SERIES PARALLEL RESISTOR COMBINATIONS
UP TO NOW WE HAVE STUDIED CIRCUITS THAT
CAN BE ANALYZED WITH ONE APPLICATION OF
KVL(SINGLE LOOP) OR KCL(SINGLE NODE-PAIR)
WE HAVE ALSO SEEN THAT IN SOME SITUATIONS
IT IS ADVANTAGEOUS TO COMBINE RESISTORS
TO SIMPLIFY THE ANALYSIS OF A CIRCUIT
NOW WE EXAMINE SOME MORE COMPLEX CIRCUITS
WHERE WE CAN SIMPLIFY THE ANALYSIS USING
THE TECHNIQUE OF COMBINING RESISTORS…
… PLUS THE USE OF OHM’S LAW
SERIES COMBINATIONS
PARALLEL COMBINATION
G p  G1  G2  ...  G N
FIRST WE PRACTICE COMBINING RESISTORS
3k
SERIES
6k||3k
(10K,2K)SERIES
6k || 12k  4k
5k
3k
12k
3k || 6k  2k
12k
6k || (4k  2k )
12k || 12k  6k
If things get confusing…
EXAMPLES COMBINATION SERIES-PARALLEL
9k
If the drawing gets confusing…
Redraw the reduced circuit
and start again
18k || 9k  6k
RESISTORS ARE IN SERIES IF THEY CARRY
EXACTLY THE SAME CURRENT
6k  6k  10k
RESISTORS ARE IN PARALLEL IF THEY ARE
CONNECTED EXACTLY BETWEEN THE SAME TWO
NODES
AN “INVERSE SERIES PARALLEL COMBINATION”
Given the final value
Find a proper combination
SIMPLE CASE
VR MUST BE 600mV WHEN I  3A
ONLY 0.1 RESISTORS ARE AVAILABLE
REQUIRED R 
.6V
 0.2  R  0.1  0.1
3A
NOT SO SIMPLE CASE
VR MUST BE 600mV WHEN I  9A
ONLY 0.1 RESISTORS ARE AVAILABLE
REQUIRED R 
.6V
 0.0667 
9A
R
EFFECT OF RESISTOR TOLERANCE
NOMINAL RESISTOR VALUE : 2.7k
RESISTOR TOLERANCE : 10%
RANGES FOR CURRENT AND POWER?

10
NOMINAL CURRENT : I 
 3.704 mA
2.7
NOMINAL POWER : P 
10
2
2.7
_
 37.04 mW
10
 3.367 mA
MINIMUM POWER(VI min ) : 33.67 mW
1.1 2.7
10

 4.115 mA MAXIMUM POWER : 41.15 mW
0.9  2.7
MINIMUM CURRENT : I min 
MAXIMUM CURRENT : I max
THE RANGES FOR CURRENT AND POWER ARE DETERMINED BY THE TOLERANCE
BUT THE PERCENTAGE OF CHANGE MAY BE DIFFERENT FROM THE PERCENTAGE
OF TOLERANCE. THE RANGES MAY NOT EVEN BE SYMMETRIC
CIRCUIT WITH SERIES-PARALLEL RESISTOR COMBINATIONS
THE COMBINATION OF COMPONENTS CAN REDUCE
THE COMPLEXITY OF A CIRCUIT AND RENDER IT
SUITABLE FOR ANALYSIS USING THE BASIC
TOOLS DEVELOPED SO FAR.
COMBINING RESISTORS IN SERIES ELIMINATES
ONE NODE FROM THE CIRCUIT.
COMBINING RESISTORS IN PARALLEL ELIMINATES
ONE LOOP FROM THE CIRCUIT
GENERAL STRATEGY:
•REDUCE COMPLEXITY UNTIL THE CIRCUIT
BECOMES SIMPLE ENOUGH TO ANALYZE.
•USE DATA FROM SIMPLIFIED CIRCUIT TO
COMPUTE DESIRED VARIABLES IN ORIGINAL
CIRCUIT - HENCE ONE MUST KEEP TRACK
OF ANY RELATIONSHIP BETWEEN VARIABLES
4k || 12k 12k
FIRST REDUCE IT TO A SINGLE LOOP CIRCUIT
SECOND: “BACKTRACK” USING KVL, KCL OHM’S
6k
I3
V
OHM' S : I 2  a
6k
KCL : I1  I 2  I3  0
OHM'S : Vb  3k * I3
…OTHER OPTIONS...
6k || 6k
KCL : I 5  I 4  I 3  0
OHM' S : VC  3k * I 5
I1 
12V
12k
Va 
3
(12)
39
12
I3
4  12
Vb  4k * I 4
I4 
2k || 2k  1k
VOLTAGE DIVIDER : VO 
LEARNING
BY DOING
1k
(3V )  1V
1k  2k
1k  1k  2k
CURRENT DIVIDER : I O 
1k
(3 A)  1A
1k  2k
AN EXAMPLE OF “BACKTRACKING”
 1.5mA
I1  3mA
Vxz  6V
3V
 1.5mA
1mA
VO  36V
3V
 0.5mA
A STRATEGY. ALWAYS ASK: “WHAT ELSE CAN I
COMPUTE?”
Vb  6k * I 4
I3 
Vb
3k
I 2  I3  I 4
Va  2k * I 2
Vxz  Va  Vb
V
I 5  xz
4k
I1  I 2  I 5
VO  6k * I1  Vxz  4k * I1
FIND VO
60k
V1 6V
FIND VS
V1  60k * 0.1mA
0.15mA 
2V
30k || 60k  20k
STRATEGY : FIND V1
USE VOLTAGE DIVIDER
6V
9V

