THE Electric POWER RECEIVED BY THEELEMENT IS 6[W]HOW DO WE RECOGNIZE IF AN ELEMENTSUPPLIES OR RECEIVES POWER?

5 VIP=IN GENERAL =21)(),(12ttdxxpttw19 PASSIVE SIGN CONVENTION( )POWER RECEIVED IS POSITIVEWHILE POWERSUPPLIED IS CONSIDERED NEGATIVEA CONSEQUENCE OF THIS CONVENTION IS THATTHE REFERENCE DIRECTIONS FOR CURRENT ANDVOLTAGE ARE NOT INDEPENDENT -- IF WE ASSUME PASSIVE ELEMENTSab +abVabIababIVP=IF VOLTAGE AND CURRENTARE BOTH POSITIVE THECHARGES MOVE FROM HIGH TO LOW VOLTAGE AND THE COMPONENTRECEIVES ENERGY--IT ISA PASSIVE ELEMENTab +abVGIVEN THE REFERENCE POLARITY ababIIF THE REFERENCE DIRECTION FOR CURRENTIS GIVEN +THIS IS THE REFERENCE FOR POLARITYREFERENCE DIRECTION FOR CURRENTab +abVVVab10 =EXAMPLETHE ELEMENT RECEIVES 20W OF IS THE CURRENT? abISELECT REFERENCE DIRECTION BASED ONPASSIVE SIGN CONVENTION abababIVIVW)10(][20 ==][2 AIab =A220 Voltage(V)Current A - A'S1S2positivepositivesupplies receivespositivenegativereceives suppliesnegativepositivereceives suppliesnegativenegativesupplies receivesS2S1We must examine the voltage across the componentand the current through it0,0<>ABABIV1 SON 0,0''''>>BABAIV2 SON AA BB ''''21 BABASABABSIVPIVP==UNDERSTANDING PASSIVE SIGN CONVENTION0,0''''> <BABAIVS2 ON +VI21 CHARGES RECEIVE BATTERY SUPPLIES ENERGYCHARGES LOSE BATTERY RECEIVES THE ENERGYWHAT WOULD HAPPEN IF THE CONNECTIONS ARE REVERSEDIN ONE OF THE BATTERIES?

6 22 Determine whether the elementsare supplying or receiving powerand how =WP8 =SUPPLIES POWERVVab2=2A 2abIA= 4PW= ABSORBS POWER231212 AIVV4,121212 ==AIVV2,41212==Determine the amount of power absorbed or supplied by the elements ?24 SELECT VOLTAGE REFERENCE POLARITYBASED ON CURRENT REFERENCE DIRECTION +)5(][20 AVWAB = ][4 VVAB = +IVW =)5(][40][8AI =25 SELECT HERE THE CURRENT REFERENCE DIRECTIONBASED ON VOLTAGE REFERENCE POLARITYA2 )2(][401 AVW =][201VV =IVW = ])[10(][50][5AI =WHICH TERMINAL HAS HIGHER VOLTAGE AND WHICH IS THE CURRENT FLOW DIRECTION26+- +V24 +V6 +V18A2A2123P1 = 12WP2 = 36WP3 = -48W)2)(6(1 AVP=)2)(18(2 AVP=)2)(24()2)(24(3 AVAVP = =IMPORTANT: NOTICE THE POWER BALANCE IN THE CIRCUITCOMPUTE POWER ABDORBED OR SUPPLIED BY EACH ELEMENTT ellegen s theorem: the sum of the powers absorbed by all elements in an electrical network is zero. Another statement of this theorem is that the power supplied in a network is exactly equal to the power ELEMENTSPASSIVE ELEMENTSINDEPENDENT SOURCESVOLTAGE DEPENDENTSOURCESCURRENTDEPENDENTSOURCES?

7 ,,, rgFORUNITS28 EXERCISES WITH DEPENDENT SOURCESOVFIND][40 VVO=OIFINDmAIO50=29 DETERMINE THE POWER SUPPLIED BY THE DEPENDENT SOURCES][40V][80])[2])([40(WAVP = =TAKE VOLTAGE POLARITY REFERENCETAKE CURRENT REFERENCE DIRECTION][160])[44])([10(WAVP = =30 POWER ABSORBED OR SUPPLIED BY EACHELEMENT][48)4)(12(1 WAVP==][48)2)(24(2 WAVP==][56)2)(28(3 WAVP==][8)2)(4()2)(1(WAVAIPxDS = = =][144)4)(36(36 WAVPV = =NOTICE THE POWER BALANCE31 USE POWER BALANCE TO COMPUTE IoW12 ))(6(OI)9)(12( )3)(10( )8)(4( )11)(28( ][1 AIO=POWER BALANCE


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