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Circuit analysis: DC Circuits (3 cr)Fall 2009 / Class AS09
Vesa Linja-aho
Metropolia
October 8, 2010
The slides are licensed with CC By 1.0.http://creativecommons.org/licenses/by/1.0/
Slideset version: 1.1
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 1 / 125
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 2 / 125
1. lecture
About the Course
Lecturer: M.Sc. Vesa Linja-aho
Lectures on Mon 11:00-14:00 and Thu 14:00-16:30, room P113
To pass the course: Home assignments and final exam. The exam ison Monday 12th October 2009 at 11:00-14:00.
All changes to the schedule are announced in the Tuubi-portal.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 3 / 125
1. lecture
The Home AssignmentsThere are 12 home assignments.Each assignment is graded with 0, 0,5 or 1 points.To pass the course, the student must have at least 4 points from theassignments.Each point exceeding the minimum of 4 points will give you 0,5 extrapoints in the exam.In the exam, there are 5 assignments, with maximum of 6 points each.To pass the exam, you need to get 15 points from the exam.All other grade limits (for grades 2-5) are flexible.
Example
The student has 8 points from the home assignments. He gets 13 pointsfrom the exam. He will pass the exam, because he gets extra points fromthe home assignments and his total score is (8− 4) · 0,5 + 13 = 15 points.
However, is one gets 8 of 12 points from the home assignments, he usuallygets more than 13 points from the exam :-).
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1. lecture
The Course Objectives
From the curriculum:
Learning outcomes of the course unit
Basic concepts and basic laws of electrical engineering. Analysis of direct current(DC) circuits.
Course contentsBasic concepts and basic laws of electrical engineering, analysis methods,controlled sources. Examples and exercises.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 5 / 125
1. lecture
The Course Schedule1 The basic quantities and units. Voltage source and resistance.
Kirchhoff’s laws and Ohm’s law.2 Conductance. Electric power. Series and parallel circuits. Node.
Ground.3 Current source. Applying the Kirchhoff’s laws to solve the circuit.
Node-voltage analysis.4 Exercises on node-voltage analysis.5 Source transformation.6 Thevenin equivalent and Norton equivalent.7 Superposition principle.8 Voltage divider and current divider.9 Inductance and capacitance in DC circuits.10 Controlled sources.11 Recap.12 Recap.
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1. lecture
The Course is Solid Ground for Further Studies in
Electronics
The basic knowledge on DC circuits is needed on the courses CircuitAnalysis: Basic AC-Theory, Measuring Technology, Automotive
By studying this course well, studying the upcoming courses will be
easier!
The basics of DC circuits are vital for automotive electronics engineer, justlike the basics of accounting are vital for an auditor, and basics of strengthof materials are vital for a bridge-building engineer etc.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 7 / 125
1. lecture
What is Not Covered on This Course
The basic physical characteristics of electricity is not covered on thiscourse. Questions like ”What is electricity?” are covered on the courseRotational motion and electromagnetism.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 8 / 125
1. lecture
Studying in Our School
You have an opportunity to learn on the lectures. I can not force youto learn.
You have more responsibility on your learning than you had invocational school or senior high school.
1 cr ≈ 26,7 hours of work. 3 cr = 80 hours of work. You will spend39 hours on the lectures.
Which means that you should use about 40 hours of your own timefor studying!
If I proceed too fast or too slow, please interject me (or tell me byemail).
Do not hesitate to ask. Ask also the ”stupid questions”.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 9 / 125
1. lecture
What Is Easy and What Is Hard?
Different things are hard for different people. But my own experienceshows that
DC analysis is easy, because the math involved is very basic.
DC analysis is hard, because the circuits are not as intuitive as, forexample, mechanical systems are.
Studying your math courses well is important for the upcoming courses oncircuit analysis. For example, in AC circuits analysis you have to usecomplex arithmetics.
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1. lecture
Now, Let’s Get into Business
Any questions on the practical arrangements of the course?
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1. lecture
Electric Current
Electric current is a flow of electric charge.
The unit for electric current is the ampere (A).
The abbreviation for the quantity is I .
One may compare the electric current with water flowing in a pipe (socalled hydraulic analogy).
The current always circulates in a loop: current does not compressnor vanish.
The current in a wire is denoted like this:
I = 2mA
-
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1. lecture
Kirchhoff’s Current Law
As mentioned on the previous slide, the current can not vanishanywhere.
Kirchhoff’s Current Law (or: Kirchhoff’s First Law)
At any area in an electrical circuit, the sum of currents flowing into thatarea is equal to the sum of currents flowing out of that area.
I1 = 3mA-
I2 = 2mA-
I3 = 1mA6
If you draw a circle in any place in the circuit, you can observe that thereis as the same amount of current flowing into the circle and out from thecircle!
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1. lecture
Be Careful with Signs
One can say: ”The balance of my account -50 euros” or equally ”Iowe 50 euros to my bank”.
One can say: ”The profit of the company was -500000 euros” orequally ”The loss of the company was 500000 euros”.
If you measure a current with an ammeter and it reads −15mA, byreversing the wires of the ammeter it will show 15mA.
The sign of the current shows the direction of the current. The twocircuits below are exactly identical.
I1 = 3mA-
I2 = 2mA-
I3 = 1mA6
Ia = −3mA
Ib = −2mA
I3 = 1mA6
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1. lecture
Voltage
The potential difference between two points is called voltage.
The abbreviation for the quantity is U.
