Notes Ch. 18-20 - geeSf(‘1{137'9’1 IE CIRCUIT ELEMENTS...

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Unformatted text preview: geeSf (‘1'?{137'9’1 IE CIRCUIT ELEMENTS, VOLTAGE SOURCES, & 1 CAPACITANCE NOW DISCUSS ELECTRICAL CIRCUITS. (VERY IMPORTANT FOR “ENGINEERS") CIRCUITS ARE MADE OF ELEMENTS LIKE l. BATTERIES 2. RESISTORS 3. CAPACITORS 4. INDUCTORS. WE NEED TO LEARN ABOUT THE CHARGE DEPOSITED ON ELEMENTS, THE CURRENTS THROUGH ELEMENTS AND THE ELECTRICAL POTENTIAL AT POINTS IN THE CIRCUIT. INTRODUCE SOME TERMINOLOGY. TWO POINTS IN A CIRCUIT, “A” AND “B", WILL BE AT ELECTRICAL POTENTIALS, VA AND VB . , VA THE POTENTIAL DIFFERENCE . BETWEEN A AND B IS VA — VB = AVAB. 0 THE POTENTIAL DIFFERENCE VB BETWEEN B AND A IS V3 — VA = AVBA. ALL POTENTIALS ARE MEASURED IN “VOLTS”. CIRCUITS DON’T HAVE TO BE EITHER SERIES OR 3 PARALLEL SUCH CIRCUITS ‘ ARE DIFFICULT TO ANALYZE AND SHOULD BE AVOIDED. VOLTAGE (POTENTIAL) DIFFERENCE IS “ZERO” OVER THE LENGTH OF THE CONNECTING WIRES. ,__,¥E - IF 2 NODES ARE CONNECTED Lainlnj » TOGETHER WITH A WIRE THEY I \ ARE “SHORTED” TOGETHER. THEY HAVE SAME VOLTAGE. NOW LOOK AT SOME SPECIFIC CIRCUIT ELEMENTS. INDEPENDENT VOLTAGE SOURCES THESE ARE SOURCES OF ELECTROMOTIVE FORCE (EMF). (EMF IS NOT A FORCE.) i EXAMPLES ARE: j . BATTERIES } . POWER GENERATORS } . “PHOTOELECTRIC CELLS" THESE OBJECTS MAINTAIN A CONSTANT POTENTIAL DIFFERENCE ACROSS THEMSELVES REGARDLESS OF THE AMOUNT OF CURRENT DRAWN. THEY ARE SOURCES OF POWER. EVERY CIRCUIT ELEMENT IS AN OBJECT WITH TWO 2 WIRES, OR LEADS, ATTACHED TO ITS ENDS. EACH ELEMENT CAN HAVE A POTENTIAL DIFFERENCE ACROSS IT. VA ~.~ V3 A CIRCUIT IS MADE BY CONNECTING LEADS TOGETHER. TWO PRIMARY TYPES OF CONNECTIONS: 1. SERIES CONNECTIONS. AN ELECTRON MOVING THRU CIRCUIT WOULD GO FIRST THRU ONE ELEMENT AND THEN THRU NEXT ELEMENT. 2. PARALLEL CONNECTIONS. AN ELECTRON MOVING THRU $ CIRCUIT WOULD GO THRU ONE {3 ELEMENT AND NOT THRU OTHERS. .7- .n . .-.... ZT- FE mun-- amt—n. A DOT MARKS A CONNECTION OF TWO OR MORE LEADS. INDEPENDENT VOLTAGE SOURCES CONNECTED IN 4 SERIES HAVE AN EQUIVALENT VOLTAGE THAT IS THE SUM OF THEIR EMF’S. V1 V2 V3 A+._ +._ +._a A GOOD CIRCUIT DOES NOT HOOK + TO + OR - TO —. V1 Vz V3 A+._ +.—- _.