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

FIRST WE PRACTICE COMBINING RESISTORS

6k||3k

(10K,2K)SERIES

SERIESk3

kkk 412||6

k12k3

k5

kkk 612||12

kkk 26||3

)24(||6 kkk

12kIf things get confusing…

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

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

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

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

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

kkk 12||2

VVkk

kV

O1)3(

21

1

:DIVIDER VOLTAGE

kkk 211

AAkk

kI

O1)3(

21

1

:DIVIDER CURRENT

LEARNING BY DOING

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

OV FIND

DIVIDER VOLTAGEUSE

FIND :STRATEGY1

V

1V

k60

kkk 2060||30

+

-

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

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

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

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 ...

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

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