In circuit theory, it is insignificant how the potential difference isgenerated (chemically, by induction etc.).
The unit of voltage is the volt (V).
One may compare the voltage with a pressure difference in hydraulicsystem, or to a difference in altitude.
Voltage is denoted with an arrow between two points.
+
−
12V U = 12V
?
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1. lecture
Kirchhoff’s voltage law
The voltage between two points is the same, regardless of the pathchosen.
This is easy to understand by using the analogy of differences inaltitude. If you leave your home, go somewhere and return to yourhome, you have traveled uphill as much as you have traveled downhill.
Kirchhoff’s Voltage Law (or: Kirchhoff’s Second Law)
The directed sum of the voltages around any closed circuit is zero.
− +1,5V
− +1,5V
− +1,5V
4,5Vr r
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1. lecture
Ohm’s law
Resistance is a measure of the degree to which an object opposes anelectric current through it.
The larger the current, the larger the voltage – and vice versa.
The abbreviation of the quantity is R and the unit is (Ω) (ohm).
The definition of resistance is the ratio of the voltage over theelement divided with the current through the element. R = U/I
U = RI
R
U -
I-
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1. lecture
Definitions
Electric circuit A system consisting of compontents, in where electriccurrent flows.
Direct current (DC) The electrical quantities (voltage and current) areconstant (or nearly constant) over time.
Direct current circuit An electric circuit, where voltages and currents areconstant over time.
Example
In a flashlight, there is a direct current circuit consisting of abattery/batteries, a switch and a bulb. In a bicycle there is an alternatingcurrent circuit (dynamo and bulb).
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 18 / 125
1. lecture
An alternate definition for direct current
One may define also that direct current means a current, which does notchange its direction (sign), but the magnitude of the current can vary overtime. For example, a simple lead acid battery charger outputs a pulsatingvoltage, which varies between 0 V ... ≈ 18 V. This can be also called DCvoltage.
Agreement
On this course, we define DC to mean constant voltage and current. Themagnitude and sign are constant over time.
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1. lecture
A simple DC circuit
A light bulb is wired to a battery. The resistance of the filament is10Ω.
+
−
12V @@
I =?-
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1. lecture
A simple DC circuit
A light bulb is wired to a battery. The resistance of the filament is10Ω.
+
−
12V 10Ω
I =?-
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 20 / 125
1. lecture
A simple DC circuit
A light bulb is wired to a battery. The resistance of the filament is10Ω.
+
−
12V 10Ω
I =?-
12V
?
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 20 / 125
1. lecture
A simple DC circuit
A light bulb is wired to a battery. The resistance of the filament is10Ω.
+
−
12V 10Ω
I = 1,2A-
12V
?
U = RI
I = U
R= 12V
10Ω = 1,2A
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 20 / 125
1. lecture
Homework 1 (released 31st Aug, to be returned 3rd Sep)
The homework are to be returned at the beginning of the next lecture.
Remember to include your name and student number.
Homework 1
Find the current I .
+
−
1,5V
+
−
1,5V
R = 20Ω
I?
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 21 / 125
2. lecture
Homework 1 - Model solution
Homework 1
Find the current I .
+
−
1,5V
+
−
1,5V
R = 20Ω
I?
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2. lecture
Homework 1 - Model solution
Homework 1
Find the current I .
+
−
1,5V
+
−
1,5V
R = 20Ω
I?U1
?
U2
?UR
?
U1 + U2 − UR = 0 ⇔ UR = U1 + U2
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 22 / 125
2. lecture
Homework 1 - Model solution
Homework 1
Find the current I .
+
−
1,5V
+
−
1,5V
R = 20Ω
I?U1
?
U2
?UR
?
U1 + U2 − UR = 0 ⇔ UR = U1 + U2
U = RI ⇒ UR = RI ⇒ I =UR
R=
U1 + U2
R=
1,5V + 1,5V
20Ω= 150mA
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 22 / 125
2. lecture
Conductance
Resistance is a measure of the degree to which an object opposes anelectric current through it.
The inverse of resistance is conductance. The symbol forconductance is G and the unit is Siemens (S).
Conductance measures how easily electricity flows along certainelement.
For example, if resistance R = 10Ω then conductance G = 0,1S.
G = 1R
U = RI ⇔ GU = I
G = 1R
U -
I-
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2. lecture
Electric Power
In physics, power is the rate at which work is performed.
The symbol for power is P and the unit is the Watt (W).
The DC power consumed by an electric element is P = UI
I-
U -
If the formula outputs a positive power, the element is consumingpower from the circuit. If the formula outputs a negative power, theelement is delivering power to the circuit.
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2. lecture
Electric Power
Energy can not be created nor destroyed
The power consumed by the elements in the circuit = the power deliveredby the elements in the circuit.
+
−
E R
I6 I?
I = U
R
PR = UI = U U
R= U2
R
PE = U · (−I ) = U −U
R= −
U2
R
The power delivered by the voltage source is consumed by the resistor.
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2. lecture
Series and Parallel Circuits
Definition: series circuit
The elements are in series, if they are connected so that the same currentflows through the elements.
Definition: parallel circuit
The elements are in parallel, if they are connected so that there is thesame voltage across them.
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2. lecture
Series and Parallel Circuits
Series circuit
I-
I-
Parallel Circuit
U -
U -
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2. lecture
Resistors in series and in parallel
In series
R1 R2
⇐⇒
R = R1 + R2
In parallel
R1
R2
⇐⇒
R = 11R1
+ 1R2
Or, by using conductances: G = G1 + G2.
