5EPC0 Circuits: 1.1
Lectures for 5EPC0 & 5LFK0
Not sure which you
did / should register for?
Come up and talk in the break!
1
2023-2024 v2
Course instructors
Kevin Williams
Responsible lecturer
Classes 1, 2, 6, 7
Yuqing Jiao
Lecturer
Classes 3, 4, 5, 8
And many technicians and teaching assistants
Bert Stolk
Lab skills responsible
Contact us by email:
circuits@tue.nl 2
Agenda for week 1
1. Course
Logistics, learning, lectures and labs
2. Sign conventions
Short maths quiz; Voltage, current, power and Ohm’s Law
3. Elements and nodes
Kirchoff’s Current Law
4. Simplifying circuits
Kirchoff’s Voltage Law
We will take 15 minutes breaks every hour – do come up to ask questions
3
Working with methods for any type of circuit
Applicable to
• computer microchips with
billions of transistors
• consumer electronics with
tens of microchips
• electrical systems with
transformers
Circuit designers need
methods that scale
• many node circuits
• passive components
4
Circuits for wired and wireless systems
3kHz
VLF
30kHz
300kHz
LF
3MHz
MF
300MHz
HF
3GHz
UHF
30GHz
300GHz
SHF
EHF
Direct currents for powering and controlling
Alternating signals for carrying information in wired and wireless systems
Transients that occur in digital systems
Again – same methods applying to all these technologies
5
The basis for many electrical engineering courses
The first course of many in circuits in the Bachelor College
Common techniques for circuits in security, internet, power systems, transport and more
Need idealized representations of real systems
6
Circuits in practice
Five types of components: sources, resistors, inductors, capacitance, switches
DC, AC, magnetically-coupled and switched circuits
7
Circuits: 5EPC0 and 5ECA0 and 5LFK0
5EPC0 Circuits
5LFK0 Circuit Analysis
5ECA0 Circuits (-2022)
5ECTS Bachelor course
2.5ECTS Masters course
Legacy course
Default for EE and AU
Includes labs
Default for homologation No longer examined, and
e.g. Embedded Systems the syllabus changed
No labs
5EPCO is the new name for the Q1 circuits course for EE and AU students
560 of you are registered for 5EPC0
8
Circuits: 5EPC0 and 5ECA0 and 5LFK0
Circuits
Credits
Measurement labs
5EPC0
for EE and AU
5EPC
for other faculties
5
Two on campus
Homelabs at location of choice
Two on campus
Homelabs on campus
5LFK0
homologation
5ECA0
no longer examined
2.5
No labs
0
No course
5
9
Blended learning
Tuesday lectures
Lab skills weeks 2-7
Self study
10
Problem classes
Home labs weeks 2-7
Weekly tests
Canvas home page
• Study guide
• News
• Sign-up
• Syllabus
• Homework
assignments
• Self-study
exercises
11
Canvas modules page
Lecture slides
Videos
Extra content
- self-study
- tutorial
Lab materials
Double selfstudy in week 1
12
Catch up materials
Materials on topics that
you may have missed at
school
Conceptual understanding
not essential, but it can
really help!
Main topics that people
miss are complex numbers
and linear algebra but this
should be in maths also
13
Self-study exercises
Self-study circuit problems as PDF
file
• Questions, final numerical
answer and also, a model
answer
• Some longer format questions,
more typical of exam style
questions after week 2
• Not graded. Provided as a
learning resource
Example from week 6 using
complex numbers in the solution
14
Weekly homework quizzes
Weekly online quiz
• Ensures you are keeping
up as each week builds on
previous content
• Graded and you have until
the end of the weekend to
finish
• Numerical values are
randomised for fair
assessment
15
Instruction classes on Friday afternoons
Friday 13:30 – 17:00 for weeks 1-3 and 5-8 inclusive
(excludes MomenTUm in week 4)
Multiple rooms so check Canvas and Discord
Multiple experienced Teaching Assistants and the Lecturer
for that week
Peer learning opportunity: TAs not giving answers but can
help you to understand the methods. Ask for help when
struggling on the method or concepts!