VS  20k * 0.15mA  6V

20k
6V
120k
THIS IS AN INVERSE PROBLEM
WHAT CAN BE COMPUTED?
20k
+
-
I1 
0.05mA
V1

20k
(12)  6V
20k  20k

12V
VOLTAGE DIVIDER
VO 
20k
V1
20k  40k
SERIES
PARALLEL
http://www.wiley.com/college/irwin/0470128690/animations/swf/D2Y.swf
Y   TRANSFORMA TIONS
THIS CIRCUIT HAS NO RESISTOR IN
SERIES OR PARALLEL
IF INSTEAD
OF THIS
WE COULD
HAVE THIS
THEN THE CIRCUIT WOULD
BECOME LIKE THIS AND
BE AMENABLE TO SERIES
PARALLEL TRANSFORMATIONS
Rab  R2 || ( R1  R3 )
 Y
Rab  Ra  Rb
Y 
Ra R1
Rb R1
R2 ( R1  R3 )
R
R
1
2


R

Ra  Rb 
3
Ra 
R
R
Ra
b
3
R1  R2  R3
R1  R2  R3
Rb R2
RR

 R2  b 1
Rc R1
Rc
REPLACE IN THE THIRD AND SOLVE FOR R1
R2 R3
R ( R  R2 ) Rb 
Ra Rb  Rb Rc  Rc Ra
Rb  Rc  3 1
R1  R2  R3
R

1
R1  R2  R3
Rb
R3 R1
Rc 
R R  Rb Rc  Rc Ra
R1  R2  R3
R2  a b
R1 ( R2  R3 )
Rc
Rc  Ra 
 Y
R1  R2  R3
R R  Rb Rc  Rc Ra
R3  a b
Ra
SUBTRACT THE FIRST TWO THEN ADD
TO THE THIRD TO GET Ra
Y 
LEARNING EXAMPLE: APPLICATION OF WYE-DELTA TRANSFORMATION
c
R1

R3
R2
12k  6k
12k  6k  18k
R1 R2
Ra 
R1  R2  R3
Rb 
R2 R3
R1  R2  R3
Rc 
R3 R1
R1  R2  R3
 Y
a
b
a
c
DELTA CONNECTION
b
COMPUTE IS
REQ  6k  3k  9k  || (2k  6k )  10k
IS 
12V
 1.2mA
12k
ONE COULD ALSO USE A
WYE - DELTA TRANSFORMATION ...
LEARNING EXAMPLE
CONVERT THIS Y INTO A DELTA?
SHOULD KEEP THESE TWO NODES!
IF WE CONVERT TO Y INTO A DELTA THERE ARE SERIES PARALLEL REDUCTIONS!
R1 
Ra Rb  Rb Rc  Rc Ra
3 *12k  *12k 

 36k 
12k 
Rb
R2 
Ra Rb  Rb Rc  Rc Ra
Rc
R3 
Ra Rb  Rb Rc  Rc Ra
Ra
Y 
36k
4mA
36k
36k
THE RESULTING
CIRCUIT IS A
CURRENT DIVIDER
12k

12k V

O
CIRCUIT AFTER PARALLEL RESISTOR
REDUCTION
36k ||12k  9k
4mA
36k
IO

9k
VO

IO 
36k
8
 4mA  mA
36k  18k
3
8
VO  9k   I O  9k   mA  24V
3
NOTICE THAT BY KEEPING
THE FRACTION WE PRESERVE
FULL NUMERICAL ACCURACY
WYE DELTA