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2. lecture
Resistors in series and in parallel
The formulae on the previous slide can be applied to an arbitrarynumber of resistors. For instance, the total resistance of five resistorsin series is R = R1 + R2 + R3 + R4 + R5.
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2. lecture
Voltage Sources in Series
The voltages can be summed like resistances, but be careful withcorrect signs.
Voltage sources in parallel are inadmissible in circuit theory. There cannot be two different voltages between two nodes at the same time.
− +E1
+ −
E2
− +E3
r r
⇐⇒
− +E = E1 − E2 + E3
r r
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 30 / 125
2. lecture
What Series and Parallel Circuits are NOT
Just the fact that two components seem to be one after the other,does not mean that they are in series.
Just the fact that two components seem to be side by side, does notmean that they are in parallel.
In the figure below, which of the resistors are in parallel and which arein series with each other?
+
−
E1 +
−
E2R3
R1 R2
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 31 / 125
2. lecture
What Series and Parallel Circuits are NOT
Just the fact that two components seem to be one after the other,does not mean that they are in series.
Just the fact that two components seem to be side by side, does notmean that they are in parallel.
In the figure below, which of the resistors are in parallel and which arein series with each other?
+
−
E1 +
−
E2R3
R1 R2
SolutionNone! E1 ja R1 are in series and E2 ja R2 are in series. Both of these serial circuits are inparallel with R3. But no two resistors are in parallel nor in series.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 31 / 125
2. lecture
Terminal and Gate
A point which provides a point of connection to external circuits iscalled a terminal (or pole).
Two terminals form a gate.
An easy example: a car battery with internal resistance.
+
−
E
RS
b
b
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2. lecture
Node
A node means an area in the circuit where there are no potentialdifferences, or alternatively a place where two or more circuitelements meet.
A ”for dummies” –way to find nodes in the circuit: put your pen on awire in the circuit. Start coloring the wire, and backtrack when yourpen meets a circuit element. The area you colored is one node.
How many nodes are there in the circuit below?
+
−
E
R1 R3 R5
R2 R4 R6
I-
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2. lecture
Ground
One of the nodes in the circuit can be appointed the ground node.
By selecting one of the nodes to be the ground node, the circuitdiagram usually appear cleaner.
The car battery is connected to the chassis of the car. Therefore it isconvenient to handle the chassis as the ground node.
When we say ”the voltage of this node is 12 volts” it means that thevoltage between that node and the ground node is 12 volts.
+
−
E
R1 R3 R5
R2 R4 R6
I-
r
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2. lecture
Ground
The ground node can be connected to the chassis of the device or itcan be leave not connected to the chassis.
Therefore, the existence of the ground node does not mean that thedevice is ”grounded”.
The circuit on the previous slide can be presented also like this:
+
−
E
R1 R3 R5
R2 R4 R6
I-
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2. lecture
Homework 2 (released 3rd Sep, to be returned 7th Sep)
Homework 2
Find the current I .
+
−
E
R1 R3 R5
R2 R4 R6
I-
R1 = R2 = R3 = R4 = R5 = R6 = 1Ω E = 9V
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 36 / 125
3. lecture
Homework 2 - Model solution
Homework 2
Find the current I .
+
−
E
R1 R3 R5
R2 R4 R6
I-
R1 = R2 = R3 = R4 = R5 = R6 = 1Ω E = 9V
R5 ja R6 are in series. The total resistance of the serial connection isR5 + R6 = 2Ω.
Furthermore, the serial connection is in parallel with R4. Theresistance of this parallel circuit is 1
11+ 1
2
Ω = 23 Ω.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 37 / 125
3. lecture
Solution continues
R3 is in series with the parallel circuit calculated on the previous slide.The resistance for this circuit is R3 +
23 Ω = 5
3 Ω.
And the serial connection is in parallel with R2. The resistance for theparallel circuit is 1
( 53)−1+ 1
1
= 58 Ω.
Lastly, R1 is in series with the resistance computed in the previousstep. Therefore, the total resistance seen by voltage source E is58 Ω+ R1 =
138 Ω.
The current I is computed from Ohm’s law I = E138Ω= 72
13 A ≈ 5,5A.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 38 / 125
3. lecture
The Current Source
The current source is a circuit element which delivers a certaincurrent throught it, just like the voltage source keeps a certainvoltage between its nodes.
The current can be constant or it can vary by some rule.
J6
R
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3. lecture
The Current Source
If there is a current source in a wire, you know the current of thatwire.
J = 1A6
R1
I = 1A-
R2
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3. lecture
Applying Kirchhoff’s Laws Systematically to the CircuitWhen solving a circuit, it is highly recommended to use a systematicmehtod to find the voltages and/or currents. Otherwise it is easy to endup with writing a bunch of equations which can not be solved. Onesystematic method is called the nodal analysis:
1 Name each current in the circuit.2 Select one node as the ground node. Assign a variable for each
voltage between each node and ground node.3 Write an equation based on Kirchhoff’s current law for each node
(except the ground node).4 State the voltage of each resistor by using the node voltage variables
in step 2. Draw the voltage arrows at the same direction you used forthe current arrows (this makes it easier to avoid sign mistakes).
5 State every current by using the voltages and substitute them into thecurrent equations in step 2.
6 Solve the set of equations to find the voltage(s) asked.7 If desired, solve the currents by using the voltages you solved.Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 41 / 125
3. lecture
Example
Find the current I .
+
−
E1 +
−
E2R3
R1 R2
I?