Additional option to use discord channel for peer learning
Server link: https://discord.gg/tCRkKTtnCn
16
Instruction classes on Friday afternoons
Week
Free choice for rooms
1 Gemini-Noord 1.610 Atlas 4.215 Atlas 4.225 Atlas 6.225
2 Gemini-zuid 3A.12
Atlas 4.215 Atlas -1.820 Matrix 1.340
3 Gemini-zuid 3A.12
Atlas 4.215 Atlas 4.225 Gemini-zuid 3A.10
4
No teaching due to momentum
5 Gemini-Noord 1.710 Atlas 4.215 Atlas 4.225 Atlas 6.225
6 Gemini-Noord 1.710 Atlas 4.215 Atlas 4.225 Atlas 6.225
7 Gemini-Noord 1.610 Atlas 4.215 Atlas 4.225 Atlas 6.225
8 Matrix 1.340
Atlas 4.215 Atlas 4.225 Matrix 1.345
No Teaching Assistant
Lecturer present
17
Learning through reflection
Competence
4. You can do it
automatically without
having to think
3. Become competent
through practice, experience,
but can take lots of effort
1. Unaware which
skills need to be
learned
Lectures
Consciousness
2. Aware of
challenge some idea
of what is needed
Practice
Reflection
adapted from http://www.learning-to-see.co.uk/phases-of-learning
Explicitly used repetition and reflection to improve lab-work
18
Keep asking; Keep on top of the work
Coursework is critical to
understanding
• Each week builds on the
previous week
• Increasingly challenging
material
• Find out early if you do not
understand something!
• Do not leave it until exams
as already too late.
The boiling frog problem
19
The curriculum
Other editions are also fine
1
Contents
Ohm, Kirchoff, sources
Chapter
1&2
2
Nodal and loop analysis
3&5
3
DC circuit analysis
5
4
Capacitors, inductors
6 & 10
5
Time domain transients
7
6
AC circuit techniques
8
7
Magnetic coupling
8 & 10
8
Revision
20
Syllabus change in 2023 for 5EPC0
1
Contents
Ohm, Kirchoff, sources
Chapter
1&2
2
Nodal and loop analysis
3&5
3
DC circuit analysis
5
4
Capacitors, inductors
6 & 10
AC methods taught in two weeks
instead of one
5
Time domain transients
7
6
AC circuit techniques
8
No more frequency domain analysis or
Bode diagrams
7
Magnetic coupling
8 & 10
8
Revision
Proactive encouragement to use calculators:
simultaneous equations from week 2 and
complex numbers from week 6
Transformers added in weeks 4 & 7
21
Grading
Theory
- Canvas coursework
- Written examination
Laboratories
- Technical skills (TV)
5EPC0 (5 ECTS)
5LFK0 (2.5 ECTS)
✓
✓
(20%)
(50%)
✓
✓
✓
(30%)
(20%)
(80%)
Note: These courses share the same lecture classes, coursework, exam
Course materials posted in 5EPC0
22
On-campus laboratory classes
Senior lab instructor team:
Rainer
van Dommele
Michel
van Eerd
Frans
Huijskens
Antonia
Almeida
Bert
Stolk
As well as researchers, doctoral students and masters students
23
On-campus laboratory classes: Timeslot D
Wednesdays 13:30-17:15 10.09 17.09 24.09 01.10 08.10 15.10 22.10
Groups
A
B
C
A
B
C
X
Fridays 8:45-12:30
Groups
12.09 19.09 26.09 03.10 10.10 17.10 24.10
D
E
F
D
E
F
X
Fridays 13:30-17:15
Groups
12.09 19.09 26.09 03.10 10.10 17.10 24.10
G
H
I
G
H
I
X
Sessions in weeks 2 through 7 separated by 3 weeks
Register yourself to a group in Canvas with 2 students per set-up
Floor 10 in Flux: Rooms 10.070 and 10.072 and group floor soldering lab
24
Lab registration in canvas
5EPC0:
Labs are
compulsory!
5LFK0:
No labs!
Home Labs challenge: Collecting myDAQ kits
EE & AT students should have the kits: See email from Sjoerd Hulshof
5 August and reminder from Bert Stolk 23 August
Other faculties: Special sessions with loan kits on campus – register for
26
extra lab session(s)
Home Labs challenge: myDAQ kit support
Recommended to perform first on campus lab before starting homelab
Most students able to complete in ten hours
But start early in case of unexpected delays!
Discord channel and circuits@tue.nl for peer learning and help from
Teaching Assistant
Lecturer support in Tuesday lecture breaks and Friday instruction classes
Strongly encouraged to perform this lab in pairs and small groups
Assessment quiz in Canvas
27
5EPC0 Circuits: Break
Short maths quiz next.
