FOWLER CHAPTER 5
LECTURE 6 MULTIPLE LOAD
CIRCUITS
MULTIPLE LOAD CIRCUITS
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
IT  I1  I 2  I 3
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
RT  R1  R2  R3  ....RN
R1 5Ω
R2 10Ω
90 VDC
EXAMPLE:
R3 30Ω
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
RT  R1  R2  R3  5  10  30  45
VOLTAGE IN SERIES CIRCUITS
P.101
VOLTAGE IS DIVIDED UP ACROSS EACH LOAD.
KIRCHOFF’S VOLTAGE LAW: THE SUM OF VOLTAGE DROPS AROUND THE CIRCUIT
PATH = THE SOURCE VOLTAGE.
VT  V1  V2  V3
VR1  IR1  2 A  5  10V
R1 5Ω
VR 2  IR2  2 A 10  20V , VOLTAGE DROP IS 2  R1
VR 3  IR3  2 A  30  60V , VOLTAGE DROP IS 6  R1
2A
VT  10V  20V  60V  90V
THIS CONFIRMS KIRCHOFF VOLTAGE LAW.
R2 10Ω
90 VDC
R3 30Ω
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).
-
P.101
+
a
b
R1
-
-
I
VDC
R2
+
c
+
MEASURING SERIES VOLTAGE. F.5-7 P-102
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
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
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
0V,0W
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
VRN
RN
 VT 
RT
VOLTAGE DIVIDER EQ
R1
35Ω
R1
VR1  VT 
R1  R 2  R3
35
VR1  90 
35  70  45
VR1  21V
70Ω
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.
R2
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 ↓
M
120VAC
NOT EFFICIENT SINCE R WASTES
ENERGY AS HEAT.↑
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
12V
MPT OCCURS WHEN RB=R1=33Ω
PR1=PRB= I x V=2AX12=24W
SEE T 5-1 P.108
RL 3Ω
LOAD
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.
IT
P.108
I1
V
I2
R1
I3
R2
R3
CURRENT IN PARALLEL CIRCUITS
P.109
IT=IR1+IR2+IR3 : TOTAL CURRENT IS THE SUM OF THE BRANCH CURRENTS.
A
B
IT
IR1
V
R1
IR2
R2
IR3
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
3A
5A
2A
10A=10 A
NODE
3A
A
B
IT
IA
IR1
V
7A
R1
AT NODE A:
IT  I A  I R1
AT NODE B:
I A  I B  I R2
IB
IR2
IR3
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.
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
A
11mA
A
10mA
R2 10KΩ
A
1mA
R1 RESISTANCE IS 10 TIMES LESS THEN R2, BUT HAS 10 TIMES THE CURRENT FLOW THRU IT.
RESISTANCE IN PARALLEL CIRCUITS
THE TOTAL RESISTANCE OF A PARALLELCIRCUIT IS ALWAYS<THAT OF THE
LOWEST BRANCH RESISTANCE.
P.110
HOW IS THIS POSSIBLE?
V
R
I T  10V
IT 
10V
R1
10KΩ
10000
I T  0.001A  1`mA
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
10V
R1
10KΩ
R2
SO
100KΩ
RT  VT
IT
 10V
1.1mA
 10
0.0011
 9091  9.09 K
RT IS < THE LOWEST BRANCH RESISTANCE.
P.113
10V
R1
10KΩ
R2
100KΩ
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
R3
1KΩ
EXAMPLE 5-3 P-112
R1=20Ω R2=30Ω R3=60Ω
20Ω
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
30Ω
60Ω
MEASURING RESISTANCE IN PARALLEL
P.112
MUST DISCONNECT THE RESISTOR BEING MEASURED FROM THE LOAD TO READ THE
INDIVIDUAL RESISTOR RESISTANCE,IF YOU DON’T, YOU WILL MEASURE RT
R1
R1
RESISTOR LEAD DISCONNECTED
R2
R2
Ω
Ω
R3
R3
CURRENT DIVIDER FORMULA FOR TWO PARALLEL RESISTORS P.114
2A
P.114
IR2
IT
IR1
VT
R1
IRI=ITxR2/(R1+R2)
R2
47Ω
22Ω
IR2=ITxR1/(R1+R2)
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
P.115
IS THE RECIPROCAL OF RESISTANCE,IS THE ABILITY TO CONDUCT
CURRENT.