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3. lecture
Example
Find the current I .
+
−
E1 +
−
E2R3
R1 R2
I?
I1-
I2
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 42 / 125
3. lecture
Example
Find the current I .
+
−
E1 +
−
E2R3
R1 R2
I?
I1-
I2
U3
?
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3. lecture
Example
Find the current I .
+
−
E1 +
−
E2R3
R1 R2
I?
I1-
I2
U3
?
I = I1 + I2
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3. lecture
Example
Find the current I .
+
−
E1 +
−
E2R3
R1 R2
I?
I1-
I2
U3
?
I = I1 + I2
E1 − U3- E2 − U3
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 42 / 125
3. lecture
Example
Find the current I .
+
−
E1 +
−
E2R3
R1 R2
I?
I1-
I2
U3
?
I = I1 + I2
E1 − U3- E2 − U3
U3
R3=
E1 − U3
R1+
E2 − U3
R2
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 42 / 125
3. lecture
Example
Find the current I .
+
−
E1 +
−
E2R3
R1 R2
I?
I1-
I2
U3
?
I = I1 + I2
E1 − U3- E2 − U3
U3
R3=
E1 − U3
R1+
E2 − U3
R2=⇒ U3 = R3
R2E1 + R1E2
R1R2 + R2R3 + R1R3
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 42 / 125
3. lecture
Example
Find the current I .
+
−
E1 +
−
E2R3
R1 R2
I?
I1-
I2
U3
?
I = I1 + I2
E1 − U3- E2 − U3
U3
R3=
E1 − U3
R1+
E2 − U3
R2=⇒ U3 = R3
R2E1 + R1E2
R1R2 + R2R3 + R1R3
I =U3
R3=
R2E1 + R1E2
R1R2 + R2R3 + R1R3
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 42 / 125
3. lecture
Some Remarks
There are many methods for writing the circuit equations, and thereis no such thing as ”right” method.
The only requirement is that you follow Kirchhoff’s laws and Ohm’slaw1 and you have an equal number of equations and unknowns.
If there is a current source in the circuit, it will (usually) make thecircuit easier to solve, as you then have one unknown less to solve.
By using conductances instead of resistances, the equations look alittle cleaner.
1Ohm’s law can only be utilized for resistors. If you have other elements, you
must know their current-voltage equation.Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 43 / 125
3. lecture
Another Example
+
−
E1 +
−
E2R3
R1 R2
R4
R5I1-
I2-
I3?
I5-
I4?
I1 = I2 + I3
I2 = I4 + I5
U3
?
U4
?
E1 − U3
R1=
U3 − U4
R2+
U3
R3ja
U3 − U4
R2=
U4
R4+
U4 − E2
R5
G1(E1−U3) = G2(U3−U4)+G3U3 ja G2(U3−U4) = G4U4+G5(U4−E2)
Two equations, two unknowns → can be solved. Use conductances!
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 44 / 125
3. lecture
Some Remarks
There are many other methods available too: mesh analysis, modifiednodal analysis, branch current method . . .
If there are ideal voltage sources in the circuit (=voltage sourceswhich are connected to a node without a series resistance), you needone more unknown (the current of the voltage source) and one moreequation (the voltage source will determine the voltage between thenodes it is connected to).
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 45 / 125
3. lecture
Homework 3 (released 7th Sep, to be returned 10th Sep)
Homework 3a)
Find the current I4.
Homework 3b)
Verify your solution by writing down all the voltages and currents to thecircuit diagram and checking that the solution does not contradict Ohm’sand Kirchhoff’s laws.
J6 R1
− +E
R4
R2 R3 R5
I4?R1 = R2 = R3 = R4 = R5 = 1Ω E = 9V J = 1A
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4. lecture
Homework 3 - Model Solution
Homework 3a)
Find current I4.
Homework 3b)
Verify your solution by writing down all the voltages and currents to thecircuit diagram and checking that the solution does not contradict Ohm’sand Kirchhoff’s laws.
J6 R1
− +E
R4
R2 R3 R5
I4?R1 = R2 = R3 = R4 = R5 = 1Ω E = 9V J = 1A
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 47 / 125
4. lecture
Solution
J6 R1
− +E
R4
R2 R3 R5
I4?
U2
?
U3
?
I-
R1 = R2 = R3 = R4 = R5 = 1Ω E = 9V J = 1A
First we write two current equations and one voltage equation. Theconductance of the series circuit formed by R4 ja R5 is denoted with G45.
J = U2G2 + I
I = U3G3 + U3G45
U2 + E = U3
By substituting I from the second equation to the first equation and thensubstituting U2 from the third equation, we get
J = (U3 − E )G2 + U3(G3 + G45)
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4. lecture
By substituting the component values, we get
U3 = 4V
Therefore the current I is 4V · 1S = 4A.
From the voltage equation U2 + E = U3 we can solve U2 = −5V,therefore the current through R2 is 5A upwards.
The current I is therefore 1A+5A = 6A, of which 4A goes throughR3:n and the remaining 2A goes through R4 ja R5.
There is no contradiction with Kirchhoff’s laws and therefore we canbe certain that our solution is correct.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 49 / 125
4. lecture
Example 1
Find I and U.
+
−
E1 +
−
E2
− +E3
R1
R2
J6I?
U
?
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4. lecture
Example 1
Find I and U.
+
−
E1 +
−
E2
− +E3
R1
R2
J6I?
U
?
