ECE 3183 – EE Systems
Chapter 2 – Part A
Parallel, Series and General Resistive
Circuits
Chapter 2
Resistive Circuits
1. Solve circuits (i.e., find currents and voltages
of interest) by combining resistances in series
and parallel.
2. Apply the voltage-division and current-division
principles.
3. Solve circuits by the node-voltage technique.
4. Solve circuits by the mesh-current technique.
5. Find Thévenin and Norton equivalents.
6. Apply the superposition principle.
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-2
FIRST GENERALIZATION: MULTIPLE SOURCES
 v R1 
 v2 
+ -

v5

v3

i(t)
+
-

+
-
R1
+
-

v1

R2
Voltage sources in series can be
algebraically added to form an
equivalent source.
vR2


+ -
KVL
 v4 
We select the reference direction to
move along the path.
Voltage drops are subtracted from rises
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-3
FIRST GENERALIZATION: MULTIPLE SOURCES
veq = ?
R1
veq
ECE 3183 – Chapter 2 – Part A
+
-
R2
CHAPTER 2 A-4
SECOND GENERALIZATION: MULTIPLE RESISTORS
APPLY KVL
TO THIS LOOP
v R  Ri i 
i
VOLTAGE DIVISION FOR MULTIPLE RESISTORS
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-5
SECOND GENERALIZATION: MULTIPLE RESISTORS
FIND I ,Vbd , P (30k )
APPLY KVL
TO THIS LOOP
LOOP FOR Vbd
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-6
SECOND GENERALIZATION: MULTIPLE RESISTORS
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-7
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-8
THE CONCEPT OF EQUIVALENT CIRCUITS
THIS CONCEPT WILL OFTEN BE USED TO SIMPLFY
THE ANALYSIS OF CIRCUITS. WE INTRODUCE IT
HERE WITH A VERY SIMPLE VOLTAGE DIVIDER
i
vS
R1
i
vS
+
-
R2
i
R1  R2
+
-
vS
R1  R2
AS FAR AS THE CURRENT IS CONCERNED BOTH
CIRCUITS ARE EQUIVALENT. THE ONE ON THE
RIGHT HAS ONLY ONE RESISTOR
SERIES COMBINATION OF RESISTORS
R1
ECE 3183 – Chapter 2 – Part A
R2

R1  R2
CHAPTER 2 A-9
THE DIFFERENCE BETWEEN ELECTRIC
CONNECTION AND PHYSICAL LAYOUT
SOMETIMES, FOR PRACTICAL CONSTRUCTION
REASONS, COMPONENTS THAT ARE ELECTRICALLY
CONNECTED MAY BE PHYSICALLY FAR APART
IN ALL CASES THE RESISTORS ARE
CONNECTED IN SERIES
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-10
SUMMARY OF BASIC VOLTAGE DIVIDER
VR1 = ?
VR2 = ?
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-11
SUMMARY OF BASIC VOLTAGE DIVIDER
EXAMPLE: VS  9V , R1  90k, R2  30k
VOLUME
CONTROL?
R1  15k 
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-12
SUMMARY OF BASIC VOLTAGE DIVIDER
A “PRACTICAL” POWER APPLICATION
HOW CAN ONE REDUCE THE LOSSES?
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-13
THE “INVERSE” VOLTAGE DIVIDER
R1

VS
+
-
R2
VO

VOLTAGE DIVIDER
VO 
R2
VS
R1  R2
ECE 3183 – Chapter 2 – Part A
"INVERSE" DIVIDER
VS 
R1  R2
VO
R2
CHAPTER 2 A-14
THE “INVERSE” VOLTAGE DIVIDER
COMPUTE VS
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-15
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-16
Current Division
R2
v
i1 

itotal
R1 R1  R2
R1
v
i2 

itotal
R2 R1  R2
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-17
FIND Vx, i3
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-18
Find i1, i2, and i3
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-19
SERIES AND 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
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-20
SERIES AND PARALLEL RESISTOR COMBINATIONS
SERIES COMBINATIONS
1
1
1
1


