SERIES PARALLEL RESISTOR...
Transcript of SERIES PARALLEL RESISTOR...
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SERIES PARALLEL RESISTOR COMBINATIONS
UP TO NOW WE HAVE STUDIED CIRCUITS THATCAN BE ANALYZED WITH ONE APPLICATION OFKVL(SINGLE LOOP) OR KCL(SINGLE NODE-PAIR)
WE HAVE ALSO SEEN THAT IN SOME SITUATIONSIT IS ADVANTAGEOUS TO COMBINE RESISTORS TO SIMPLIFY THE ANALYSIS OF A CIRCUIT
NOW WE EXAMINE SOME MORE COMPLEX CIRCUITSWHERE WE CAN SIMPLIFY THE ANALYSIS USINGTHE TECHNIQUE OF COMBINING RESISTORS…
… PLUS THE USE OF OHM’S LAW
SERIES COMBINATIONS
PARALLEL COMBINATION
Np GGGG ...21
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FIRST WE PRACTICE COMBINING RESISTORS
6k||3k
(10K,2K)SERIES
SERIESk3
kkk 412||6
k12k3
k5
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kkk 612||12
kkk 26||3
)24(||6 kkk
12kIf things get confusing…
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EXAMPLES COMBINATION SERIES-PARALLEL
k9
kkk 69||18
kkk 1066
RESISTORS ARE IN SERIES IF THEY CARRYEXACTLY THE SAME CURRENT
RESISTORS ARE IN PARALLEL IF THEY ARECONNECTED EXACTLY BETWEEN THE SAME TWONODES
If the drawing gets confusing…Redraw the reduced circuit and start again
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AN “INVERSE SERIES PARALLEL COMBINATION”
AVAILABLE ARERESISTORS ONLY
WHEN600mV BE MUST
1.0
3AIVR
2.03
6.
A
VR REQUIRED 1.01.0R
AVAILABLE ARERESISTORS ONLY
WHEN600mV BE MUST
1.0
9AIVR
0667.09
6.
A
VR REQUIRED
R
SIMPLE CASE
NOT SO SIMPLE CASE
Given the final valueFind a proper combination
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EFFECT OF RESISTOR TOLERANCE
10% :TOLERANCE RESISTOR
2.7k : VALUERESISTORNOMINAL
RANGES FOR CURRENT AND POWER?
mAI
mAI
115.47.29.0
10
367.37.21.1
10
max
min
:CURRENT MAXIMUM
:CURRENT MINIMUM
THE RANGES FOR CURRENT AND POWER ARE DETERMINED BY THE TOLERANCE BUT THE PERCENTAGE OF CHANGE MAY BE DIFFERENT FROM THE PERCENTAGEOF TOLERANCE. THE RANGES MAY NOT EVEN BE SYMMETRIC
mAI 704.37.2
10
:CURRENTNOMINAL
_
mWP 04.377.2
102
:POWERNOMINAL
:POWER MAXIMUM
:)POWER(VI MINIMUM min
mW
mW
15.41
67.33
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CIRCUIT WITH SERIES-PARALLEL RESISTOR COMBINATIONS
COMBINING RESISTORS IN SERIES ELIMINATESONE NODE FROM THE CIRCUIT.COMBINING RESISTORS IN PARALLEL ELIMINATESONE LOOP FROM THE CIRCUIT
THE COMBINATION OF COMPONENTS CAN REDUCETHE COMPLEXITY OF A CIRCUIT AND RENDER ITSUITABLE FOR ANALYSIS USING THE BASIC TOOLS DEVELOPED SO FAR.
GENERAL STRATEGY: •REDUCE COMPLEXITY UNTIL THE CIRCUIT BECOMES SIMPLE ENOUGH TO ANALYZE.•USE DATA FROM SIMPLIFIED CIRCUIT TOCOMPUTE DESIRED VARIABLES IN ORIGINALCIRCUIT - HENCE ONE MUST KEEP TRACKOF ANY RELATIONSHIP BETWEEN VARIABLES
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FIRST REDUCE IT TO A SINGLE LOOP CIRCUITk12kk 12||4
k6
kk 6||6
k
VI
12
121 )12(
93
3
aV
SECOND: “BACKTRACK” USING KVL, KCL OHM’S
k
VI a
62 :SOHM'
0321 III :KCL
3*3 IkVb :SOHM'
3I
…OTHER OPTIONS...