+B Vaquiualcnt 2 V1 + V2 + V! Veqm’valent ,= VI + V2 ‘ V3 NEVER CONNECT TWO EMF’S V1 TOGETHER IN PARALLEL!!! m CAPACITORS V 2 A CAPACITOR CAN STORE ELECTRIC CHARGE AND CAN HAVE A POTENTIAL DIFFERENCE ACROSS ITSELF IN A STATIC SITUATION . HOWEVER, IN A CIRCUIT ITS CHARGE AND'POTENTIAL DIFFERENCE CAN VARY, UNLIKE AN “INDEPENDENT VOLTAGE SOURCE”. A CAPACITOR CAN BE THOUGHT , OF AS TWO PARALLEL PLATES a SEPARATED BY SOME DISTANCE, “mm, rap-Hear mp Pl!!! h- ham , dune +Q ONE PLATE HAS CHARGE +Q AND THE OTHER PLATE HAS CHARGE —Q. aha-g: _Q THE TOTAL (NET) CHARGE IS ZERO. AN E“ FIELD EXISTS BETWEEN THE TWO PLATES. 5 7 EXAMPLE: FLAT PARALLEL PLATE CAPACITOR. s Hem-it: fiddlin- THE PLATES ARE AT TWO TOTAL CHARGE ; “Q”, SEPARATION DISTANCE = “d”, DIFFERENT POTENTIALS. , A “1 AREA = “A”, CHARGE DENSITY = “0'” = Q/A CAREFUL-BAD NOTATION: AV E V USE GAUSS‘S LAW ON _. TOP PLATE TO FIND E. A Gun-“an V IS PROPORTIONAL TO E .L "m“ WHICH IS PROPORTIONAL TO Q. ASSUME E = 0 Mot“ 1F Q —> 0, E —+ 0, V -* 0- OUTSIDE THE CAPACITOR inwgnn‘on SYMBOLS FOR CAPACITOR: + I E — <1> = FLUX THRU SURFACE = IE‘ - d§ = Engam DEFINITION OF “CAPACITANCE : f E" . d S: = EA 2 g : LQI I 0' 'V‘ E=§z=§ N S: M = 1 FARAD I S U IT 1V0” V=Vto,,—mem=—I}§E-dl=ngd:c=Ed A FARAD IS A VERY LARGE CAPACITANCE. 91 THE TYPICAL CAPACITOR HAS UNITS OF . V = 60 A MICROFARADS (10-6 F) NANOFARADS (10-9 F) C' = 3 = fl = 3:14 PICOFARADS (10-12 F) PUT IN SOME NUMBERS: THE CAPACITANCE OF AN OBJECT IS DETERMINED A = 0.1 7712, d = 1 mm = 10-3 m, so = 9 x 10-12 F032! ENTIRELY BY ITS GEOMETRY. __ _ oA _ 9x10-12 10—1 CAPACITANCE DOES NOT DEPEND UPON ITS CHARGE. C _ 3 _ ET _ i—MAL—l C = 9 X 10'1“ F m 10‘9 F z 1 NANOFARAD. ANOTHER EXAMPLE: SINGLE CONDUCTING SPHERE. 7 , ANOTHER EXAMPLE: 2 CONCENTRIC SPHERES a “CAPACITOR” IS SPHERE AND POINT AT INFINITY. PUT GAUSSIAN SPHERE BETWEEN Ram AND Rom. V=47rleo% #:473502ng 0 =8 V021) : 413cc R21 ' C = W V(R0) = 47350 % 0 = 4% R . V = V(Rz) — WHO) PUT IN NUMBERS FOR THE EARTH’S SURFACE. = 4135,, (1% _ %) 2 4‘3; ($225311) R = 6.4 x 106 METERS . C = 3 : 47m, (Fit—51%) C = 4m. R = 47r(9 x 10*”)(64 x 106) PUT IN SOME NUMBERS: R1 = 0.9 cm, R0 = 1 cm C = 73 X 10—4 FARADS C = 47r (9 x 10-12) (0.9 x 10-2) (10-2) / (0.1 x 10-2) 0 = 126 X 10-14 F = 1.26 x 10-12 F = 1.26 PICOFARAD NOTICE THAT IF R, z R0 WITH R0 — R, = d THEN 47rR1R0 a: 47rR2 z AREA OF SPHERE C’ = EDA/d = CAPACITANCE OF FLAT, PARALLEL PLATE CAPACITOR. CAPACITORS IN CIRCUITS: 9 1) CONNECT A BATTERY TO A CAPACITOR. BATTERY SUPPLIES VOLTAGE V. BI y _' CAPACITOR HAS V ACROSS IT LEADS. CHARGES ON THE CAPACITOR PLATES: :l: q. q=CV 2) CAPACITORS IN PARALLEL Tam-I EACH CAPACITOR HAS SAME VOLTAGE = V. TOTAL CHARGE ON ALL CAPACITORS IS SUM OF INDIVIDUAL CHARGES. + +q . B V Qtotal = 2i ‘15 = 91 + <12 + Q3 "I filo“l 08411;“;th = EQUIVALENT CAPACITANCE 5%} Cg=fli9i93=$l+9¢+¥§=01+02+03 parallel _ Ceqm'ualent — 2 Ci I NUMERICAL EXAMPLES: 11 GIVEN 3 CAPACITORS: 01 = 5 pf, 02 = 10 pf, 03 = 15 pf FIND (:6qu WHEN CONNECTED IN PARALLEL. 