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CIRCUITS 1 DEVELOP TOOLS FOR THE ANALYSIS AND DESIGN OF BASIC LINEAR ELECTRIC CIRCUITS A FEW WORDS ABOUT ANALYSIS USING MATHEMATICAL MODELS BASIC STRATEGY USED IN ANALYSIS MATHEMATICAL ANALYSIS DEVELOP A SET OF MATHEMATICAL EQUATIONS THAT REPRESENT THE CIRCUIT - A MATEMATICAL MODEL LEARN HOW TO SOLVE THE MODEL TO DETERMINE HOW THE CIRCUIT WILL BEHAVE IN A GIVEN SITUATION THIS COURSE TEACHES THE BASIC TECHNIQUES TO DEVELOP MATHEMATICAL MODELS FOR ELECTRIC CIRCUITS THE MODELS THAT WILL BE DEVELOPED HAVE NICE MATHEMATICAL PROPERTIES. IN PARTICULAR THEY WILL BE LINEAR WHICH MEANS THAT THEY SATISFY THE PRINCIPLE OF SUPERPOSITION Model y Tu Principle of Superposition T (1 u1 2 u2 ) 1T ( u1 ) 2T ( u2 ) THE MATHEMATICS CLASSES - LINEAR ALGEBRA, DIFFERENTIAL EQUATIONS- PROVIDE THE TOOLS TO SOLVE THE MATHEMATICAL MODELS FOR THE FIRST PART WE WILL BE EXPECTED TO SOLVE SYSTEMS OF ALGEBRAIC EQUATIONS 12V1 9V2 4V3 8 4V1 16V2 V3 0 2V1 4V2 6V3 20 LATER THE MODELS WILL BE DIFFERENTIAL EQUATIONS OF THE FORM dy y f dt d2y dy df 4 8 y 3 4f dt dt dt 2 3 ELECTRIC CIRCUIT IS AN INTERCONNECTION OF ELECTRICAL COMPONENTS The concept of node is extremely important. We must learn to identify a node in any shape or form 2 TERMINALS COMPONENT a b NODE characterized by the current through it and the voltage difference between terminals NODE L R1 R2 vS TYPICAL LINEAR CIRCUIT vO + - C LOW DISTORTION POWER AMPLIFIER BASIC CONCEPTS LEARNING GOALS •System of Units: The SI standard system; prefixes •Basic Quantities: Charge, current, voltage, power and energy •Circuit Elements: Active and Passive http://physics.nist.gov/cuu/index.html Information at the foundation of modern science and technology from the Physics Laboratoryof NIST Detailed contents 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 SI DERIVED BASIC ELECTRICAL UNITS ONE AMPERE OF CURRENT CARRIES ONE COULOMB OF CHARGE EVERY SECOND. ACs 1 COULOMB 6.28 1018 (e) (e) IS THE CHARGE OF ONE ELECTRON VOLT IS A MEASURE OF ENERGY PER CHARGE. TWO POINTS HAVE A VOLTAGE DIFFERENCE OF ONE VOLT IF ONE COULOMB OF CHARGE GAINS ONE JOULE OF CHARGE WHEN IT IS MOVED FROM ONE POINT TO THE OTHER. V J C OHM IS A MEASURE OF THE RESISTANCE TO THE FLOW OF CHARGE. THERE IS ONE OHM OF RESISTENCE IF IT IS REQUIRED ONE VOLT OF ELECTROMOTIVE FORCE TO DRIVE THROUGH ONE AMPERE OF CURRENT V A IT IS REQUIRED ONE WATT OF POWER TO DRIVE ONE AMPER OF CURRENT AGAINST AN ELECTROMOTIVE DIFFERENCE OF ONE VOLTS W V A CURRENT AND VOLTAGE RANGES 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)? 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 A PHYSICAL ANALOGY THAT HELPS VISUALIZE ELECTRIC CURRENTS IS THAT OF WATER FLOW. CHARGES ARE VISUALIZED AS WATER PARTICLES EXAMPLE EXAMPLE 0 t 0 i (t ) 2 t e mA t 0 q(t ) 4 103 sin(120 t )[C ] i (t ) 4 103 120 cos(120 t ) [ A] i (t ) 0.480 cos(120 t ) [mA ] FIND THE CHARGE THAT PASSES DURING IN THE INTERVAL 0<t<1 1 q(t ) q e 0 2 x 1 1 1 1 dx e 2 x e 2 ( e 0 ) 2 2 2 0 1 q (1 e 2 ) 2 Units? FIND THE CHARGE AS A FUNCTION OF TIME q(t ) t t i ( x)dx e 2 x dx t 0 q(t ) 0 t 1 t 0 q(t ) e 2 x dx (1 e 2t ) 2 0 And the units for the charge?... DETERMINE THE CURRENT Charge(pC) Here we are given the charge flow as function of time. 30 20 10 10 10 1012 10 1012 C 9 m 10 10 (C / s) 3 s 2 10 0 1 2 3 4 5 6 Time(ms) Current(nA ) To determine current we must take derivatives. PAY ATTENTION TO UNITS 40 30 20 10 10 20 1 2 3 4 5 6 Time(ms) CONVENTION FOR CURRENTS THE DOUBLE INDEX NOTATION IT IS ABSOLUTELY NECESSARY TO INDICATE THE DIRECTION OF MOVEMENT OF CHARGED PARTICLES. 