4
34
*4
124
12
IkV
II
b
5
345
*3
0
IkV
III
C
:SOHM'
:KCL
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kkk 12||2
VVkk
kV
O1)3(
21
1
:DIVIDER VOLTAGE
kkk 211
AAkk
kI
O1)3(
21
1
:DIVIDER CURRENT
LEARNING BY DOING
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AN EXAMPLE OF “BACKTRACKING”
A STRATEGY. ALWAYS ASK: “WHAT ELSE CAN ICOMPUTE?”
4*6 IkV
b
k
VI b
33
mA1
432III
mA5.1
2*2 IkV
a
V3
V3
baxzVVV
VVxz
6
k
VI xz
45
mA5.1
521III
mAI 31
11*4*6 IkVIkV
xzO
VVO 36
mA5.0
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OV FIND
DIVIDER VOLTAGEUSE
FIND :STRATEGY1
V
1V
k60
kkk 2060||30
+
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1Vk20
k20
V12
Vkk
k6)12(
2020
20
V6
DIVIDER VOLTAGE
1
4020
20V
kk
kV
O
V2
SV FIND
THIS IS AN INVERSE PROBLEMWHAT CAN BE COMPUTED?
V6
mA05.0mA15.0
mAkV 1.0*601
k
VI
120
61
VmAkVS
615.0*20
V9
SERIESPARALLEL
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TIONSTRANSFORMAY
THIS CIRCUIT HAS NO RESISTOR IN SERIES OR PARALLEL
IF INSTEADOF THIS
WE COULDHAVE THIS
THEN THE CIRCUIT WOULDBECOME LIKE THIS ANDBE AMENABLE TO SERIESPARALLEL TRANSFORMATIONS
http://www.wiley.com/college/irwin/0470128690/animations/swf/D2Y.swf
http://www.wiley.com/college/irwin/0470128690/animations/swf/D2Y.swf
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Y
baab RRR
)(|| 312 RRRRab
321
312 )(
RRR
RRRRR ba
321
213 )(
RRR
RRRRR cb
321
321 )(
RRR
RRRRR ac
SUBTRACT THE FIRST TWO THEN ADDTO THE THIRD TO GET Ra
Y
RRR
RRR
RRR
RRR
RRR
RRR
c
b
a
321
13
321
32
321
21
a
b
b
a
R
RRR
R
R
R
R 13
3
1 c
b
c
b
R
RRR
R
R
R
R 12
1
2
REPLACE IN THE THIRD AND SOLVE FOR R1
Y
R
RRRRRRR
R
RRRRRRR
R
RRRRRRR
a
accbba
c
accbba
b
accbba
3
2
1
Y
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LEARNING EXAMPLE: APPLICATION OF WYE-DELTA TRANSFORMATION
SI COMPUTE DELTA CONNECTION
a b
c
a b
c
kkkkkkREQ 10)62(||936
Y
RRR
RRR
RRR
RRR
RRR
RRR
c
b
a
321
13
321
32
321
21
1R
2R
3Rkkk
kk
18612
612
mAk
VIS 2.1
12
12
ONE COULD ALSO USE A WYE - DELTA TRANSFORMATION ...
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LEARNING EXAMPLE
SHOULD KEEP THESE TWO NODES!
CONVERT THIS Y INTO A DELTA?
IF WE CONVERT TO Y INTO A DELTA THERE ARE SERIES PARALLEL REDUCTIONS!
Y
R
RRRRRRR
R
RRRRRRR
R
RRRRRRR
a
accbba
c
accbba
b
accbba
3
2
1
3*12 *1236
12
k kk
k
4mA 36k
36k
36k
12k
12kO
V
THE RESULTINGCIRCUIT IS A CURRENT DIVIDER
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4mA 36k
36 ||12 9k k k
OV
9k
CIRCUIT AFTER PARALLEL RESISTORREDUCTION
OI
36 84
36 18 3O
kI mA mA
k k
89 9 24
3O O
V k I k mA V NOTICE THAT BY KEEPINGTHE FRACTION WE PRESERVEFULL NUMERICAL ACCURACY
WYE DELTA