5 pf Isl eggs/talent = 01 + 02 + 03 A 10 Pf C§=5pf+10pf+15pf=3oPICOFARAD 15pf FIND Cequmlm WHEN CONNECTED IN SERIES. __.__1 —L+L+_1_ g —— cserges — C] 02 03 A 3 equivalent K K [E 1 —l _1_ L—ll 5 10 55—5+Ia+15‘3o pf pf 15” 05 = g = 2.73 PICOFARADS parallel Ceqm'ualent > Cindiuidual capacitors ‘ series Cequiualent < Cindividual capacitars I CAPACITORS IN SERIES. T . 1 10 EACH CAPACITOR, Cl, 02, 03 HAS THE SAME CHARGE. V11. _ _ _ _l EXAMPLE: LOOK AT BOTTOM PLATE OF 01 AND TOP PLATE OF 02. 3+ THEY HAVE NO NET CHARGE INITIALLY. THEY HAVE NO NET CHARGE FINALLY. THEREFORE, (—41) + (+q2) = 0 l¢11|=|112|=|113|E 4 FROM C = Q/V WE GET qI = CIVI: Q2 = C2V2, 113 = 03V3, ' " equivalent = q/VT = q/(V1+V2+V:3+...) I _ 1 i L @fi;*fi+®+%+ ENERGY STORAGE IN CAPACITORS 12 SLOWLY MOVE + CHARGE FROM PLATE “B" TO “A”. MA START WITH THE PLATES UNCHARGED AND CONTINUE MOVING + CHARGE UNTIL THE CAPACITOR HAS CHARGE +Q ON “A” AND ——Q ON “B”. V ov POTENTIAL DIFFERENCE BETWEEN “A” AND “B” IS V. AT MIDDLE STAGE, WITH CHARGE :I: q ON PLATES V01) = % TO TRANSFER THE NEXT CHARGE dq WE ‘DO WORK WUS = V01) d4 = % dq TOTAL AMOUNT OF WORK WE DO CHARGING CAPACITOR oq lqzo 1Q2 VVtataI=/dw=/o EdQ=§Eo=§Er THIS WORK IS THE TOTAL AMOUNT OF ENERGY STORED IN THE CAPACITOR. POTENTIAL ENERGY OF CHARGES = STORED ENERGY : WORK DONE BY US. 2 22 PE: %= 0" =§QV2=§QV l 2 c NIH EXAMPLE: 0' = 1 FARAD, V = 5 VOLTS 13 PE. = écv2 = g (1) (52) = 12.5 IOULES ENOUGH ENERGY TO RAISE 1.25 kg 1 In ON EARTH. ANOTHER EXAMPLE: 0 = 10-6 FARADS, V = 5,000 VOLTS PE. = §CV2 = § (10-6) (5 x 103)2 = 12.5 JOULES SAME. FOR A PARALLEL PLATE CAPACITOR C = GOA/d. TO GET A LARGE CAPACITANCE YOU WANT LARGE AREA “A” AND SMALL PLATE SEPARATION “d". BUT TO HOLD OFF A LARGE VOLTAGE WITHOUT BREAKING DOWN “d” NEEDS TO BE LARGE. LET’S RELATE THE ENERGY IN THE CAPACITOR TO E: 1E|=g —) EA=0A=9 6° 6° Q=EAe° AND V=Ed SUBSTITUTE THESE INTO P.E. = %Q v PE. = 2% (EAR) (Ed) = % 60E? (Ad) Ad IS CAPACITOR VOLUME. SO ENERGY/VOLUME, I.E. ENERGY DENSITY, WITHIN AN E FIELD, IS RE. _ m = = %€OEZ SO FAR WE HAVE DONE FIELDS ONLY IN VACUUM. ‘5 NOW INTRODUCE NON-CONDUCTING MATERIALS INTO THE SPACE. DIELECTRICS NON-CONDUCTORS (INSULATORS) CONSIST OF ATOMS AND MOLECULES WHICH CONTAIN CHARGED PARTICLES. THESE