Please log in:
https://socrative.com/
Click on student login
Enter room number : F948NTZ
28
5EPC0 Circuits: 1.2
Short maths quiz next.
Please log in:
https://socrative.com/
Click on student login
Enter room number : F948NTZ
29
https://socrative.com/
Socrative for in-class quizzes
Please log in during the break
https://socrative.com/
Click on student login
Enter room number :
F948NTZ
30
1.2 Sign conventions
1. Maths check
Sign conventions
2. Voltage
3. Current
4. Power
31
Calculators
5EPC0 is not strict but other courses may be
• Before buying a new calculator, check other
courses
Recommended functions include:
• Complex number arithmetic
• Solving simultaneous equations
(up to four real-valued equations)
• Scientific should be enough
A calculator is not a substitute for showing your
methods!
32
Prior knowledge
We need to use some mathematical techniques, so be sure to be on
top of this:
• Creating and solving simultaneous equations for circuits with many
loops and nodes
• Working with complex numbers when we have alternating signals
• Using the solutions to differential equations when we change the
state of switches
33
How is your linear algebra (simultaneous equations)
I know matrix algebra
Can solve simulation equations with the subtraction method
Can solve simultaneous equations by the substitution method
Can recognise simultaneous equations
Don’t know simultaneous equations
34
Solving many equations together
We need methods to solve many
equations simultaneously (at the same time)
But this is a circuits course rather than
a computing course, so we use calculators
Opportunities to practice from week 2
35
Do you know the imaginary number
j = (–1)1/2
Can already do algebra with complex numbers
Not a problem if I have a calculator
Can do trigonometry with right-angle triangles
Can add and subtract vectors
What is a complex number?
36
Solving equations with complex numbers
Complex numbers are
super-useful in all forms of
engineering
Also super-simple on a
calculator
But get practising before the
exams!
Really handy to understand
them conceptually as well!
37
Voltage, current and power
Which definitions and images would you associate with voltage, current and power
• The flow of charged particles moving through a conductor
• The rate of transfer of electrical energy within a circuit
• The difference in electric potential between two points
38
Voltage = Potential Difference
Giga
GV
109
Mega
MV
106
kilo
kV
103
milli
mV
10-3
micro
µV
10-6
nano
nV
10-9
The difference in electric potential between two points
39
*
Voltage as a driver for current
I
Potential
energy
Wmech = mgh
v
Voltage drives the flow of charged particles (current) through a path
Positive current defined to flow from high potential to low potential
Electrical
energy
Welec = qV
40
Voltage and current with many paths
Height potential differences and
path resistances define the current
Ohm’s Law relates
potential difference and current
Voltages fixed at the nodes on each side of the resistive elements
41
Referencing
V
+ symbol at node A
– symbol at node B
V = V+ – V – = VAB = 2
When the circuit is too complex to know the signs for
the voltages, we need to define the sign of the variables
Then use algebra to work out the actual values
42
Referencing
V
V
+ symbol at node A
– symbol at node B
V = V+ – V – = VAB = – 5
V = V+ – V – = VAB = 2
=> V – – V+ = VBA = 5
43
Referencing
V
+ symbol at node A
– symbol at node B
V = V+ – V – = VAB = 2
V
V = V+ – V – = VAB = – 5
=> V – – V+ = VBA = 5 = – V
V
Now we change referencing
V = V+ – V – = VBA = 5
44
Referencing
V
V
V
Note VAB = – VBA because we reversed the referencing
VA – VB = VAB
VB – VA = VBA
45
V
V
V
(d) don’t know
Which circuit shows the highest value for VAB? Use Socrative.
46
Referencing
VAB
VAB
(a)
VBA
 VAB = –5V
(b)
(c)
Remember to change the sign when you reverse the referencing!
47
Current
The charge moving per second
48
Current and flow of electrons
• Current is the flow of electrons. The number of electrons
passing per unit time
• As electrons move through circuit elements, their
electrical potential (voltage) changes
• Supplies like batteries increase electrical potential energy
• Loads such as resistors reduce electrical potential energy
• Cannot directly see electrons, so using analogies
Size : Sub-atomic
Strength: Charge of 1.6x10-19C
Mass: Not really: 9.1x10-31 Kg
Potential: All over the place
Speed: Super quick
49
Current: Open and short circuits
No possibility for current flow so
all water (charge) stored in
reservoir (battery)
Removing resistance allows
current (water particles or charge)
to flow in the circuit
Current is the flow of charge in the element (between nodes)
50
Current: Completing the circuit
1) Our charge flows
down as a current,
experiencing resistance
from the hillside, trees
3) But nature
has a pump to
get water back
up to storage in
the sky
2) Stops when it gets to
the sea level
Remember that current will be zero later on if the circuit is not completed.