 ...
Gs G1 G2
GN
PARALLEL COMBINATION
G p  G1  G2  ... GN
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-21
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-22
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-23
FIRST WE PRACTICE COMBINING RESISTORS
FIND RAB
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-24
FIRST WE PRACTICE COMBINING RESISTORS
(10K,2K)SERIES = 12K
3k
SERIES
6k||3k
(12K||6K) = 4K
(4K,2K)SERIES = 6K
We need to re-draw!
5k
12k
3k
RAB = 5K
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-25
EXAMPLES COMBINATION SERIES-PARALLEL
9k
AN EXAMPLE WITHOUT REDRAWING
12 k
6k || (4k  2k )
12k || 12k  6k
3k || 6k  2k
18k || 9k  6k
RAB  3k
6k  6k  10k
ECE 3183 – Chapter 2 – Part A
RAB  22k
CHAPTER 2 A-26
EXAMPLES COMBINATION SERIES-PARALLEL
RESISTORS ARE IN SERIES IF THEY CARRY
EXACTLY THE SAME CURRENT (SHARE ONE COMMON NODE)
RESISTORS ARE IN PARALLEL IF THEY HAVE THE SAME
VOLTAGE ACROSS THEM AND ARE CONNECTED EXACTLY
BETWEEN THE SAME TWO NODES
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-27
Strategy for analyzing circuits with series and
parallel combinations of resistors:
• Systematically reduce the resistive network so that the
resistance seen by the source is represented by a single
resistor.
• Determine the source current for a voltage source or the
source voltage for a current source.
• Expand the network, apply Ohm’s law, KVL, KCL,
voltage division, and current division to determine all
currents and voltages in the network
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-28
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-29
Circuit analysis example
Find Io in the circuit shown.
6 k
6 k
3 k
12 V
2 k
4 k
Io
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-30
FIND VO
60k
V1 6V
FIND VS
 2V
30k || 60k  20k
STRATEGY : FIND V1
USE VOLTAGE DIVIDER
9V
V1  60k * 0.1mA
0.15mA 
6V

0.05mA
I1 
THIS IS AN INVERSE PROBLEM
WHAT CAN BE COMPUTED?
VS  20k * 0.15mA  6V
20k

+
-
20k
V1

12V

20k
(12)  6V
20k  20k
VOLTAGE DIVIDER
20k
VO 
V1
20k  40k
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-31
6V
120k
Circuits with dependent sources
• When writing the KVL and/or KCL equations for
the network, treat the dependent source as though
it were an independent source.
• Write the equations that specify the relationship of
the dependent source to the controlling
parameters.
• Solve the equations for the unknowns. Be sure that
the number of linearly independent equations
matches the number of the unknowns.
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-32
KCL TO THIS NODE. THE
DEPENDENT SOURCE IS JUST
ANOTHER SOURCE
FIND VO
A PLAN:
IF V_s IS KNOWN V_0 CAN BE DETERMINED USING VOLTAGE DIVIDER.
TO FIND V_s WE HAVE A SINGLE NODE-PAIR CIRCUIT
THE EQUATION FOR THE CONTROLLING
VARIABLE PROVIDES THE ADDITIONAL EQUATION
ALGEBRAICALLY, THERE ARE TWO UNKNOWNS
AND JUST ONE EQUATION
SUBSTITUTION OF I_0 YIELDS
VOLTAGE DIVIDER
* / 6k  5VS  60 VO 
ECE 3183 – Chapter 2 – Part A
4k
2
VS  (12)V
4k  2k
3
CHAPTER 2 A-33
Problem solving strategy:
Circuit with dependent sources
For the network shown, what is the resulting ratio , Vo /Vs ?
RS
Ro
+
VS
ECE 3183 – Chapter 2 – Part A
Rin
Vin
_
+
 Vin
RL
Vo
_
CHAPTER 2 A-34
Problem solving strategy:
Circuit with dependent sources
For the network shown, what is the resulting ratio , Vo /Vs ?
RS
Ro
+
VS
 Rin
Vin  
 RS  Rin

 VS

ECE 3183 – Chapter 2 – Part A
Rin
Vin
_
+
 Vin
RL
Vo
_
(voltage divider for resistors in series)
CHAPTER 2 A-35
Problem solving strategy:
Circuit with dependent sources
For the network shown, what is the resulting ratio , Vo /Vs ?
RS
Ro
+
VS
 Rin
Vin  
 RS  Rin

 VS

Rin
Vin
_
 Vin
RL
Vo
_
(voltage divider for resistors in series)
 R 
 R  R 
Vo   L    Vin     in  L  VS
 Ro  RL 
 RS  Rin  Ro  RL 
ECE 3183 – Chapter 2 – Part A
+
(voltage divider for
resistors in series)
CHAPTER 2 A-36
Problem solving strategy:
Circuit with dependent sources
For the network shown, what is the resulting ratio , Vo /Vs ?
RS
Ro
+
VS
 Rin
Vin  
 RS  Rin

 VS

Rin
Vin
_
+
 Vin
RL
Vo
_
(voltage divider for resistors in series)
 R 
 R  R 
Vo   L    Vin     in  L  VS
 Ro  RL 
 RS  Rin  Ro  RL 
(voltage divider for
resistors in series)
 Rin  RL 
Vo
 


VS
R

R
R

R
in  o
L 
 S
ECE 3183 – Chapter 2 – Part A
CHAPTER 2 A-37