CHARGED PARTICLES ' ‘ ‘ RESPOND TO, AND MODIFY, .1" § 0 AN APPLIED E FIELD. I ‘\ :1 I 1. POLAR MATERIALS: . E = U HAVE A PERMANENT ELECTRIC DIPOLE MOMENT. E I THE MOLECULES ROTATE IN g THE APPLIED PIELD AND “if REDUCE THE E A LOT. E 9e 0 2. NON-POLAR ATOMS AND MOLECULES: HAVE NO PERMANENT ELECTRIC DIPOLE MOMENT, BUT A DIPOLE MOMENT IS INDUCED BY THE APPLIED E FIELD TO REDUCE THE EFFECTIVE FIELD A LITTLE. DEFINE “PERMITTIVITY” OF MATERIAL: e = K60. I; E DIELECTRIC CONSTANT. K > 1 FOR MATERIALS. ., . w rc = 1 FOR VACUUM. THIS RESULT IS GOOD IN GENERAL. u ENERGY DENSITY OF E FIELD IS U(F) = §eoE2m SHOW THAT IT WORKS FOR CASE OF CHARGED SPHERE. EARLIER WE FOUND Cam": = 47m, R. THE STORED POTENTIAL ENERGY IS _ l _ 1 Q _ 1 2 PE' _ ZQV _ E Q 41m, R _ 87mg % LET’S FIND TOTAL ENERGY IN E BY INTEGRATING THE ENERGY DENSITY OVER ALL VOLUME. USE SPHERICAL COORDINATE SYSTEM. _ l 2 EW-mrg __, E2=(1)2% 41m, 1 Utah! = [lemme dU(T) = [all Space EEDEZ dz) 1 1 2 1r 00 T ’I‘ =_2-(‘F€o) 6°02 02 /—11 f3 Ma 2 =% (L) EoQ2 (2") (2) med—z 47re° -2 (—% H 87m, _ 1 Q2 WEE AGREES!!! EXAMPLE: 1“ ADD DIELECTRIC TO FLAT PARALLEL PLATE CAPACITOR PLATE AREA = “A”, TOTAL PLATE CHARGE = «nq CHARGE DENSITY a 0' = q /A “-53 WITH NO DIELECTRIC GAUSS’S LAW IS: £L__ fig] ‘1 ”A ‘ ‘7 I ‘1 <I>=/E-dS=—=— —> E:— THE VOLTAGE DIFFERENCE IS V = Ed = a' d/eo. 60 e0 6° THE CAPACITANCE Is 0 = q/V = aA/(a d/e.) = eoA/d FILL CAPACITOR WITH DIELECTRIC. Ematerial : ICED. CHARGE ON PLATES IS STILL iq. THERE IS ALSO SURFACE CHARGE 1 ON THE DIELECTRIC MATERIAL: iq’ +1] SURFACE CHARGE IS NEXT TO - q PLATE CHARGE. —q’ SURFACE CHARGE IS NEXT TO + q PLATE CHARGE. E0 IS FIELD FROM PLATE (:l:q) CHARGES. 11 IT POINTS “DOWN”. Ebmmd IS FROM CHARGES (iq’) ON DIELECTRIC SURFACE. IT POINTS “UP”. NET E FIELD IS Em, = E, + Emm MACNITUDE OF Em; IS [E’fiml = [Eel/n. DIRECTION OF Em, IS “DOWN”. |Eol > |Em,,d|. GAUSS’S LAW Now GIVES anclosed=q_ql EDA 00A_ 4 5° 60 1c 60 K: can <I>=/E-d§= 5—1 g=q—q’ —+ ql=q( ) K} ’5 VOLTAGE ACROSS CAPACITOR IS FOUND USING Efgnal. Vfinal = Efinal d = Eo d/K = Vo/n. VOLTAGE ACROSS CAPACITOR IS REDUCED FROM VALUE IN VACUUM. CAPACITANCE OF CAPACITOR WITH DIELECTRIC IS _ ._ _ 30.