51
Current and electronic charge
Current can be calculated
from the rate of change in
charge with time
i(t) (A) 2A
t (ms)
52
Current and electronic charge
Current can be calculated
from the rate of change in
charge with time
i(t) (A) 2A
4/3A
–5/2A
t (ms)
53
Power = Voltage x Current
Voltage defined as energy required per unit of charge to move charge through element
J/C
w work done to move charge in units Joules
q charge in units Coulombs
J/s
Calculate power by multiplying the energy to move charge per unit charge (voltage)
by the rate of change in charge with time (current)
54
Power: Sources and loads
Switch open: Open circuit
Converting energy to and from
electrical energy
• Battery is a source with stored
chemical energy
• Bulb is a load that dissipates
heat and light when current
flows
Chemical energy converted to
electrical energy
Electrical energy converted to
heat and light
55
Power: Absorbing and Supplying
Switch closed : Short circuit
Battery supplying:
• Current I from negative to positive.
• Electrons at higher potential
Bulb absorbing:
• Current I from positive to negative
• Electrons lose electrical energy.
Current I should always be drawn with
an arrow for direction for consistent
circuit equations
56
Passive sign convention for power
Supply
Absorber
Source: Current goes
from –ve terminal
to +ve terminal
P = (– I)(+V) = – IV
Negative power for source!
Load: Current goes
from +ve terminal
to – ve terminal
P = (+I)(+V) = + IV
Positive power for load!
Sign of the power tells us whether an element supplies or absorbs.
In complex circuits, we need to calculate whether elements supply or absorb
57
Resistance, Ohm’s Law and signs
Terminal voltage
V+
Current defined by
voltage difference
and the resistance
to the current
Use + and – signs to
denote the measurement
referencing
V+ – V– = I R
Resistance R
Ohm’s Law
Current I
V–
Terminal voltage
Ohm’s Law has referencing:
If voltage value is positive, current value is positive through referencing
58
Voltage referencing in measurements
V defined as
voltage at +
relative to
voltage at –
+
V
–
Current I
Current arrow shows the defined direction of current I
Voltage and current values may be positive or negative
Must define the referencing and direction for consistent variable use
I defined as
flowing from
node + to
node at –
For resistance
positive I flows
from high to low
59
Using Ohm’s law
The resistance defines the voltage for a known current
= 10 k
i
V=IR
Current flows into positive terminal of resistor and out of negative terminal
VS = i R = (+0.4m) (+10k) = (+0.4x10–3) (+10x103) = +4V
Ideal resistor values are always positive and real.
60
Absorbing energy
Absorbing energy :
positive current entering positive terminal of
element X and leaving negative terminal
+
X
I = 2 A [Coulomb/second]
Welec = qV = 2/1 x 3 = 6 J every second
Pelec = 2 x 3 V = 6 W
Positive power in absorbing element
Electrical energy converted to another form
61
Supplying energy
Supplying energy :
positive current leaving positive terminal
Equivalent to negative current entering
+
X
I = –2 A [Coulomb/second]
Welec = qV = – 2 x 1 x 3 = – 6 J every second
Pelec = – 2 x 3 V = – 6 W
Negative power for supply element
Watch signs and arrows carefully for referencing of values
Use sign of power to identify whether absorbing or supplying
62
*
Identifying sources and load
Is element X a source or a load?
+4A
+
–4V
X
Is the circuit to the left of
terminals A and B a source or a
load?
Method: For the element define
P = IV where
I flows from + to – and
V = V+ – V–
Let P = (V+– V – ) (I) where I flows from + to –
P = (–4V)( –4A ) = + 16W
The circuit element X is a load
63
5EPC0 Circuits: Break
64
5EPC0 Circuits: 1.3
65
1.3 Kirchoff’s current law
• Sign convention check
• Circuit elements
• Kirchoff’s current law for one node
• Node pairs
• Current dividers
66
What is current I in the resistor
– 10 mA
+ 10 mA
– 16 mA
+ 16 mA
Don’t know
4V
400
Hint: Use the arrows and signs to decide whether the current is positive
Check consistency with passive sign convention; Recall V = IR
67
I = 4 / 400 = 10mA
Is the element X shown on the right absorbing or supplying energy
Absorbing energy
Supplying energy
Neither
Don’t know
X
For an absorbing element positive current flows from high potential to low potential
68
Is the element X absorbing or supplying power?
