# FOWLER CHAPTER 5 LECTURE 6 MULTIPLE LOAD CIRCUITS

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FOWLER CHAPTER 5 LECTURE 6 MULTIPLE LOAD CIRCUITS

P.99 MULTIPLE LOAD CIRCUITS - + R1 - - R2 VT + + IT + - R3
IF YOU HAVE TWO OR MORE LOADS, THEY CAN BE IN SERIES, PARALLEL OR SERIES-PARALLEL. SUBSCRIPTS: R1, R2, ……RN INCIDATES # OF DIFFERENT LOAD RESISTORS. V1 , V2,……..V INCIDATES # OF DIFFERENT VOLTAGES. II , I2 ,………IN INCIDATES # OF DIFFERENT CURRENTS. VT: TOTAL VOLTAGE FOR CIRCUIT. IT : TOTAL CURRENT FOR CIRCUIT. POWER : PT = P1 +P2 …….PN TOTAL POWER FOR CIRCUIT. SERIES CIRCUITS: ONE PATH FOR CIRCUIT FLOW. P.99 - + R1 - - R2 VT + + IT + - R3 CURRENT IS THE SAME IN EACH ELEMENT: IT=I1=I2=I3

IN SERIES CIRCUITS, CURRENT IS THE SAME IN EACH ELEMENT.
P.100 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. 2A R1 5Ω R2 10Ω 90 VDC R3 30Ω EXAMPLE: THERE ARE TWO WAYS TO FIND THE TOTAL RESISTANCE FOR THIS CIRCUIT. USE OHM’S LAW V=IR, R=V/I =90V/2A =45Ω OR

P.101 VOLTAGE IN SERIES CIRCUITS R1 5Ω 2A
VOLTAGE IS DIVIDED UP ACROSS EACH LOAD. P.101 KIRCHOFF’S VOLTAGE LAW: THE SUM OF VOLTAGE DROPS AROUND THE CIRCUIT PATH = THE SOURCE VOLTAGE. R1 5Ω 2A THIS CONFIRMS KIRCHOFF VOLTAGE LAW. R2 10Ω 90 VDC R3 30Ω

P.101 VOLTAGE DROP AND POLARITY POLARITY - + a b R1 - - I VDC R2 + + c
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). - + P.101 a b R1 - - I VDC R2 + + c

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

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 V R1 12 V R2 R2 V 0V R3 CIRCUIT PATH IS BROKEN V 0V

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

VOLTAGE DIVIDER EQ. CAN BE USED TO FIND VOLTAGE ACROSS ANY RESISITOR IN A SERIES CIRCUIT.( RN)
FROM EXAMPLE 5-2 P-105 FIND VR1, P.105 VOLTAGE DIVIDER EQ R1 35Ω 70Ω R2 90V 45Ω R3 IF DOMINANT RESISTOR IS SO MUCH GREATER THEN THE OTHERS IN THE CIRCUIT,THE EFFECTS OF THE SMALLER RESISTOR CAN BE IGNORED.

APPLICATIONS OF SERIES CIRCUITS F5-13,14 P-106
R IS A VARIABLE SERIES RESISTOR(RHEOSTAT) R AS R↓ MOTOR SPEED ↑ AS R↑ MOTOR SPEED ↓ NOT EFFICIENT SINCE R WASTES ENERGY AS HEAT.↑ M 120VAC S 24V L LIGHT INTENSITY CONTROL F.5-14 P-106 WORKS WELL FOR DC ONLY

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 RB 3Ω SOURCE RL 3Ω LOAD 12V MPT OCCURS WHEN RB=R1=33Ω PR1=PRB= I x V=2AX12=24W SEE T 5-1 P.108

P.108 PARALLEL CIRCUITS IT I1 I2 I3 R3 V R1 R2
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. IT P.108 I1 I2 I3 R3 V R1 R2

P.109 CURRENT IN PARALLEL CIRCUITS
IT=IR1+IR2+IR3 : TOTAL CURRENT IS THE SUM OF THE BRANCH CURRENTS. P.109 A B IT IR1 IR3 IR2 V R1 R2 R3

P.109 5A 3A 2A NODE 3A 7A A IT IA IB IR1 IR2 IR3 V R1 R2 R3
KIRCHHOFF’S CURRENT LAW:: THE SUM OF CURRENTS ENTERING A JUNCTION=SUM OF CURRENTS LEAVING A JUNCTION. (NODE) P.109 CURRENT GOING INTO NODE=CURRENT GOING OUT NODE. 3A+5A+2A=7A+3A 10A=10 A 5A 3A 2A NODE 3A 7A A B AT NODE A: IT IA IB AT NODE B: IR1 IR2 IR3 V R1 R2 R3

KIRCHHOFF’S CURRENT LAW
KIRCHHOFF’S VOLTAGE LAW

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.

