Chapter 1 Basic Electric Circuit Concepts

Transcription

1 Chapter 1 Basic Electric Circuit Concepts 1

2 BASIC CONCEPTS LEARNING GOALS System of Units: The SI standard system Systeme International unit (=International System of Units) ( 法 ) 國 際 單 位 制 Basic Quantities: Charge, current, voltage, power and energy Circuit Elements: Active and Passive 2

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5 Detailed contents Information at the foundation of modern science and technology from the Physics Laboratoryof NIST Values of the constantsand related information Searchable bibliographyon the constants In-depth information on the SI, the modern metric system Guidelinesfor the expression of uncertainty in measurement About this reference. Feedback. Privacy Statement / Security Notice - NIST Disclaimer 5

6 SI DERIVED BASIC ELECTRICAL UNITS 6

7 CURRENT AND VOLTAGE RANGES 7

8 Strictly speaking current is a basic quantity and charge is derived. However, physically the electric current is created by a movement of charged particles. What is the meaning of a negative value for q(t)? PROBLEM SOLVING TIP IF THE CHARGE IS GIVEN DETERMINE THE CURRENT BY DIFFERENTIATION IF THE CURRENT IS KNOWN DETERMINE THE CHARGE BY INTEGRATION q(t) A PHYSICAL ANALOGY THAT HELPS VISUALIZE ELECTRIC CURRENTS IS THAT OF WATER FLOW. CHARGES ARE VISUALIZED AS WATER PARTICLES 8

9 EXAMPLE EXAMPLE q(t) q( t) = sin(120π t)[ C] i (t) = π cos(120π t) [A] i( t) = 0.480π cos(120π t)[ ma] q i e ( t) = 2t 0 t < 0 ma t 0 FIND THE CHARGE THAT PASSES DURING IN THE INTERVAL 0<t<1 1 = 2 x e dx = e x = e ( e ) q 1 2 = (1 e 2 ) Units? FIND THE CHARGE AS A FUNCTION OF TIME q( t) = t i( x) dx = t e 2 x dx t 0 q( t) = 0 t > 0 q( t) = t 0 e 2 x And the units for the charge?... dx = 1 2 (1 e 2t 9 )

10 DETERMINE THE CURRENT Here we are given the charge flow as function of time. Charge(pC) m = C 9 = ( C / s) s Time(ms) To determine current we must take derivatives. PAY ATTENTION TO UNITS Current(nA) Time(ms) 10

11 CONVENTION FOR CURRENTS IT IS ABSOLUTELY NECESSARY TO INDICATE THE DIRECTION OF MOVEMENT OF CHARGED PARTICLES. A POSITIVE VALUE FOR THE CURRENT INDICATES FLOW IN THE DIRECTION OF THE ARROW (THE REFERENCE DIRECTION) A NEGATIVE VALUE FOR THE CURRENT INDICATES FLOW IN THE OPPOSITE DIRECTION THAN THE REFERENCE DIRECTION a THE UNIVERSALLY ACCEPTED CONVENTION IN ELECTRICAL ENGINEERING IS THAT CURRENT IS FLOW OF POSITIVE CHARGES. AND WE INDICATE THE DIRECTION OF FLOW FOR POSITIVE CHARGES -THE REFERENCE DIRECTIONa a THE DOUBLE INDEX NOTATION IF THE INITIAL AND TERMINAL NODE ARE LABELED ONE CAN INDICATE THEM AS SUBINDICES FOR THE CURRENT NAME 5A I A b ab = 5 3A I ab = 3A 3A I ba = 3A b b POSITIVE CHARGES FLOW LEFT-RIGHT I = a a 3A I ab = 3A 3A I ba = 3A b b POSITIVE CHARGES FLOW RIGHT-LEFT ab I ba 11

12 a I = 2A I 2A b I I cb ab c = = 4A 3A This example illustrates the various ways in which the current notation can be used 12