Absorbing
Supplying
Don’t know
X
69
Idealised circuit elements: sources and loads
Sources and loads come in all sorts of shapes and sizes
In circuit analysis, we need a way to describe these concisely and mathematically
Idealised sources combined with additional circuit elements to represent real sources
Loads created from combinations of idealised circuit elements such as resistors (today),
capacitors and inductors (lecture 4-7 onwards), diodes and transistors not in this course
70
Ideal independent sources
Ideal voltage source
Any current I possible
Voltage VAB is
always the same
No external influence
71
Ideal independent sources
Ideal voltage source
Any current I possible
Voltage VAB is
always the same
No external influence
Ideal current source
Any voltage VAB possible
Current i(t) is
always the same
No external influence72
Dependent (or controlled) sources
The front panel dial on an oven controls uses the voltage on a variable
resistor to control the signal to the heater. Can be current or voltage.
73
Dependent sources
controlling circuit
dependent circuit
X
An ideal source can be controlled by another (part of the) circuit
We can express this mathematically with a simple expression and analyse in the same way
P = V I = (10 Vs ) Io = (10 . 4) . 2 = 80W
Positive currents into positive terminals – power absorbed
74
Dependent sources: Representing amplifiers
voltage
amplifier
transimpedance
amplifer
transconductance
amplifer
current
amplifier
In the Circuits course, we will describe amplifiers with ideal controlled sources
75
Circuit terminology
Elements
• One independent
current source
• Two independent
voltage source
• Five resistors
• Eight elements
Currents different for each branch as value
depends on element and potential difference
Nodes
• Five
numbered
points in
circuit
• Connection of
two or more
elements
76
Equi-potentials : Same electrical potential
1
Node as an
equi-potential
Potential is
changing only in
the elements
3
2
4
Ideal wires have
zero resistance
5
For circuit analysis, we only have one voltage value in a wire because there is no
resistance in the wire. Nodes have unique voltages.
77
How many elements
(resistors and sources)?
3
4
6
8
Don’t know
78
How many nodes?
3
4
6
8
Don’t know
Remember – if there is a wire connecting two elements,
it is one node because the potential is the same
79
A node is a point of connection of two or more connecting elements.
How many nodes?
3
4
5
8
Don’t know
Hint – think carefully about how many nodes there are on the bottom line
80
Analysis at a node: Kirchoff’s current law
Node has one potential (voltage)
value, but many flows (currents) coming
in and out
For a road junction, the number of
cars coming in equals cars going out
81
Analysis at a node: Kirchoff’s current law (KCL)
μ = 10–6
Electrons in = electrons out
Current in = current out
Incoming currents (source) are
negative by convention
–5+3+2=0
I=0
82
Kirchoff’s current law (KCL) for one node
1
leaving
leaving
leaving
entering
Positive sign for
currents leaving the node
1
entering
Negative sign for
current entering into node
Sum of currents equals zero for conservation of energy
83
Single node-pair circuit example
KCL at node 2
2
– i(t) + i1(t) + i2(t) = 0
Apply Ohm’s law
1
i(t) = v(t) ( R1–1 + R2–1 )
We can combine resistor values to make a simpler equivalent circuit
84
Combining parallel resistors
2
KCL at node 2
– i(t) + i1(t) + i2(t) = 0
Apply Ohm’s law
1
v(t) / i(t) =
( R1–1 + R2–1 )–1
R1 R2
= R + R = RP
2
1
Because both branches all have same voltage
85
Combining parallel resistors

i = v / R1 + v / R2
P
i = v / RP
So 1 / R1 + 1 / R2 = 1 / RP
And for more resistors this generalizes to  1 / R = 1 / RP
86
Current division
Now consider i1(t) through R1:
The voltage terms cancel because v(t) is the same for both branches
Currents are divided independently of the voltage
Worth learning the current divider equation as well
87
How to combine parallel current sources? What is current I ?