I3
J = UG2 + I3
I3 = I + (E1 − E2)G1
U = E1 + E3
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4. lecture
Example 2
Find U2 and I1.
+
−
E1 +
−
E2 +
−
E3
J1
-
J2
RU2
I1
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4. lecture
Example 2
Find U2 and I1.
+
−
E1 +
−
E2 +
−
E3
J1
-
J2
RU2
I1
I1 = (E1 − E3)G + J1
E2 + U2 = E3
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 50 / 125
4. lecture
How To Get Extra Exercise?
There are plenty of problems with solutions available athttp://users.tkk.fi/~ksilvone/Lisamateriaali/lisamateriaali.
For example, you can find 175 DC circuit problems athttp://users.tkk.fi/~ksilvone/Lisamateriaali/teht100.pdf
At the end of the pdf file you can find the model solutions, so you cancheck your solution.
If you are enthusiastic, you can install and learn to use a circuitsimulator:http://www.linear.com/designtools/software/ltspice.jsp
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Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 52 / 125
4. lecture
Example 3
Find U4.
+
−
E R2 R4
R1 R3
U4
?
U2
?
(E − U2)G1 = U2G2 + (U2 − U4)G3
(U2 − U4)G3 = G4U4
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4. lecture
Homework 4 (released 10th Sep, to be returned 14th Sep)
Homework 4
Find the voltage U1. All resistors have the value 10Ω, E = 10V jaJ = 1A.
J6
R1 R3
R2
R4
+
−
EU1
?
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5. lecture
Homework 4 - Model Solution
Homework 4
Find the voltage U1. All resistors have the value 10Ω, E = 10V jaJ = 1A.
J6
R1 R3
R2
R4
+
−
EU1
?
U2
U3
I-
U1 − U2-
?U2 − U3
J = U1G1 + (U1 − U2)G2
(U1 − U2)G2 = (U2 − U3)G3 + I
G3(U2 − U3) + I = U3G4
U2 − U3 = E
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5. lecture
Solution
J = U1G1 + (U1 − U2)G2
(U1 − U2)G2 = EG3 + I
G3E + I = U3G4
U2 − U3 = E
I is solved from the third equation and substituted into the secondequation, then U3 is solved from the equation and substituted.
J = U1G1 + (U1 − U2)G2
(U1 − U2)G2 = EG3 + (U2 − E )G4 − G3E
1 = 0,2U1 − 0,1U2
0,1U1 − 0,1U2 = 0,1U2 − 1
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5. lecture
Solution
1 = 0,2U1 − 0,1U2
0,1U1 − 0,1U2 = 0,1U2 − 1
Which is solved
U1 = 10
U2 = 10
Therefore the voltage U1 is 10 Volts. This is easy to verify with a circuitsimulator.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 56 / 125
5. lecture
Circuit Transformation
1 An operation which transforms a part of the circuit into an internallydifferent, but externally equally acting circuit, is called a circuit
transformation.
2 For example, combining series resistors or parallel resistors into oneresistor, is a circuit transformation. Combining series voltage sourcesinto one voltage source is a circuit transformation too.
3 On this lecture, we learn dealing with parallel current sources and thesource transformation, with which we can transform a voltage sourcewith series resistance into a current source with parallel resistance.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 57 / 125
5. lecture
An Example of a Circuit Transformation
Two (or more) resistors are combined to a single resistor, which acts justlike the original circuit of resistors.
Resistors in series
R1 R2
⇐⇒
R = R1 + R2
Resistors in parallel
R1
R2
⇐⇒
R = 11R1
+ 1R2
Or, by using conductance: G = G1 + G2.
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5. lecture
Current Sources in Parallel
One or more current sources are transformed into single current source,which acts just like the original parallel circuit of current sources.
Current sources in parallel
J16
J26
J3
? b
b
⇐⇒ 6
J = J1 + J2 − J3
b
b
Just like connecting two or more voltage sources in parallel, connectingcurrent sources in series is an undefined (read: forbidden) operation incircuit theory, just like divide by zero is undefined in mathematics. Therecan not be two currents in one wire!
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 59 / 125
5. lecture
The Source Transformation
A voltage source with series resistance acts just like current source
with parallel resistance.
The source transformation
+
−
E
R
b
b
⇐⇒
J6
R
b
b
E = RJ
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 60 / 125
5. lecture
Important
Note that an ideal voltage or current source can not be transformedlike in previous slide. The voltage source to be transformed must haveseries resistance and the current source must have parallel resistance.
The resistance remains the same, and the value for the source isfound from formula E = RJ, which is based on Ohm’s law.
The source transform is not just a curiosity. It can save from manylines of manual calculations, for example when analyzing a transistoramplifier.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 61 / 125
5. lecture
Rationale for the Source Transformation
The source transformation
+
−
E
R I-
U
?
E
R
6R
I-
U
?
In the figure left:
I =E − U
RU = E − RI
In the figure right:
I =E
R−
U
R=
E − U
RU = (
E
R− I )R = E − RI
Both the circuits function equally.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 62 / 125
5. lecture
Example
Solve U.
+
−
E1
R1
R3
R2
+
−
EU
?
The circuit is transformed
J16
R1 R3R2
J26
And we get the result:
U =J1 + J2
G1 + G2 + G3
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 63 / 125
5. lecture
A Very Important Notice!
The value of the resistance remains the same, but the resistor is notthe same resistor! For instance, in the previous example the currentthrough the original resistor is not same as current through thetransformed resistor!