13 CONVENTIONS FOR VOLTAGES ONE DEFINITION FOR VOLT TWO POINTS HAVE A VOLTAGE DIFFERENTIAL OF ONE VOLT IF ONE COULOMB OF CHARGE GAINS (OR LOSES) ONE JOULE OF ENERGY WHEN IT MOVES FROM ONE POINT TO THE OTHER + a 1C b IF THE CHARGE GAINS ENERGY MOVING FROM a TO b THEN b HAS HIGHER VOLTAGE THAN a. IF IT LOSES ENERGY THEN b HAS LOWER VOLTAGE THAN a DIMENSIONALLY VOLT IS A DERIVED UNIT VOLT JOULE = = COULOMB N m A s VOLTAGE IS ALWAYS MEASURED IN A RELATIVE FORM AS THE VOLTAGE DIFFERENCE BETWEEN TWO POINTS IT IS ESSENTIAL THAT OUR NOTATION ALLOWS US TO DETERMINE WHICH POINT HAS THE HIGHER VOLTAGE 13

14 THE + AND - SIGNS DEFINE THE REFERENCE POLARITY V IF THE NUMBER V IS POSITIVE POINT A HAS V VOLTS MORE THAN POINT B. IF THE NUMBER V IS NEGATIVE POINT A HAS V LESS THAN POINT B POINT A HAS 2V MORE THAN POINT B POINT A HAS 5V LESS THAN POINT B 14

15 THE TWO-INDEX NOTATION FOR VOLTAGES INSTEAD OF SHOWING THE REFERENCE POLARITY WE AGREE THAT THE FIRST SUBINDEX DENOTES THE POINT WITH POSITIVE REFERENCE POLARITY V AB = 2V V AB = 5V V BA = 5V V = AB V BA 15

16 ENERGY VOLTAGE IS A MEASURE OF ENERGY PER UNIT CHARGE CHARGES MOVING BETWEEN POINTS WITH DIFFERENT VOLTAGE ABSORB OR RELEASE ENERGY THEY MAY TRANSFER ENERGY FROM ONE POINT TO ANOTHER BASIC FLASHLIGHT Converts energy stored in battery to thermal energy in lamp filament which turns incandescent and glows EQUIVALENT CIRCUIT The battery supplies energy to charges. Lamp absorbs energy from charges. The net effect is an energy transfer Charges gain energy here Charges supply Energy here 16

17 ENERGY VOLTAGE IS A MEASURE OF ENERGY PER UNIT CHARGE CHARGES MOVING BETWEEN POINTS WITH DIFFERENT VOLTAGE ABSORB OR RELEASE ENERGY WHAT ENERGY IS REQUIRED TO MOVE 120[C] FROM POINT B TO POINT A IN THE CIRCUIT? THE CHARGES MOVE TO A POINT WITH HIGHER VOLTAGE -THEY GAINED (OR ABSORBED) ENERGY THE CIRCUIT SUPPLIED ENERGY TO THE CHARGES V AB = 2V W V = W = VQ = 240J Q 17

18 EXAMPLE A CAMCODER BATTERY PLATE CLAIMS THAT THE UNIT STORES 2700mAHr AT 7.2V. WHAT IS THE TOTAL CHARGE AND ENERGY STORED? CHARGE THE NOTATION 2700mAHr INDICATES THAT THE UNIT CAN DELIVER 2700mA FOR ONE FULL HOUR 3 C Q = S 3 = [ C] 3600 s Hr 1Hr TOTAL ENERGY STORED THE CHARGES ARE MOVED THROUGH A 7.2V VOLTAGE DIFFERENTIAL J W = Q[ C] V = C 4 = [ J ] 3 7.2[ J ] ENERGY AND POWER 2[C/s] PASS THROUGH THE ELEMENT EACH COULOMB OF CHARGE LOSES 3[J] OR SUPPLIES 3[J] OF ENERGY TO THE ELEMENT THE ELEMENT RECEIVES ENERGY AT A RATE OF 6[J/s] THE ELECTRIC POWER RECEIVED BY THE ELEMENT IS 6[W] IN GENERAL P =VI w( t2, t1) HOW DO WE RECOGNIZE IF AN ELEMENT SUPPLIES OR RECEIVES POWER? 18 t = 2 t 1 p( x) dx