– 1A
0A
1A
3A
Don’t know
1A
2A
==
If the two circuits are equivalent, what is the new current I
Hint: Think of cars moved along a road.
I
88
Summarizing Kirchoff for parallel elements
i(t)
R1
R2
RN
2

i1(t)
i2(t)
i3(t)
iP(t)
–
RP
v(t)
iN(t)
+
1
From Kirchoff’s current law at 2 i (t) = i (t) – i (t) – i (t) +i (t) = i(t)
P
1
2
3
N
Noting the same voltage
between node 2 and node
1
89
Combining parallel elements : example
Equivalent circuits can allow calculation of branch currents. Consider IL
2

IS
RS
1
We want to calculate the load current IL.
Method:
First determine the equivalent current IS and
The equivalent resistance RS on the left of the terminals
Then use the current divider equation
90
Combining parallel elements : example
Equivalent circuits can allow calculation of branch currents. Consider IL
2

1
1. Combine all parallel
resistors to give 4k
2. Combine parallel
currents to give 1mA
– 1 + 4 – 2 = 1mA
91
Combining parallel elements : example
Using KCL at the top terminal to calculate the load current IL
Method 1:
KCL at the +ve terminal
+
1 + V /4k + IL = 0
where V = IL.12k
–
 1 + 3IL + IL = 0
 IL = – ¼ m
= –0.25mA
v
Method 2: the current divider equation
92
5ECA0 Circuits : Break
93
5ECA0 Circuits : 1.4
94
Kirchoff’s voltage law & Welsh hill walking
If complete the circuit, we start and finish with same potential energy
95
 Wmech = 0
 V = 0
Hill walking analogy
a
Stretch
Height climbed
Cumulative
Starting point
0m
a to b
– 100m
– 100m
b to c
– 300m
– 400m
c to d
+ 500m
+ 100m
d to e
+ 100m
+ 200m
e to a
– 200m
0m
e
d
b
c
H=0
Hab + Hbc + Hcd + Hde + Hea = 0
The direction – the referencing – needs to be correct
to avoid the carpark changing height during the walk!
96
Kirchoff’s voltage law for one loop
V=0
Same applies for electrical potential
Summing voltage drops for each
element around the loop gives zero net
change in voltage
Vab+Vbc+Vcd+Vde+Vef+Vfa = 0
Only one potential at any given node so
voltage at a will be same after adding all
voltage drops in the loop
97
Kirchoff’s voltage law for one loop
V=0
Vab +Vbc+Vcd +Vde+Vef +Vfa=0
98
*
Kirchoff’s voltage law for one loop
V=0
Vab +Vbc+Vcd +Vde+Vef +Vfa=0
Substitution
VR1 –5 +VR2 –15 +VR3 –30=0
Rearrangement:
VR1+VR2+VR3 =50
99
Derive KVL for the loop a-b-e-f-a
VR1+VR4 = –40V
VR1+VR4 = –8V
VR1+VR4 = +8V
VR1+VR4 = +40V
Don’t know
Reminder:  V = 0 = Vab + Vbe + Vef + Vfa
100
Voltage dividers
Calculating voltages
KVL : – v + vR1 + vR2 = 0
 v = vR1 + vR2
101
Voltage dividers
Calculating voltages
KVL : – v + vR1 + vR2 = 0
 v = vR1 + vR2
Ohm: vR1 = R1 i but KVL shows
the same current flows in R1 and R2:
vR2 =
R2 i
v = (R1 + R2 ) i
102
Voltage dividers
Calculating voltages
KVL : – v + vR1 + vR2 = 0
 v = vR1 + vR2
Ohm: vR1 = R1 i but KVL shows
the same current flows in R1 and R2:
vR2 =
R2 i
=
v = (R1 + R2 ) . i
103
A recipe for single-loop circuits
• Define a loop current i(t) : there is only one current for a single loop
• We will define loop currents to flow in a clockwise direction
• Use Ohm’s law to define voltages across resistor
• Apply KVL to the single loop circuit
104
Parallel and serial combinations
KVL: Same current through series connected resistors so resistances add
105
Parallel and serial combinations
KVL: Same current through series connected resistors so resistances add
KCL: Same voltage across parallel resistors so inverse of resistance adds
Techniques commonly used to simplify circuit analysis
106
Combining serial elements
vN(t)
v(t)
v3(t)

v2(t)
v1(t)
v(t) = v1(t) + v2(t) + ... vN(t)
RS = R1+R2+R3+R4+R5 + ... RN
Voltages and resistances may be added on a branch for analysis
107
Mixed serial and parallel circuit simplication
Systematic combination of serial and parallel resistors to determine the
terminal resistance for a resistive network
108
Mixed serial and parallel circuit simplication
10
Rp =
10 + (3) ||(6) =
6
1
10+3x6/(3+6) =
10+2 = 12k
2
Group as one parallel (6k parallel to 1k+2k) and add one serial resistor (10k)
and reduce to the equivalent 12k value
109
Equivalent circuits: Reducing resistive networks
Serial combination
RS = R1 + R2
12k
Parallel combination
1/RP = 1/R1 + 1/R2
 RP = (R1R2)/(R1+R2)
Next step is to combine the rightmost parallel 6k and 12k values.