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5. lecture
Homework 5 (released 14th Sep, to be returned 17th Sep)
Homework 5
Find current I by using source transformation. J1 = 10A, J2 = 1A,R1 = 100Ω, R2 = 200Ω ja R3 = 300Ω.
J16
R1 R3
R2
J26
I-
This is easy and fast assignment. If you find yourself writing many lines ofequations, you have done something wrong.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 65 / 125
6. lecture
Homework 5 - Model Solution
Homework 5
Find current I by using source transformation. J1 = 10A, J2 = 1A,R1 = 100Ω, R2 = 200Ω ja R3 = 300Ω.
J16
R1 R3
R2
J26
I-
+
−
R1J1
R1 R3R2
+
−
R3J2
I-
I =R1J1 − R3J2
R1 + R2 + R3=
1000V − 300V
600Ω=
7
6A ≈ 1,17A.
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6. lecture
Thevenin’s Theorem and Norton’s Theorem
So far we have learned the following circuit transformations: voltagesources in series, current sources in parallel, resistances in parallel andin series and the source transformation.
Thevenin’s theorem and Norton’s theorem relate to circuittransformations too.
By Thevenin’s and Norton’s theorems an arbitrary circuit constistingof voltage sources, current sources and resistances can be transformedinto a single voltage source with series resistance (Thevenin’sequivalent) or a single current source with parallel resistance(Norton’s equivalent).
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6. lecture
Thevenin’s Theorem and Norton’s Theorem
Thevenin’s Theorem
An arbitrary linear circuit with two terminals is electrically equivalent to asingle voltage source and a single series resistor, called Thevenin’sequivalent.
Norton’s Theorem
An arbitrary linear circuit with two terminals is electrically equivalent to asingle current source and a single parallel resistor, called Norton’sequivalent.
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6. lecture
Calculating the Thevenin Equivalent
+
−
E
R1
R2
b
b
⇐⇒ +
−
ET
RT
b
b
The voltage ET in the Thevenin equivalent is solved simply by calculatingthe voltage between the terminals. For solving RT, there are two ways:
By turning off all independent (= non-controlled) sources in thecircuit, and calculating the resistance between the terminals.
By calculating the short circuit current of the port and applyingOhm’s law.
Independent source is a source, whose value does not depend on any othervoltage or current in the circuit. All sources we have dealt with for now,have been independent. Controlled sources covered later in this course.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 69 / 125
6. lecture
Calculating the Thevenin Equivalent
+
−
E
R1
R2
b
b
⇐⇒ +
−
ET
RT
b
b
The voltage at the port is found by calculating the current through theresistors and multiplying it with R2. The voltage at the port, called alsothe idle voltage of the port, is equal to ET.
ET =E
R1 + R2R2
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6. lecture
Calculating the Thevenin Equivalent
There are two ways to solve RT. Way 1: turn off all the (independent)sources, and calculate the voltage at the port. A turned-off voltage sourceis a voltage source, whose voltage is zero, which is same as just a wire:
R1
R2
b
b
⇐⇒
RT
b
b
Now it is easy to solve the resistance between the nodes of the port: R1 jaR2 are in parallel, and therefore the resistance is
RT =1
G1 + G2=
R1R2
R1 + R2.
This method is usually simpler than the other way with short circuitcurrent!
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 71 / 125
6. lecture
Solving RT by Using the Short-circuit CurrentThere are two ways to solve RT. Way 2: short-circuit the port, andcalculate the current through the short-circuit wire. This current is calledthe short-circuit current:
+
−
E
R1
R2
b
b
⇐⇒ +
−
ET
RT
b
b
IK? IK?
The value for the short-circuit current is
IK =E
R1
and the resistance RT is (by applying Ohm’s law to the figure on the right):
RT =ET
IK=
ET
E
R1
=E
R1+R2R2
E
R1
=R1R2
R1 + R2
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6. lecture
Norton Equivalent
The Norton equivalent is simply a Thevenin equivalent, which has beensource transformed into a current source and parallel resistance (or viceversa). The resistance has the same value in both equivalents. The valueof the current source is the same as the short-circuit current of the port.
JN6
RN
b
b
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6. lecture
Example 1
Calculate the Thevenin equivalent. All component values = 1.
J16
R1 R2
− +E
b
b
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6. lecture
Example 2
Calculate the Thevenin equivalent. All component values = 1.
+
−
E R2
J6R3
b
bR1
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6. lecture
Homework 6 (released 17th Sep, to be returned 21th Sep)
Homework 6
Calculate the Thevenin equivalent. All the component values are 1. (Everyresistance is 1Ω and the current source is J1 = 1A.)
J16
R1 R3
R2
b
b
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7. lecture
Homework 6 - Model Solution
Homework 6
Calculate the Thevenin equivalent. All the component values are 1. (Everyresistance is 1Ω and the current source is J1 = 1A.)
J16
R1 R3
R2
b
b
First we solve the voltage ET. This can be done by using applying sourcetransformation:
+
−
J1R1
R1
R3
R2
b
b
ET = J1R1R1+R2+R3
R3 =13 V
?
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 77 / 125
7. lecture
Solution
Next we solve the resistance RT of the Thevenin equivalent. The easiestway to do it is to turn off all the sources and calculate the resistancebetween the output port. (The other way is to find out the short-circuitcurrent.) The resistance can be solver either from the original or thetransformed circuit. Let’s use the transformed circuit and turn off thevoltage source:
R1
R3
R2
b
bRT = 1
1R1+R2
+ 1R3
= 23 Ω
Now the resistors R1 ja R2 are in series, and the series circuit is in parallelwith R3. Now we know both ET ja RT and we can draw the Theveninequivalent (on the next slide).