— Cfinal — 7;;— Voq/g— EV — EC. CAPACITANCE IS INCREASED RELATIVE TO VACUUM. geese (unwell /? INTRODUCTORY CIRCUIT ANALYSIS 1 WE HAVE DISCUSSED THE CHARGING OF CAPACITORS. THIS LEADS TO DISCUSSION OF “CURRENTS” IN CIRCUITS. “ELECTRIC CURRENT”, E “I”, IS THE MOVEMENT OF ELECTRIC CHARGES WITH TIME. dQ COULOMBS I = -— “$— : dt —> UNITS SECOND _ AMPERES 1 AMPERE = 1 COULOMB/SECOND = 1 AMP FOR CHARGES MOVING THROUGH A REGION OF SPACE THE “CURRENT DENSITY” IS DEFINED AS J = CURRENT DENSITY = CURRENT/AREA = I/A. UNITS = AMPERES/METER? IF n = NUMBER OF CHARGE CARRIERS PER VOLUME = CARRIER DENSITY AND < 2) >= AVERAGE DRIFT VELOCITY AND q = CHARGE PER CARRIER THEN J=qn<v>. SUMMARY. Ema: Eo/n —» SMALLER. sz= Vo/n —+ SMALLER Cfinal_ -— I500 —-) LARGER 0": an (L1; —) —+ SMALLER THAN 0,, WHAT IS THE POTENTIAL ENERGY STORED IN THE CAPACITOR WITH THE DIELECTRIC? ORIGINALLY, P.E.o = qu/Z. 1 1 V0 qu P.E'.u P, ,i =_ in =_ _=_=__ Efm‘ 2T0“, 2ch 2!: n THE STORED ENERGY IS REDUCED. WHERE DID THE ENERGY GO? THE DIELECTRIC IS PULLED INTO THE CAPACITOR. IF IT DOES NOT HIT THE WALLS THERE WILL BE 18 KINETIC ENERGY IN THE MOTION OF THE DIELECTRIC. KE. = PE0 — REM, INTUITIVE PICTURE OF WHY DIELECTRIC IS PULLED INTO CAPACITOR. THE DIELECTRIC SURFACE CHARGES ATTRACT THE CAPACITOR PLATE CHARGES GIVING AN ATTRACTIVE FORCE BETWEEN THE DIELECTRIC AND CAPACITOR. An W “I'm“ location are: A THE NUMBER OF CHARGE CARRIERS WHICH PASS A REFERENCE POINT IN TIME dt IS dN = nA < 'u > dt SO THE CHARGE PASSING THE REFERENCE POINT IS dQ=qu=an<v>dt AND THE CURRENT IS 1:3 = W < v >. EXAMPLE: COPPER WIRE HT“! I = 2 AMPS OF CURRENT m DIAMETER = 2 mm. T; ASSUME n = l/ATOM <—7 FIND < v >. p = DENSITY = 9 x 103 Icy/m3 MOLAR MASS = 63. 5 GRA MJM=OLE 6 35 x 10 2 leg/mole n = #atmm/m3 = ‘—LJ—<2—LJ * SSS—6:22.72“ "W” = 8.5 X IOzaatoms/ma q =ELECTRCN CHARGE = 1.6 X 10-19 COULOMBS A = AREA OF WIRE = «R2 =' 7r x 10‘6 m2 I = 2 AMPS = nqA < U >—)'< 1) >= 4.7 ><10‘5 m/s. m QUESTION: HOW DOES THE CURRENT DEPEND UPON THE APPLIED ELECTRIC FIELD? FORCE = 13 = qE = mg AS E INCREASES < v > WILL INCREASE. FOR MANY MATERIALS < v > o< E so THE CURRENT DENSITY f o< E, —» J 2 CE. 0 E “CONDUCTIVITY” OF MATERIAL. SUCH A MATERIAL IS “OHMIC” (GEORGE OHM). “RESISTIVITY” = 1/0 = l/CONDUCTIVITY. p THE VOLTAGE, OR POTENTIAL DIFFERENCE, ACROSS A PIECE OF MATERIAL OF LENGTH “1” IS V = El ' FOR OHMIC MATERIAL V = (J/a) l = Jpl USING “I” INSTEAD OF “2.1 VIA J = I/A V = Ipl/A = IR WITH R= “RESISTANCE” = pl/A. RESISTIVITY VERSUS TEMPERATURE 5 THE RESISTIVITY IN MOST MATERIALS INCREASES SLIGHTLY WITH TEMPERATURE. IT IS LIKE A “FRICTION”. THE COLLISIONS BETWEEN THE CHARGE CARRIERS AND THE ATOMS INCREASE. THE < v > DECREASES. —» J = qn < 11 > DECREASES. p(T) J = E/p w; p INCREASES. p0 _ T — T APPROXIMATE: p(T) = p0 [1+ (1 (T — To)]. ( °) p0 = I RESISTIVITY AT ROOM TEMPERATURE = To. FROM R = pl/A —> R(T) = R0[1+ a (T — To)]. EXAMPLE: LIGHT BULBS'USUALLY BURN OUT WHEN FIRST SWITCHED ON. LOWER TEMPERATURE —) LOWER “R” —) LARGER “I" SUPERCONDUCTIVITY LC To T FOR SOME MATERIALS p —> O SUDDENLY BELOW A “CRITICAL TEMPERATURE” E Tc. QUANTUM MECHANICAL EFFECT. OHM’S LAW 4 OHM’S LAW IS V = IR OR J 2 0E. UNIT FOR RESISTANCE = OHM. UNIT FOR CONDUCTIVITY = MHO. OHM SPELLED BACKWARDS. , 10 V ‘ 30 OHM IOHM=IC=IVOLT/IAMPERE. - EXAMPLE: 10 VOLT BATTERY WITH 30 OHM RESISTOR. CURRENT IS I = V/R = 10 W30 (2 = 0.33 AMPERES. IF THE RESISTOR IS 5 mm LONG WITH 4 mm DIAMETER, FIND THE RESISTIVITY: R = pz/A.—-> p: RA/l —;i 2 >(300)(7r(2><10 m))=7.5§<10‘89m p= 5x10‘3m ,.,. a = CONDUCTIVITY =1/p = 1.33 x 107 OHM-l m-l SEMICONDUCTORS 6 FREE THE CURRENT IN SEMICONDUCTORS, LIKE SILICON, INCREASES WITH TEMP. ”ENERGY GAP” THERE IS AN “ENERGY GAP” WHICH , ’ , / THE ELECTRONS MUST GET OVER TO BE “FREE” AND CONDUCT. THE PROBABILITY OF AN ELECTRON GETTING OVER THE GAP T INCREASES WITH TEMPERATURE. THE NUMBER OF CHARGE CARRIERS INCREASES WITH TEMPERATURE. A SEMICONDUCTOR ACTS LIKE INSULATOR AT LOW TEMPERATURE AND CONDUCTOR AT HIGH TEMP. RESISTOR COLOR CODING RESISTORS ARE LABELED WITH “COLORED BANDS”. (NUMBER NUMBER) X Iomuliipli" COLOR NUMBER MULTIPLYING FACTOR 10 BLACK 0 BROWN 1 101 RED 2 102 ORANGE 3 103 YELLOW 4 104 GREEN 5 105 BLUE 6 105 VIOLET 7 107 GRAY 8 108 . WHITE 9 109 . I THE ACCURACY IS CONVEY ED BY A 4‘” BAND. COLOR TOLERANCE GOLD 5% SILVER 10% ABSENT 20% COLOR CODING EXAMPLE: 7 RED GREEN ORANGE GOLD 2 5 3 —) 103 5% RESULT: 25000 9:}: 5% —} BE‘I WEEN 23750 9 AND 26250 9. NOW RETURN TO CIRCUITS. RESISTORS IN SERIES: TOTAL POTENTIAL DIFFERENCE FROM ”A” TO “B" VtatalEVT=V1+V2+V3 SAME CURRENT GOES THRU ALL THREE RESISTORS. [T = [1 = 12 = 13 FROM OHM’S LAW: View: = V1 + V2+V3 = [131+12Rz+1333 , = 1(R1+R2+R3) = [Rtotal A 'v RT = Requivalent = R1 + R2 + R3 FOR n RESISTORS IN SERIES series _ ‘ equivalent “ .2 R1 EXAMPLE: 3 RESISTORS: R1 = 59, R2 = 109, R3 = 159 9 FIND R253? Rserl'e: = R1+R2 + R3 = 30 9 equiv FIND R5335“ 1 1 1 _1__I+ 1 + 1 W =§1+E+Ea‘5n 109 159 equiv 3 [H _ Emfi=fin—- Rseries '> EACH Rd equiv Rparallel < EACH Ri equiv RESISTORS IN PARALLEL 5 POTENTIAL DIFFERENCE BETWEEN “A” AND “B" IS THE SAME AS ACROSS EACH RESISTOR. VAB = V] = V2 = Va _ parallel VAR - I 10301 Requiualent VI=IIR1=VAB=V V2=I2R2=VAB=V V5=13R3=VAB=V 11 = V/R1 12 = V/Rz 13 = V/Ra R1 R2 R3 _ V — Requiualent 1 _ 1 1 1 fl$fl_E+E+E