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FOWLER CHAPTER 5 LECTURE 6 MULTIPLE LOAD CIRCUITS.

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Presentation on theme: "FOWLER CHAPTER 5 LECTURE 6 MULTIPLE LOAD CIRCUITS."— Presentation transcript:

1 FOWLER CHAPTER 5 LECTURE 6 MULTIPLE LOAD CIRCUITS

2 MULTIPLE LOAD CIRCUITS IF YOU HAVE TWO OR MORE LOADS, THEY CAN BE IN SERIES, PARALLEL OR SERIES-PARALLEL. SUBSCRIPTS: R 1, R 2, ……R N INCIDATES # OF DIFFERENT LOAD RESISTORS. V 1, V 2,……..V INCIDATES # OF DIFFERENT VOLTAGES. I I, I 2,………I N INCIDATES # OF DIFFERENT CURRENTS. V T: TOTAL VOLTAGE FOR CIRCUIT. I T : TOTAL CURRENT FOR CIRCUIT. POWER : P T = P 1 +P 2 …….P N TOTAL POWER FOR CIRCUIT. SERIES CIRCUITS: ONE PATH FOR CIRCUIT FLOW. CURRENT IS THE SAME IN EACH ELEMENT: I T =I 1 =I 2 =I 3 R1R1 R3R3 R2R2 ITIT VTVT P.99

3 THERE ARE TWO WAYS TO FIND THE TOTAL RESISTANCE FOR THIS CIRCUIT. 1.USE OHM’S LAW V=IR, 2. R=V/I 3. =90V/2A 4.=45Ω 5.OR 2A 90 V DC R 1 5 Ω R 2 10 Ω R 3 30 Ω IN SERIES CIRCUITS, CURRENT IS THE SAME IN EACH ELEMENT. AN AMPERE METER PLACED AT ANY POINT IN THE CIRCUIT WILL GIVE THE SAME CURRENT READING. TO FIND TOTAL RESISTANCE IN A SERIES CIRCUIT, ADD EACH RESISTANCE TOGETHER. EXAMPLE: P.100

4 VOLTAGE IN SERIES CIRCUITS VOLTAGE IS DIVIDED UP ACROSS EACH LOAD. KIRCHOFF’S VOLTAGE LAW: THE SUM OF VOLTAGE DROPS AROUND THE CIRCUIT PATH = THE SOURCE VOLTAGE. THIS CONFIRMS KIRCHOFF VOLTAGE LAW. 2A 90 V DC R 1 5 Ω R 2 10 Ω R 3 30 Ω P.101

5 VOLTAGE DROP AND POLARITY POTENTIAL ENERGY DROP ACROSS A RESISTOR IS KNOW AS A VOLTAGE DROP. SOURCE VOLTAGE PROVIDES THE ENERGY. LOAD RESISTANCE VOLTAGE CONVERTS IT TO ANOTHER FORM. (HEAT IN THIS CASE) POLARITY IN THIS CASE INDCIATES THE DIRECTION OF CURRENT FLOW. CURRENT THRU A LOAD MOVES FROM – TO + (ENERGY CONVERTED FROM ELECTRICAL TO ANOTHER FORM). R1 R V DC ab c I P.101

6 MEASURING SERIES VOLTAGE. F.5-7 P-102 CHOOSE CORRECT FUNCTION, RANGE AND POLARITY ON THE METER. MEASURE VOLTAGE DROP ACROSS EACH RESISTOR. R1 R3 R2 ITIT V VTVT

7 FINDING A OPEN IN A SERIES CIRCUITS F.5-8 P-103 IF ANY PART OF THE CIRCUIT IS OPEN CURRENT STOPS,VOLTAGE,POWER ARE REMOVED FROM ALL THE LOADS. 0V 12 V V V V R2R2 R1 R3R3 0V R2R2 CIRCUIT PATH IS BROKEN

8 FINDING SHORTS IN SERIES CIRCUITS F5-9 P-104 WHEN ONE LOAD IS SHORTED OUT, OTHERS MAY CONTINUE TO OPERATE. IN THIS EXAMPLE LAMPS L1 AND L3 CURRENT AND VOLTAGE INCREASE 50%. THIS INCREASED PWR.TO THE LAMPS WILL LIKELY BURN THEM OUT. SHORT 1A 10V,10W L1L1 30V L2 L3L3 10V,10W 1.5A 15V,25W L1L1 30V L2 L3L3 0V,0W 15V,25W