G=1/R
G IS MEASURED IN SIEMENS
SEE APPENDIX C
USEFUL FORMULA’S
FOR SERIES CIRCUITS
IT=I1=I2…..=In
P.474-475
R1
VT=V1+V2…..+Vn
RT=R1+R2…..+Rn
VT
PT=P1+P2……+Pn
VOLTAGE DIVIDER EQ.
TO FIND VOLTAGE ACROSS ONLY ONE
RESISITOR IN A SERIES CIRCUIT.
VT  RN
VRn 
RT
RN
____________________________________________________________________________________________________________________
FOR PARALLEL CIRCUITS
IT=I1+I2…..+In
VT=V1=V2…..=Vn
RT=1/(1/R1+1/R2…+1/Rn)
PT=P1+P2…….+Pn
V
R1
R1  R2
RT 
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
CURRENT DIVIDER FORMULA
FOR 2 PARALLEL RESISTORS
I  R2
I T  R1
I R1  T
I 
R1  R2 R 2 R1  R2
R2
IT
I2
A
IT
I1
A
I2
SERIES-PARALLEL CIRCUIT
R1
SERIES PARALLEL CIRCUITS P.117
15Ω
V
R2
R3
30Ω
20Ω
FIND RT FOR THIS CIRCUIT
FIRST SOLVE FOR THE PARALLEL PAIR R2,R3
R2,3=R2XR3/(R2+R3)
=20X30/(20+30)
R1
15Ω
V
=600/50
=12Ω
R2,R3
12Ω
NOW FIND RT FOR THE SERIES CIRCUIT
RT=R1+R2,3
V
RT=27Ω
=15Ω+12Ω
=27Ω
FIND RT FOR THIS SERIES PARALLEL CIRCUIT
R1
40Ω
V
R3
FIND SERIES RESISTANCE FOR R1 + R2 FIRST.
22Ω
R1,2=R1+R2
R2
60Ω
=40Ω+60Ω
=100Ω
R1,2
100Ω
V
R3
22Ω
NEXT FIND RT FOR THIS PARALLEL CIRCUIT.
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
R1
VR1=60V
FINDVR3: VR2=VR3+VR4
A
VR3=VR2-VR4
VT
I1
R2
40V
100V
I2
R3
VR3=40V-30V
VR3=10V
USING KIRCHHOFF’S CURRENT LAW,FIND IR1,IR2,IR3
R4
SINCE .6A FLOWS THRU NODE B, 0.6A FLOWS THRU
30V R1(THIS LOOP IS A SERIES CIRCUIT)
IT=.6A
B
FIND I1, AT NODE A
I2=.2A
IT=I1+I2
I1
IT
A I2
I1=IT-I2
I1=0.6A-0.2A=0.4A
VOLTAGE DIVIDERS AND REGULATORS P-121
USE VOLTAGE DIVIDER EQ. TO FIND
VOLTAGE’S AT POINTS A,B,C.
VRN=VTRN/RT
C +50V
FIND Va=VtXRa/Rt
R1
=50VX2KΩ/10KΩ
5KΩ
=10V
B
VT
+25V
FIND Vb=Vtx(Ra+Rb)/Rt
50V
R2
3KΩ
`
R3
2KΩ
=50Vx(2KΩ+3KΩ)/10KΩ
=25V
A
+10V
FIND Vc=Vtx(Ra+Rb+Rc)/Rt
=50Vx(2KΩ+3KΩ+5KΩ)/10KΩ
=50V
YOU TUBE: Voltage divider tutorial http://www.youtube.com/watch?v=XxLKfAZrhbM
PROBLEM: PLACE 5KΩ LOAD RESISTOR HERE;
Rb=5kΩX5KΩ/10KΩ=2.5KΩ
5KΩ
B
VT
50V
TWO RESISTORS IN PARALLEL
5KΩ
3KΩ
2KΩ
A
Vb=VtxRb/Rt
5KΩ
LOAD
B
2.5KΩ
=50VX2.5KΩ/7.5KΩ
=16.67V
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
%REG@ PT. A WITH NO LOAD=10-6.67/10(100)=50%
%REG@ PT. B WITH 5KΩ LOAD=25-16.67/10(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
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.