19 PASSIVE SIGN CONVENTION ( 慣 例 ) POWER RECEIVED IS POSITIVE WHILE POWER SUPPLIED IS CONSIDERED NEGATIVE a + V ab I ab b P =V ab I ab A CONSEQUENCE OF THIS CONVENTION IS THAT THE REFERENCE DIRECTIONS FOR CURRENT AND VOLTAGE ARE NOT INDEPENDENT -- IF WE ASSUME PASSIVE ELEMENTS a + V ab IF VOLTAGE AND CURRENT ARE BOTH POSITIVE THE CHARGES MOVE FROM HIGH TO LOW VOLTAGE AND THE COMPONENT RECEIVES ENERGY --IT IS A PASSIVE ELEMENT GIVEN THE REFERENCE POLARITY REFERENCE DIRECTION FOR CURRENT b THIS IS THE REFERENCE FOR POLARITY a + I ab IF THE REFERENCE DIRECTION FOR CURRENT IS GIVEN EXAMPLE + V ab b a I ab b V ab = 10V THE ELEMENT RECEIVES 20W OF POWER. WHAT IS THE CURRENT? SELECT REFERENCE DIRECTION BASED ON PASSIVE SIGN CONVENTION 20[ W ] = V I = ( 10V ) I ab ab I ab = 2[ A] ab 2A 19

20 UNDERSTANDING PASSIVE SIGN CONVENTION We must examine the voltage across the component and the current through it I S1 A A + V S2 P P S1 S 2 = V = V AB I A' B' AB I A' B' B B Voltage(V) Current A - A' S1 S2 positive positive supplies receives positive negative receives supplies negative positive receives supplies negative negative supplies receives ON S 1 V AB 0, I < 0 > AB ON S 2 V A ' ' 0, I ' ' > 0 B > A B ON S2 V A ' B' < 0, I A' B' > 0 20

21 CHARGES RECEIVE ENERGY. THIS BATTERY SUPPLIES ENERGY CHARGES LOSE ENERGY. THIS BATTERY RECEIVES THE ENERGY WHAT WOULD HAPPEN IF THE CONNECTIONS ARE REVERSED IN ONE OF THE BATTERIES? 21

22 Determine whether the elements are supplying or receiving power and how much. I ab = 4A a a V ab = 2V 2A V ab = 2V Iab = 2A P = 8W SUPPLIES POWER b P = 4W ABSORBS POWER b 22

23 Determine the amount of power absorbed or supplied by the elements? V = V, I = 4A V = V, I 2A = 23

24 I = 8[ A] + + V AB = 4[ V ] 20[ W ] = VAB (5A) SELECT VOLTAGE REFERENCE POLARITY BASED ON CURRENT REFERENCE DIRECTION 40[ W ] = ( 5V ) I 24

25 V1 = 20[ V ] 2A I = 5[ A] 40[ W ] = V1 ( 2A) SELECT HERE THE CURRENT REFERENCE DIRECTION BASED ON VOLTAGE REFERENCE POLARITY 50[ W ] = (10[ V ]) I WHICH TERMINAL HAS HIGHER VOLTAGE AND WHICH IS THE CURRENT FLOW DIRECTION 25

26 COMPUTE POWER ABDORBED OR SUPPLIED BY EACH ELEMENT + 24V 2A P 1 = (6V )(2A) + 6V 2A 1 P3 = (24V )( 2A) = ( 24V )(2A) V P1 = 12W P2 = 36W P3 = -48W P 2 = (18V )(2A) Tellegen 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 absorbed. IMPORTANT: NOTICE THE POWER BALANCE IN THE CIRCUIT 26

27 CIRCUIT ELEMENTS PASSIVE ELEMENTS VOLTAGE DEPENDENT SOURCES UNITS FOR μ, g, r, β? INDEPENDENT SOURCES CURRENT DEPENDENT SOURCES 27

28 EXERCISES WITH DEPENDENT SOURCES FIND V O V 40[ V ] O = FIND I O I O = 50mA 28

29 DETERMINE THE POWER SUPPLIED BY THE DEPENDENT SOURCES 40[ V ] P = ( 40[ V ])( 2[ A]) = 80[ W ] P = ( 10[ V ])(4 4[ A]) = 160[ W ] TAKE VOLTAGE POLARITY REFERENCE TAKE CURRENT REFERENCE DIRECTION 29

30 POWER ABSORBED OR SUPPLIED BY EACH ELEMENT P 1 = (12V )(4A) = 48[ W ] P 2 = (24V )(2A) = 48[ W ] P 3 = (28V )(2A) = 56[ W ] P DS P V = ( 1I )( 2A) = (4V )( 2A) = 8[ W ] x 36 = (36V )( 4A) = 144[ W ] NOTICE THE POWER BALANCE 30

31 USE POWER BALANCE TO COMPUTE Io 12W 6)( I ) ( 12)( 9) ( O ( 10)( 3) ( 4)( 8) ( 8 2)(11) POWER BALANCE I O =1[ A] 31