9 VOLTAGE DIVIDER EQ. CAN BE USED TO FIND VOLTAGE ACROSS ANY RESISITOR IN A SERIES CIRCUIT.( R N ) FROM EXAMPLE 5-2 P-105 FIND VR1, IF DOMINANT RESISTOR IS SO MUCH GREATER THEN THE OTHERS IN THE CIRCUIT,THE EFFECTS OF THE SMALLER RESISTOR CAN BE IGNORED. 90V 35 Ω 70 Ω 45 Ω R1R1 R2 R3 P.105 VOLTAGE DIVIDER EQ

10 APPLICATIONS OF SERIES CIRCUITS F5-13,14 P-106 R IS A VARIABLE SERIES RESISTOR(RHEOSTAT) AS R ↓ MOTOR SPEED ↑ AS R↑ MOTOR SPEED ↓ NOT EFFICIENT SINCE R WASTES ENERGY AS HEAT.↑ LIGHT INTENSITY CONTROL F.5-14 P VAC R M 24VL S WORKS WELL FOR DC ONLY

11 MAXIMUM PWR. TRANSFER (MPT) GETTING THE MAX. AMOUNT OF PWR. FROM A SOURCE TO A LOAD. MPT OCCURS WHEN THE SOURCE INTERNAL OPPOSITION TO THE CURRENT EQUALS THE LOAD’S OPPOSITION TO THE CURRENT. F.5-16 P-108 MPT OCCURS WHEN R B =R 1 =33Ω P R1 =P RB = I x V=2A X 12=24W SEE T 5-1 P.108 R B 3 Ω SOURCE R L 3 Ω LOAD 12V

12 PARALLEL CIRCUITS MULTILOAD CIRCUIT THAT HAS MORE THEN ONE CURRENT PATH. EACH PATH IS CALLED A BRANCH. EACH BRANCH HAS ITS OWN CURRENT. TOTAL CURRENT IS SPLIT ACROSS EACH BRANCH. EACH BRANCH CURRENT IS INDEPENDENT OF THE OTHERS. VR1R1 R2 R3 ITIT I1I1 I2 I3 P.108

13 CURRENT IN PARALLEL CIRCUITS I T =I R1 +I R2 +I R3 : TOTAL CURRENT IS THE SUM OF THE BRANCH CURRENTS. V R1R1 R2R2 R3R3 ITIT A I R1 I R2 I R3 B P.109

14 KIRCHHOFF’S CURRENT LAW:: THE SUM OF CURRENTS ENTERING A JUNCTION=SUM OF CURRENTS LEAVING A JUNCTION. (NODE) CURRENT GOING INTO NODE=CURRENT GOING OUT NODE. 3A+5A+2A=7A+3A 10A=10 A V R1R1 R2R2 R3R3 ITIT A I R1 I R2 I R3 B 3A 5A 2A 3A 7A NODE P.109 AT NODE A: AT NODE B: IAIA IBIB

15 KIRCHHOFF’S CURRENT LAW KIRCHHOFF’S VOLTAGE LAW

16 IF A BREAK IN CURRENT FLOW OCCURS IN A SERIES CIRCUIT LIGHTS GO OUT! IN A PARALLEL CIRCUIT, IF ONE BRANCH IS CUT OFF.THE OTHER BRANCH STILL WORKS AND THAT LAMP GLOWS.

17 SINCE THESE TWO SOLAR PANELS ARE WIRED IN PARALLEL THE CURRENT THEY PRODUCE IS DOUBLED AND THE VOLTAGE STAYS CONSTANT. I T = I FIRST PANEL + I SECOND PANEL V T = V FIRST PANEL = V SECOND PANEL

18 UNLIKE SERIES CIRCUITS, THE BRANCH WITH THE LOWEST R VALUE DOMINATES THE CIRCUIT, USING THE MOST CURRENT AND POWER FROM THE SOURCE. 10V R1 1K Ω R2 10K Ω A 11mA A 10mA A 1mA R1 RESISTANCE IS 10 TIMES LESS THEN R2, BUT HAS 10 TIMES THE CURRENT FLOW THRU IT.