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

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

P.110 RESISTANCE IN PARALLEL CIRCUITS 10V R1 10KΩ
THE TOTAL RESISTANCE OF A PARALLELCIRCUIT IS ALWAYS<THAT OF THE LOWEST BRANCH RESISTANCE. HOW IS THIS POSSIBLE? P.110 10V R1 10KΩ ADD A 2ND RESISTOR IN PARALLEL SINCE VOLTAGE IS THE SAME IN EACH BRANCH IR2=10V/100,000Ω=0.1mA FROM KIRCHHOFF’S CURRENT LAW IT=IRI+IR2 IT=1mA+0.1mA IT=1.1mA RT IS < THE LOWEST BRANCH RESISTANCE. 10V R1 10KΩ R2 100KΩ

P.113 R1 10KΩ R2 100KΩ R3 1KΩ 10V ADD A 3TH RESISTOR R3 TO THIS CIRCUIT, FOLLOW THE SAME LOGIC AS ON THE LAST SLIDE. RT=VT/IT= VT/(I1+I2+I3) ,SINCE IN=VT/RN (OHM’S LAW) RT=VT/(VT/R1+VT/R2+VT/R3) USING ALGEBRAIC MAGIC GIVES RT=1/(1/R1+1/R2+1/R3) RECIPROCAL FORMULA

RT=R1xR2/(R1+R2) RT=R/n 20Ω 30Ω 60Ω EXAMPLE 5-3 P-112
RT= 1/( 1/R1+ 1/R2+ 1/R3) RT= 1/( 1/20Ω+ 1/30Ω+ 1/60Ω) RT=1/(6/60)=10Ω IF WE HAVE ONLY TWO RESISTORS IN PARALLEL THEN WE CAN FIND RT USING; RT=R1xR2/(R1+R2) EXAMPLE 5-4 P112 GIVEN R1=27Ω R2=47Ω ,FIND RT RT=R1XR2/(R1+R2) =27X47/(27+47) =1269/74 =17.1Ω IF ALL THE RESISTORS IN A PARALLEL CIRCUIT HAVE THE SAME VALUE. YOU CAN USE THE EQUATION:` RT=R/n TO FIND THE TOTAL RESISTANCE. WHERE n =THE # OF RESISTORS IN THE CIRCUIT. EXAMPLE: GIVEN TWO 100Ω RESISTORS IN PARALLEL, FIND RT RT=R/n = 100/2 =50Ω 20Ω 30Ω 60Ω

P.112 MEASURING RESISTANCE IN PARALLEL R1 R2 R3 R1 R2 R3
MUST DISCONNECT THE RESISTOR BEING MEASURED FROM THE LOAD TO READ THE INDIVIDUAL RESISTOR RESISTANCE,IF YOU DON’T, YOU WILL MEASURE RT R1 R2 R3 Ω R1 R2 R3 Ω RESISTOR LEAD DISCONNECTED

P.114 P.115 CURRENT DIVIDER FORMULA FOR TWO PARALLEL RESISTORS P.114
IT IR2 IR1 IRI=ITxR2/(R1+R2) IR2=ITxR1/(R1+R2) R1 22Ω R2 47Ω VT FIND IR1= 2X47/(2+47) FIND IR2= 2X22/(2+47) IR1=1.36A IR2=0.897A SAVES THE TROUBLE OF HAVING TO FIND: RT USING 1/R OR: VT USING OHM’S LAW OR: IR1,IR2 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 P.115