And we can then add the 2k resistor on the top rail in series with the result
110
Equivalent circuits: Reducing resistive networks
Rp = ?
Repeat method again: This time 2k in series with parallel combination of 6k
and 12k to give 6k equivalent resistance
111
Equivalent circuits: Reducing resistive networks
Rp =
2 +(12) ||(6)=
2+(12)(6)/(12+6) =
2+4 = 6k
Repeat method again: This time 2k in series with parallel combination of 6k
and 12k to give 6k equivalent resistance
112
Equivalent circuits: Reducing resistive networks
6k
Rp = ?
6k and 6k in parallel give 3k; in parallel with 9 k to give …
113
Equivalent circuits: Reducing resistive networks
6k
Rp =
9 +(6)||(6)
=9+3 =12k
6k and 6k in parallel give 3k; in parallel with 9 k to give 12k
114
Equivalent circuits: Reducing resistive networks
RAB =
2 +(12) ||(4) =
2+48/16 =5k
2k in series with the parallel combination of 4//12 to give 3k between the
terminals. The equivalent resistance is 3k Ohm
115
Combining analysis techniques
We wish to know V0 in the above circuit
The methods we have so far are
Dividers : Voltage for series elements and current for parallel
Kirchoff : KCL for nodes and KVL for loops
116
Combining analysis techniques
Let’s analyse in two steps:
1. Use KCL at the +ve terminal to determine VS
2. Use the voltage divider at the 2k + 4k Ohm branch to determine V0
117
*
Combining techniques: KCL
4 I0
KCL at top node (positive rail)
+ 10m + Vs + Vs – 4 I0 = 0
2k+4k 3k
One equation and two unknowns VS and I0 does not yield a unique solution.
We need an additional equation that includes VS and I0
Inspection of circuit shows VS = 3 I0
Note that 4 I0 is a current and not a voltage
118
*
Combining techniques: Controlled source
4 I0
1
+ 10m + Vs + Vs – 4 I0 = 0
2k+4k 3k
Ohm’s Law :
Vs = 3 I0
 I0 = Vs / 3 (units mA)
(dropping k and specify current in mA)
2
 – 4I0 = – Vs 4/3
KCL at top node (positive rail)
Substituting 2 into
1
:
+ 10 + Vs /6 + Vs/3 – 4/3 Vs = 0
119
*
Combining techniques: Controlled source
4 I0
+ 10 + Vs /6 + Vs/3 – 4/3 Vs = 0

10 + Vs { 1/6+1/3 – 4/3 } = 0
 Vs = – 10 . (–6/5) = 12V
120
Combining techniques: voltage dividers
= 12V
The voltage VS is now known, so we can analyse each branch
independently
4k
Vo =
Vs
2k + 4k
= 8V
Many different combinations of circuit elements will require many different
techniques.
121
Week 1 :
• Resources and requirements for the Circuit course should be clear
• Importance of sign conventions for voltage, current, power
• The key circuit elements, nodes, loops and terminology refreshed
• Methods:
• Kirchoff’s Current Law for one node
• Kirchoff’s Voltage Law for one loop
• Simplifying resistive networks
• Able to use divider equations
• Already have the tools for many circuit analysis problems. Do practice these
methods as we will build on these techniques now for more complex circuits
122
5ECA0 Circuits (ch2) Class 1.4
Kirchoff
•
•
•
•
Register in Canvas for lab groups
Check Canvas maths resources
Join the instruction class on Fridays
Use the self-study examples to
practice and learn before the quiz
• On-line graded quiz by end of
weekend
123