FOR n RESISTORS IN PARALLEL l n l ~<~I equivalent ELECTRIC POWER IN CIRCUITS 1° THE WORK DONE IN MOVING A CHARGE ACROSS A POTENTIAL DIFFERENCE V IS dW = V dQ. POWER IS WORK DONE PER TIME INTERVAL: POWER = P = dW/dt = V dQ/dt = VI. ——> THE-POWER SUPPLIED BY THE BATTERY 2 P = VI. THE POWER DISSIPATED IN RESISTOR a P = VI. FOR OHMIC SYSTEM V = I R P = VI = V(V/R) = Vz/R = (IR)! = 12R UNITS FOR POWER: WATTS THE TOTAL ENERGY USED IN A HOUSE PER MONTH IS MEASURED IN KILOWATT—HOURS. POWER >< TIME —> ENERGY. EXAMPLE: 11 10 VOLT BATTERY WITH 3 RESISTORS: 5 Q, 10 Q, 15 Q A .V‘_ .‘— oy’» ' SERIES CIRCUIT R- N K 59 109 15 Requiu = R1+ R2 + R3 I 10V =59+109+159=309 OHM’S LAW: v = IR —> I = V/R = (10 V/30 O) = (1 /3) AMPS POWERMW = IV = ((1/3) AMP) 10V = 3.33 WATTS FIND POWER DISSIPATED IN RESISTORS. POWER(R1) = IV1 = [(1121) = 12121 = (1/3)2 5 = (5/9) WATTS P(R2) = IV2 = 1(IR2) = 12122 = (1/3)2 10 = (IO/9) WATTS P(R3) = IV3 = 1(1R3) = 12123 = (1/3)2 15 = (15/9) WATTS P1 + P2 + P3 = (5/9)+ (10/9)+ (15/9) = (30/9) = 3.33 WATTS THIS IS SAME AS BATTERY POWER FOUND ABOVE. NOTE. P3 > P2 > P] FOR, R3 > R2 > R1. LEARN TO ANALYZE ELECTRICAL CIRCUITS. 13 WE NEED TWO “KIRCHHOFF LAWS”. KIRCHHOFF CURRENT LAW. ALGEBRAIC SUM OF CURRENTS LEAVING A NODE IS “0”. CURRENT ENTERING NODE = CURRENT LEAVING NODE. EACH CURRENT NEEDS A DIRECTION. K.C.L = ELECTRIC CHARGE CONSERVATION. . L (_11) + (—12) + (—13) + I4 + I5 = 0 II EXAMPLE: 11 = 1A, 12 = 2A, I3 = 3A, 14 = 4A FIND 15. (—1A)+(—2A)+(—3A)+4A+I5 = o —> 15 = 6A—4A = 2AMPS. [5 IS A 2 AMPERE CURRENT IEAVINQ THE NODE. EXAMPLE: [1 = 1A, [2 = 2A, 13 = 3A, [4 = 8A FIND 15. ’ (—1A)+(—2A)+(-3A)+8A+15 = 0 —> 15 = 6A—8A = —2AMPS. 15 IS A 2 AMPERE CURRENT ENTEBDLG THE NODE. EXAMPLE: 11 SAME 10 V BATTERY AND 3 RESISTORS: 5 Q, 10 Q, 15 0. PARALLEL CIRCUIT 5 n VI = 11121 —) 11 = Vl/Rl =10V/59 = 2 AMPS . i P1 = 111/1 = (2 A)(1O V) = 20 WATTS V2 = 12R2 ——> 12 = 112/122 = 10V/Ion = 1 AMPS ' 10 V P2 = 121/2 = (1 A)(10 V) = 10 WATTS V3 = 13123 ——I 13 = V's/R3 = 10V/15D = (2/3) AMPS P3 = 131/3 = (2/3 A)(10 V) = 6.67 WATTS TOTAL POWER DISSIPATED IN RESISTORS: P1 + P2 + P3 = 36.67 WATTS Lana: = 11+ I2 + 13 = 2 +1+ 2/3 = 3.667 AMPERES PM,” = (Im) Vbam = (3.667 A)(10 V) = 36.67 WATTS POWER SUPPLIED BY BATTERY EQUALS POWER DISSIPATED IN RESISTORS. NOTE: PI > P2 > P3 FOR R3 > R2 > R1 OPPOSITE TO CASE FOR SERIES CIRCUIT. KIRCHHOFF VOLTAGE LAW 14 THE ALGEBRAIC SUM OF THE POTENTIAL DIFFERENCES AROUND ANY CLOSED LOOP OF AN ELECTRICAL CIRCUIT IS