19 RESISTANCE IN PARALLEL CIRCUITS THE TOTAL RESISTANCE OF A PARALLELCIRCUIT IS ALWAYS

20 ADD A 3TH RESISTOR R 3 TO THIS CIRCUIT, FOLLOW THE SAME LOGIC AS ON THE LAST SLIDE. R T =V T /I T = V T /(I 1 +I 2 +I 3 ), SINCE I N =V T /R N (OHM’S LAW) R T =V T /(V T /R 1 +V T /R 2 +V T /R 3 ) USING ALGEBRAIC MAGIC GIVES R T =1/(1/R 1 +1/R 2 +1/R 3 ) RECIPROCAL FORMULA 10V R 1 10K Ω R 2 100K Ω R 3 1K Ω P.113

21 EXAMPLE 5-3 P-112 R 1 =20Ω R 2 =30Ω R 3 =60Ω R T = 1/( 1/R 1 + 1/R 2 + 1/R 3 ) R T = 1/( 1/20Ω+ 1/30Ω+ 1/60Ω) R T =1/(6/60)=10Ω IF WE HAVE ONLY TWO RESISTORS IN PARALLEL THEN WE CAN FIND R T USING; R T =R 1 xR 2 /(R 1 +R 2 ) EXAMPLE 5-4 P112 GIVEN R 1 =27Ω R 2 =47Ω,FIND R T R T =R 1 X R 2 /(R 1 +R2) =27 X 47/(27+47) =1269/74 =17.1Ω IF ALL THE RESISTORS IN A PARALLEL CIRCUIT HAVE THE SAME VALUE. YOU CAN USE THE EQUATION:` R T =R/n TO FIND THE TOTAL RESISTANCE. WHERE n =THE # OF RESISTORS IN THE CIRCUIT. EXAMPLE: GIVEN TWO 100Ω RESISTORS IN PARALLEL, FIND R T R T =R/n = 100/2 =50Ω 20 Ω 30 Ω 60 Ω

22 MEASURING RESISTANCE IN PARALLEL MUST DISCONNECT THE RESISTOR BEING MEASURED FROM THE LOAD TO READ THE INDIVIDUAL RESISTOR RESISTANCE,IF YOU DON’T, YOU WILL MEASURE R T Ω Ω RESISTOR LEAD DISCONNECTED R1R1 R2R2 R3R3 R1R1 R2R2 R3R3 P.112

23 CURRENT DIVIDER FORMULA FOR TWO PARALLEL RESISTORS P.114 I RI =I Tx R 2 /(R 1 +R 2 ) I R2 =I Tx R 1 /(R 1 +R 2 ) FIND I R1 = 2X47/(2+47) FIND I R2 = 2X22/(2+47) I R1 =1.36A I R2 =0.897A SAVES THE TROUBLE OF HAVING TO FIND: R T USING 1/R OR: V T USING OHM’S LAW OR: I R1,I R2 USING OHM’S LAW APPLICATIONS OF PARALLEL CIRCUITS. HOME LIGHTING, WALL RECEPECTALES, AUTO WIRING, TV, ETC. CONDUCTANCE IS THE RECIPROCAL OF RESISTANCE,IS THE ABILITY TO CONDUCT CURRENT. G=1/R G IS MEASURED IN SIEMENS 2A VTVT R 1 22 Ω R 2 47 Ω I R2 I R1 ITIT P.114 P.115

24 USEFUL FORMULA’S FOR SERIES CIRCUITS I T =I 1 =I 2 …..=In V T =V 1 +V 2 …..+Vn R T =R 1 +R 2 …..+Rn P T =P 1 +P 2 ……+Pn VOLTAGE DIVIDER EQ. TO FIND VOLTAGE ACROSS ONLY ONE RESISITOR IN A SERIES CIRCUIT. ____________________________________________________________________________________________________________________ FOR PARALLEL CIRCUITS I T =I 1 +I 2 …..+In V T =V 1 =V 2 …..=Vn R T =1/(1/R 1 +1/R 2 …+1/Rn) P T =P 1 +P 2 …….+Pn IF ONLY TWO RESISTORS ARE IN PARALLEL, FIND R T USING: IF ALL RESISTORS ARE THE SAME: R T =R/n WHERE n=#OF RESISTORS VTVT R1R1 RNRN VR1R2 I1I1 I2I2 ITIT A I1I1 I2I2 ITIT A SEE APPENDIX C P CURRENT DIVIDER FORMULA FOR 2 PARALLEL RESISTORS

25 SERIES-PARALLEL CIRCUIT

26 SERIES PARALLEL CIRCUITS P.117 FIND R T FOR THIS CIRCUIT FIRST SOLVE FOR THE PARALLEL PAIR R 2, R 3 R 2, 3 =R 2X R 3 /(R 2 +R 3 ) =20 X 30/(20+30) =600/50 =12Ω NOW FIND R T FOR THE SERIES CIRCUIT R T =R 1 +R 2,3 = 15Ω+12Ω = 27Ω V 15 Ω R1R1 30 Ω R3R3 20 Ω R2R2 V R 1 15 Ω R2,R 3 12 Ω V R T=27 Ω

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28 R 1,2 =R 1 +R 2 =40Ω+60Ω =100Ω R T = R 1, 2 X R 3 /(R 1, 2 +R 3 ) R T =100 X 22/122 =2200/122 =18Ω FIND R T FOR THIS SERIES PARALLEL CIRCUIT 22 Ω R3R3 R 1 40 Ω R 2 60 Ω V V R 1,2 100 Ω R 3 22 Ω V R T =18 Ω FIND SERIES RESISTANCE FOR R 1 + R 2 FIRST. NEXT FIND RT FOR THIS PARALLEL CIRCUIT.