USEFUL FORMULA’S P.474-475 FOR SERIES CIRCUITS IT=I1=I2…..=In
VT=V1+V2…..+Vn RT=R1+R2…..+Rn PT=P1+P2……+Pn VOLTAGE DIVIDER EQ. TO FIND VOLTAGE ACROSS ONLY ONE RESISITOR IN A SERIES CIRCUIT. ____________________________________________________________________________________________________________________ FOR PARALLEL CIRCUITS IT=I1+I2…..+In VT=V1=V2…..=Vn RT=1/(1/R1+1/R2…+1/Rn) PT=P1+P2…….+Pn SEE APPENDIX C P R1 VT RN V R1 R2 I1 IF ONLY TWO RESISTORS ARE IN PARALLEL, FIND RT USING: IF ALL RESISTORS ARE THE SAME: RT=R/n WHERE n=#OF RESISTORS IT I2 A I1 IT I2 CURRENT DIVIDER FORMULA FOR 2 PARALLEL RESISTORS A

SERIES-PARALLEL CIRCUIT

SERIES PARALLEL CIRCUITS P.117
15Ω V R2 R3 20Ω 30Ω FIND RT FOR THIS CIRCUIT FIRST SOLVE FOR THE PARALLEL PAIR R2,R3 R2,3=R2XR3/(R2+R3) =20X30/(20+30) =600/50 =12Ω R1 15Ω V R2,R3 12Ω NOW FIND RT FOR THE SERIES CIRCUIT RT=R1+R2,3 =15Ω+12Ω =27Ω RT=27Ω V

FIND RT FOR THIS SERIES PARALLEL CIRCUIT
40Ω R3 FIND SERIES RESISTANCE FOR R1 + R2 FIRST. V 22Ω R1,2=R1+R2 =40Ω+60Ω =100Ω R2 60Ω R1,2 100Ω R3 22Ω NEXT FIND RT FOR THIS PARALLEL CIRCUIT. V RT = R1,2XR3/(R1,2+R3) RT =100X22/122 =2200/122 =18Ω V RT=18Ω

KIRCHHOFF’S LAWS IN SERIES- PARALLEL CIRCUITS F.5-30, P119
USING KIRCHHOFF’S LAW FIND VR1,VR3 GIVEN THE VAULES SHOWN ON THE SCHEMATIC FIND VR1: VT=VR1+VR2 VR1=VT-VR2 VR1=100V-40V VR1=60V FINDVR3: VR2=VR3+VR4 VR3=VR2-VR4 VR3=40V-30V VR3=10V USING KIRCHHOFF’S CURRENT LAW,FIND IR1,IR2,IR3 SINCE .6A FLOWS THRU NODE B, 0.6A FLOWS THRU R1(THIS LOOP IS A SERIES CIRCUIT) FIND I1, AT NODE A R1 A VT 100V R3 I1 R2 40V I2 R4 30V B IT=.6A I2=.2A IT=I1+I2 I1=IT-I2 I1=0.6A-0.2A=0.4A I1 IT A I2

5KΩ 50V VOLTAGE DIVIDERS AND REGULATORS P-121 C +50V R1 B +25V VT R2
USE VOLTAGE DIVIDER EQ. TO FIND VOLTAGE’S AT POINTS A,B,C. VRN=VTRN/RT C V FIND Va=VtXRa/Rt =50VX2KΩ/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 R1 5KΩ B V VT 50V R2 3KΩ ` A V R3 2KΩ YOU TUBE: Voltage divider tutorial

PROBLEM: PLACE 5KΩ LOAD RESISTOR HERE;
Rb=5kΩX5KΩ/10KΩ=2.5KΩ TWO RESISTORS IN PARALLEL 5KΩ B VT 50V 5KΩ 3KΩ Vb=VtxRb/Rt =50VX2.5KΩ/7.5KΩ =16.67V 5KΩ LOAD B 2KΩ A 2.5KΩ Va=VbxRa/Rb =16.67VX2KΩ/5KΩ =6.67V VOLTAGE DIVIDER IS NOT WORKING PROPERLY UNDER A LOAD. VOLTAGE VARATION OF A LOAD IS % OF VOLTAGE REGULATION. %REG.=VminL-VmaxL/VmaxL(100) WHERE VminL=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%

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

P.121 CATHODE - IN4735 BREAKDOWN VOLTAGE=6.2V ANODE + + 500Ω _ + 10V _
RESISTOR VOLTAGE DIVIDERS HAVE BEEN REPLACED WITH ZENER DIODES(POLARIZED DEVICE) P.121 CATHODE - IN4735 BREAKDOWN VOLTAGE=6.2V ANODE + A VOLTAGE DIVIDER CAN BE MADE WITH A RESISTOR AND A ZENER DIODE IN SERIES. + 500Ω _ + 10V _ + D1 5V RLOAD 1KΩ _ 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.