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 78 / 125
7. lecture
The Final Circuit
+
−
ET = 13 V
RT = 23 Ω
b
b
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7. lecture
Superposition Principle
A circuit consisting of resistances and constant-valued current andvoltage sources is linear.
If a circuit is linear, all the voltages and currents can be solved bycalculating the effect of each source one at the time.
This principle is called the method of superposition.
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7. lecture
Superposition Principle
The method of superposition is applied as follows
The current(s) and/or voltage(s) caused by each source is calculatedone at a time so that all other sources are turned off.
A turned-off voltage source = short circuit (a wire), a turned-offcurrent source = open circuit (no wire).
Finally, all results are summed together.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 81 / 125
7. lecture
Superposition Principle: an ExampleFind current I3 by using the superposition principle.
+
−
E1 +
−
E2R3
I3?
R1 R2
First, we turn off the rightmost voltage source:
+
−
E1 R3
I31?
R1 R2I31 =
E1
R1+1
G2+G3
1G2+G3
G3
Next, we turn off the leftmost voltage source:
+
−
E2R3
I31?
R1 R2I32 =
E2
R2+1
G1+G3
1G1+G3
G3
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7. lecture
Superposition Principle: an Example
The current I3 is obtained by summing the partial currents I31 and I32.
I3 = I31 + I32 =E1
R1 +1
G2+G3
1
G2 + G3G3 +
E2
R2 +1
G1+G3
1
G1 + G3G3
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7. lecture
When Is It Handy to Use the Superposition Principle?
If one doesn’t like solving equations but likes fiddling with the circuit.
If there are many sources and few resistors, the method ofsuperposition is usually fast.
If there are sources with different frequencies (as we learn on the ACCircuits course), the analysis of such a circuit is based on thesuperposition principle.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 84 / 125
7. lecture
Linearity and the Justification for the Superposition
Principle
The method of superposition is based on the linearity of the circuit,which means that every source affects every voltage and current witha constant factor.
This means that if there are sources E1, E2, E3, J1, J2 in the circuit,then every voltage and current is of formk1E1 + k2E2 + k3E3 + k4J1 + k5J2, where constants kn are realnumbers.
If all the sources are turned off (= 0), then all currents and voltagesin the circuit are zero. Therefore, by nullifying all sources except one,we can find out the multiplier for the source in question.
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7. lecture
Homework 7 (released 21st Sep, to be returned 24th Sep)
Homework 7
Find current I2 by using the superposition principle.
J1 = 1A R1 = 10Ω R2 = 20Ω R3 = 30Ω E1 = 5V
J16
R1 R3
R2I2
-
+
−
E1
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8. lecture
Homework 7 - Model Solution
Homework 7
Find current I2 by using the superposition principle.
J1 = 1A R1 = 10Ω R2 = 20Ω R3 = 30Ω E1 = 5V
J16
R1 R3
R2I2
-
+
−
E1
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 87 / 125
8. lecture
Ratkaisu
First, we find the effect of the current source:
J16
R1 R3
R2I21-
The voltage over R1 and R2 is the same (they are in parallel) and R2 istwice as large as R1 and therefore the current through R2 is half of thecurrent of R1. Because the total current through the resistors is J1 = 1A,the current through R1:n is 2/3A and the current through R2
isI21 = 1/3A.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 88 / 125
8. lecture
RatkaisuNext, we find out the effect of the voltage source:
R1 R3
R2I22-
+
−
E1
The resistors R1 and R2 are now in series and the total voltage over themis E = 5V, and therefore
I22 = −E
R1 + R2= −
5V
10Ω + 20Ω= −
1
6V.
The minus sign comes from the fact that the direction of the current I22 isupwards and the direction of the voltage E is downwards.Finally, we sum the partial results:
I2 = I21 + I22 =1
3A−
1
6A =
1
6A.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 89 / 125
8. lecture
Voltage Divider
R1
R2U
?
U2
?
U1 -
U1 = U R1R1+R2
ja U2 = U R2R1+R2
It is quite common in electronic circuit design, that we need areference voltage formed from another voltage in the circuit.
The formula is valid also for multiple resistors in series. Thedenominator is formed by summing all the resistances and the resistorwhose voltage is to be solved is in the numerator.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 90 / 125
8. lecture
Current Divider
R1 R2
I-
I1? I2?
I1 = I G1G1+G2
ja I2 = I G2G1+G2
The formula is valid also for multiple resistors in parallel.
The formula for current divider is not used as frequently as thevoltage divider, but it is natural to discuss it in this concept.
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8. lecture
Example 1
R4
R1 R2 R3
I1? I2? I3?
J6
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8. lecture
Example 1
R4
R1 R2 R3
I1? I2? I3?
J6
I1 = J G1G1+G2+G3
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8. lecture
Example 1
R4
R1 R2 R3
I1? I2? I3?
J6
I1 = J G1G1+G2+G3
I2 = J G2G1+G2+G3
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 92 / 125
8. lecture
Example 1
R4
R1 R2 R3
I1? I2? I3?
