1. FOWLER CHAPTER 5 LECTURE 6 MULTIPLE LOAD CIRCUITS

2. 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

3. 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

4. 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

5. 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

6. - - - - - 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.

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

8. 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

9. 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.

10. 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

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
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

12. 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

13. 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

14. 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

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. 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

18. 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.

19. 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

20. 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

21. 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

22. 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

23. P.114 P.115 CURRENT DIVIDER FORMULA FOR TWO PARALLEL RESISTORS P.114IT
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

24. 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

25. SERIES-PARALLEL CIRCUIT

26. 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

28. 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

29. 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

31. 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

32. 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%

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                                                        

36. 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.