ZERO. K.V.L = CONSERVATION OF ENERGY. EXAMPLE: COUNTER—CLOCKWISE: (—VI) +.V; + Va + (—Vz) = 0. W+W=K+W GO CLOCKWISE INSTEAD: +VI+V2+(—Va)+(—V4)=0- V1+V2=VS+V4 SAME RESULT. IT DOESN’T MATTER WHERE YOU START OR WHICH DIRECTION YOU GO. PUT IN SOME NUMBERS: V1 = 5V, V2 = 10V, V3 = 12V (—5V)+V4+12V+(—10V)= 0 —» V4 = 15— 12 = 3 VOLTS. DIFFERENT NUMBERS: V1 = 5V, V2 = 10V, V3 = 18V (—5V)+V;+18V+(—10V)= 0 —> V4 =15—18 = —3 VOLTS THE “+" END OF V4 IN THE DIAGRAM IS ACTUALLY 3 VOLTS LOWER THAN THE “—” END. IF WE PUT “GROUND” (V=0) INTO THE DIAGRAM THEN THE FIRST SET OF NUMBERS GIVES: Wzmv W=mv W=—3v FOR SECOND CASE IF V4 HAD ITS “+” AND “—" ENDS REVERSED, THEN V} WOULD HAVE EQUALED +3 VOLTS. V1=5V V2=10V V3=18V V4=3V »3 EXAMPLE: 20 VOLT BATTERY AND 5 OHM RESISTOR. I 2. NO SIGNIFICANT NODES. , 3. ONLY ONE CURRENT. MAKE IT EXIT “+" TERMINAL OF BATTERY. 4. LABEL POLARITY OF RESISTOR SO THE CURRENT ENTERS THE “+” TERMINAL AND EXITS THE “—" TERMINAL. 5. APPLY KIRCHHOFF VOLTAGE LAW. v.0“... = —20 VOLTS mm... = IR = 1(5 OHMS) ) —> (—20 V) + 1(5 9) = 0. I. CAN’T SIMPLIFY. 15 l7 6. APPLY KIRCHHOFF CURRENT LAW. NOTHING TO DO. 7. SOLVE EQUATIONS: 1 = (20 V) /(5 O) = 4 AMPERES 8. FIND POWER 12%;?“ = IV = (4 A)(20 V) = so WATTS P532?“ = [V = (—4 A)(20 V) = —80 WATTS 12.33233“ = +80 WATTS. RECIPE FOR SOLVING PROBLEMS. 16 1. SIMPLIFY THE CIRCUIT BY COMBINING SERIES OR PARALLEL ELEMENTS. 2. IDENTIFY SIGNIFICANT NODES. 3. INTRODUCE CURRENTS INTO EACH BRANCH. 4. LABEL POLARITIES OF EACH CIRCUIT ELEMENT. 5. APPLY KIRCHHOFF VOLTAGE LAW TO EACH LOOP. 6. APPLY KIRCHHOFF CURRENT LAW TO EACH LOOP. 7. SOLVE THE SIMULTANEOUS EQUATIONS. 8. FIND ALL CURRENTS, POTENTIAL DIFFERENCES AND POWERS. 9. CHECK CONSISTENCY. EXAMPLE: 20 VOLT BATTERY AND 18 3 RESISTORS: R1 = 159, R2 = 209, 123 = 609 3 1) SIMPLIFY THE CIRCUIT. . . 609 R2 AND R3 ARE IN PARALLEL. 1 l 1 1 1 I _ 20+60 _ ?— 122-3 R?” E = E + 66 = (20)<60) (20)(60) _ 1200 . 159 . R2-3=h—“= 9 20V 159 [315 ‘ 80 80 REPLACE RgAND R3 WITH 122-3. R1 IS IN SERIES WITH R24. 20 V 309 1214-3 = R1 + RH =159 +150 = 300 K, as CIRCUIT BECOMES BATTERY AND RESISTOR 121-24. 2)-7) 20V = (sum _; I = (2/3) AMPS. ALL OF THIS CURRENT GOES THROUGH R1. THE POTENTIAL DIFFERENCE ACROSS R1 IS VI = IR1 = (2/3A)(ISC) = 10 VOLTS. MUST FIND CURRENTS THROUGH R2 AND R3. APPLY K. VOLTAGE LAW FOR BATTERY, R1, AND R2. —V3+V1+V2=0—> (—20V)+(10V)+V2=0 —) V2 = 20V — 10V = 10 VOLTS. FIND 12: V2 = 12H2 —> 12 = Vz/R...
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