29 KIRCHHOFF’S LAWS IN SERIES- PARALLEL CIRCUITS F.5-30, P119 USING KIRCHHOFF’S LAW FIND V R1,V R3 GIVEN THE VAULES SHOWN ON THE SCHEMATIC FIND V R1 : V T =V R1 +V R2 V R1 =V T -V R2 V R1 =100V-40V V R1 =60V FINDV R3 : V R2 =V R3 +V R4 V R3 =V R2 -V R4 V R3 =40V-30V V R3 =10V USING KIRCHHOFF’S CURRENT LAW,FIND I R1,I R2,I R3 SINCE.6A FLOWS THRU NODE B, 0.6A FLOWS THRU R1(THIS LOOP IS A SERIES CIRCUIT) FIND I 1, AT NODE A A I1I1 I2I2 ITIT I T =I 1 +I 2 I 1 =I T -I 2 I 1 =0.6A-0.2A=0.4A V T 100V R1R1 A B I1I1 I T=.6A I 2 =.2A R 2 40V I2I2 R3R3 R 4 30V

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31 VOLTAGE DIVIDERS AND REGULATORS P-121 FIND Va=Vt X Ra/Rt =50V X 2KΩ/10KΩ =10V FIND Vb=Vtx(Ra+Rb)/Rt =50Vx(2KΩ+3KΩ)/10KΩ =25V FIND Vc=Vtx(Ra+Rb+Rc)/Rt =50Vx(2KΩ+3KΩ+5KΩ)/10KΩ =50V V T 50V C +50V B +25V A +10V R 1 5K Ω R2 3K Ω ` R3 2K Ω USE VOLTAGE DIVIDER EQ. TO FIND VOLTAGE’S AT POINTS A,B,C. VR N =VTR N /R T YOU TUBE: Voltage divider tutorial

32 PROBLEM: PLACE 5KΩ LOAD RESISTOR HERE; Rb=5kΩX5KΩ/10KΩ=2.5KΩ TWO RESISTORS IN PARALLEL Vb=VtxR b /Rt =50VX2.5K Ω /7.5K Ω =16.67V Va=VbxRa/Rb =16.67V X 2KΩ/5KΩ =6.67V VOLTAGE DIVIDER IS NOT WORKING PROPERLY UNDER A LOAD. VOLTAGE VARATION OF A LOAD IS % OF VOLTAGE REGULATION. %REG.=Vmin L -Vmax L /Vmax L (100) WHERE Vmin L =V WITH MIN. LOAD Vmax =V WITH MAX.LOAD PT. A WITH NO LOAD= /10(100)=50% PT. B WITH 5KΩ LOAD= /10(100)=50% V T 50V B A 5K Ω 3K Ω 2K Ω 5K Ω LOAD 5K Ω 2.5K Ω B

33 Voltage Divider The two resistor voltage divider is used often to supply a voltage different from that of an available battery or power supply. In application the output voltage depends upon the resistance of the load it drives. Where

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36 RESISTOR VOLTAGE DIVIDERS HAVE BEEN REPLACED WITH ZENER DIODES(POLARIZED DEVICE) A VOLTAGE DIVIDER CAN BE MADE WITH A RESISTOR AND A ZENER DIODE IN SERIES. A ZENER DIODE IS OPERATED WITH ITS POLARITY REVERSED, THIS IS CALLED REVERSE BAIS. WHEN A ZENER DIODE IS REVERSED BAISED, NO CURRENT FLOWS THRU IT. AS THE LOAD VOLTAGE INCREASES TO NEAR 5V, THE ZENER DIODE STARTS TO BREAK DOWN AND ALLOWS CURRENT TO FLOW THRU IT. THIS RESULT LETS THE LOAD VOLTAGE STAY NEAR ITS DESIRED VALUE, WHICH MAKES THIS A GOOD VOLTAGE REGULATOR. CATHODE - ANODE + IN4735 BREAKDOWN VOLTAGE=6.2V 10V + _ + + _ _ 500 Ω D 1 5V R LOAD 1K Ω P.121


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