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 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 IF THE INITIAL AND TERMINAL NODE ARE LABELED ONE CAN INDICATE THEM AS SUBINDICES FOR THE CURRENT NAME a 5A b I ab 5 A a 3A b a 3A b I ab 3 A I ab 3 A a 3A b a 3A b Iba 3 A POSITIVE CHARGES FLOW LEFT-RIGHT I ba 3 A POSITIVE CHARGES FLOW RIGHT-LEFT Iab Iba I 2 A a 2A I b I cb 4 A I ab c 3A This example illustrates the various ways in which the current notation can be used 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 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 a 1C DIMENSIONALLY VOLT IS A DERIVED UNIT VOLT JOULE N m COULOMB 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 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 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 VAB 2V VAB 5V VBA 5V VAB VBA 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 glow 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 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 VAB 2V V W W VQ 240J Q THE VOLTAGE DIFFERENCE IS 5V 5V WHICH POINT HAS THE HIGHER VOLTAGE? VAB 5V EXAMPLE A CAMCODER BATTERY PLATE CLAIMS THAT THE UNIT STORES 2700mAHr AT 7.2V. WHAT IS THE TOTAL CHARGE AND ENERGY STORED? ENERGY AND POWER 2[C/s] PASS THROUGH THE ELEMENT CHARGE THE NOTATION 2700mAHr INDICATES THAT THE UNIT CAN DELIVER 2700mA FOR ONE FULL HOUR s C Q 2700 103 3600 1Hr Hr S 9.72 103[C ] TOTAL ENERGY STORED THE CHARGES ARE MOVED THROUGH A 7.2V VOLTAGE DIFFERENTIAL J W Q[C ] V 9.72 103 7.2[ J ] C 6.998 10 [ J ] 4 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 t2 w (t 2 , t1 ) p( x )dx t1 HOW DO WE RECOGNIZE IF AN ELEMENT SUPPLIES OR RECEIVES POWER? PASSIVE SIGN CONVENTION POWER RECEIVED IS POSITIVE WHILE POWER SUPPLIED IS CONSIDERED NEGATIVE Vab a b I ab P Vab I 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 A CONSEQUENCE OF THIS CONVENTION IS THAT THE REFERENCE DIRECTIONS FOR CURRENT AND VOLTAGE ARE NOT INDEPENDENT -- IF WE ASSUME PASSIVE ELEMENTS GIVEN THE REFERENCE POLARITY Vab a b REFERENCE DIRECTION FOR CURRENT THIS IS THE REFERENCE FOR POLARITY a b I ab IF THE REFERENCE DIRECTION FOR CURRENT IS GIVEN EXAMPLE Vab 2A a b I ab Vab 10V THE ELEMENT RECEIVES 20W OF POWER. WHAT IS THE CURRENT? SELECT REFERENCE DIRECTION BASED ON PASSIVE SIGN CONVENTION 20[W ] Vab Iab (10V ) Iab I ab 2[ A] UNDERSTANDING PASSIVE SIGN CONVENTION We must examine the voltage across the component and the current through it I A S1 B Current A - A' positive positive positive negative negative positive negative negative Voltage(V) A’ V S1 supplies receives receives supplies PS1 VAB I AB S2 PS 2 VA'B ' I A'B ' B’ S2 ON S1 ON S2 receives VAB 0, I AB 0 VA B 0, I A B 0 supplies ON S2 supplies V A'B ' 0, I A'B ' 0 receives ' ' ' ' 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? DETERMINE WHETHER THE ELEMENTS ARE SUPPLYING OR RECEIVING POWER AND HOW MUCH a a Vab 2V I ab 4 A 2A I ab 2 A Vab 2V P 8W SUPPLIES POWER b P 4W ABSORBS POWER WHEN IN DOUBT LABEL THE TERMINALS OF THE COMPONENT b WHEN IN DOUBT LABEL THE TERMINALS OF THE COMPONENT 1 1 2 2 V12 12V , I12 4 A V12 4V , I12 2 A I 8[ A] VAB 4[V ] 20[W ] VAB (5 A) SELECT VOLTAGE REFERENCE POLARITY BASED ON CURRENT REFERENCE DIRECTION 40[W ] (5V ) I V1 20[V ] 40[W ] V1 (2 A) 2A I 5[ A] SELECT HERE THE CURRENT REFERENCE DIRECTION 50[W ] (10[V ]) I BASED ON VOLTAGE REFERENCE POLARITY WHICH TERMINAL HAS HIGHER VOLTAGE AND WHICH IS THE CURRENT FLOW DIRECTION COMPUTE POWER ABDORBED OR SUPPLIED BY EACH ELEMENT P1 (6V )(2 A) 2 A 6V 24V 1 + - 3 2 2A 18V P1 = 12W P2 = 36W P3 = -48W P2 (18V )(2 A) P3 (24V )(2 A) (24V )(2 A) IMPORTANT: NOTICE THE POWER BALANCE IN THE CIRCUIT CIRCUIT ELEMENTS PASSIVE ELEMENTS VOLTAGE DEPENDENT SOURCES UNITS FOR , g , r , ? INDEPENDENT SOURCES CURRENT DEPENDENT SOURCES EXERCISES WITH DEPENDENT SOURCES FIND VO VO 40[V ] FIND IO IO 50mA DETERMINE THE POWER SUPPLIED BY THE DEPENDENT SOURCES 40[V ] P (40[V ])(2[ A]) 80[W ] TAKE VOLTAGE POLARITY REFERENCE P (10[V ])(4 4[ A]) 160[W ] TAKE CURRENT REFERENCE DIRECTION POWER ABSORBED OR SUPPLIED BY EACH ELEMENT P1 (12V )(4 A) 48[W ] P2 (24V )(2 A) 48[W ] P3 (28V )(2 A) 56[W ] PDS (1I x )(2 A) (4V )(2 A) 8[W ] P36V (36V )(4 A) 144[W ] NOTICE THE POWER BALANCE USE POWER BALANCE TO COMPUTE Io 12W (12)(9) (6)( IO ) (10)(3) POWER BALANCE IO 1[ A] (4)(8) (8 2)(11)