J6
I1 = J G1G1+G2+G3
I2 = J G2G1+G2+G3
I3 = J G3G1+G2+G3
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 92 / 125
8. lecture
Example 2
R1 R2 R3
R4+
−
E
U1
W U2
W U3
W
U4
9
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 93 / 125
8. lecture
Example 2
R1 R2 R3
R4+
−
E
U1
W U2
W U3
W
U4
9
U1 = E R1R1+R2+R3+R4
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 93 / 125
8. lecture
Example 2
R1 R2 R3
R4+
−
E
U1
W U2
W U3
W
U4
9
U1 = E R1R1+R2+R3+R4
U2 = E R2R1+R2+R3+R4
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 93 / 125
8. lecture
Example 2
R1 R2 R3
R4+
−
E
U1
W U2
W U3
W
U4
9
U1 = E R1R1+R2+R3+R4
U2 = E R2R1+R2+R3+R4
U3 = E R3R1+R2+R3+R4
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 93 / 125
8. lecture
Example 2
R1 R2 R3
R4+
−
E
U1
W U2
W U3
W
U4
9
U1 = E R1R1+R2+R3+R4
U2 = E R2R1+R2+R3+R4
U3 = E R3R1+R2+R3+R4
U4 = E R4R1+R2+R3+R4
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 93 / 125
8. lecture
Homework 8 (released 24th Sep, to be returned 28th Sep)
Homework 8
Find the voltage U by applying the voltage divider formula.
E1 = 10V R1 = 10Ω R2 = 20Ω R3 = 30Ω
R4 = 40Ω R5 = 50Ω E2 = 15V
+
−
E1
R1
R2 R3
R4
U -
+
−
E2
R5
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 94 / 125
9. lecture
Homework 8 - Model Solution
Homework 8
Find the voltage U by applying the voltage divider formula.
E1 = 10V R1 = 10Ω R2 = 20Ω R3 = 30Ω
R4 = 40Ω R5 = 50Ω E2 = 15V
+
−
E1
R1
R2 R3
R4
U -
+
−
E2
R5
U2
?
U3
?
U2 = E1R2
R1+R2= 10V 20Ω
10Ω+20Ω = 623 V
U3 = E2R3
R3+R4+R5= 15V 30Ω
30Ω+40Ω+50Ω = 3,75V
U = U2 − U3 = 21112 V ≈ 2,92V.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 95 / 125
9. lecture
Inductors and Capacitors
Ri
- u
W Li
- u
W
Ci
- u
W
u = Ri u = L di
dti = C du
dt
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 96 / 125
9. lecture
Inductors and Capacitors in DC Circuit
Ri
- u
W Li
- u
W
Ci
- u
W
u = Ri u = L di
dti = C du
dt
DC voltage and current remain constant as function of time or the timederivatives of the voltage and current is zero. Therefore the voltage of aninductor and the current of a capacitor is zero in a DC circuit.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 97 / 125
9. lecture
Exception 1
The capacitor is fed with DC current so that the current has no otherroute.
J6
C
i = C du
dt⇒ J = C du
dt⇒
du
dt= J
C. The voltage of the capacitor rises at
constant speed.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 98 / 125
9. lecture
Exception 2
A constant voltage source is connected to the terminals of an inductor.
+
−
E
L
u = L di
dt⇒ E = L di
dt⇒
di
dt= E
L. The current of the inductor rises at
constant speed.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 99 / 125
9. lecture
Dealing with all other cases involving inductors and
capacitors in DC circuits
Inductors are replaced with short circuits and capacitors are replaced withopen circuits.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 100 / 125
9. lecture
Homework 9 (released 28th Sep, to be returned 1st Oct)
Homework 9
Solve the voltage U from this DC circuit.
E1 = 10V R1 = 10Ω R2 = 20Ω R3 = 30Ω
R4 = 40Ω L = 500mH C = 2F E2 = 15V
+
−
E1
R1
R2 R3
L
C
+
−
E2
R4
U
9
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 101 / 125
10. lecture
Homework 9 - Model Solution
Homework 9
Solve the voltage U from this DC circuit.
E1 = 10V R1 = 10Ω R2 = 20Ω R3 = 30Ω
R4 = 40Ω L = 500mH C = 2F E2 = 15V
+
−
E1
R1
R2 R3
L
C
+
−
E2
R4
U
9
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 102 / 125
10. lecture
Homework 9 - Model Solution
Because there are no parallel connections of inductors and voltage sourcesand no serial connections of capacitors and current sources and the circuitis a DC circuit (= constant voltages and currents), we can replace theinductors with short circuits and the capacitors with open circuits.
+
−
E1
R1
R2 R3 +
−
E2
R4
U
9
in which case we obtain U easily by applying the voltage divider formula:
U = E2R3
R3 + R4= 6
3
7V ≈ 6,4V
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 103 / 125
10. lecture
Controlled Sources
So far, all of our sources have been constant valued.
If the value of a source does not depend on any of the voltages orcurrents in the circuit, the source is an independent source. Forexample, constant valued sources and sources varying as function oftime (only) are independent sources.
If the value of a source is a function of a voltage and/or current inthe circuit, the source is a controlled source.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 104 / 125
10. lecture
Voltage Controlled Voltage Source (VCVS)
r
r
u
?+
−
e = Au
The voltage e of VCVS is dependent of some voltage u.
The multiplier A is called voltage gain.
A real-world example: an audio amplifier.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 105 / 125
10. lecture
Current Controlled Voltage Source (CCVS)
r
r
i? +
−
e = ri
The voltage e of CCVS is dependent of some current i .
The multiplier r is called transresistance.
There is no good everyday example of this source available (of coursewe can construct this kind of source by using an operational
amplifier).
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 106 / 125
10. lecture
Voltage Controlled Current Source (VCCS)
r
r
u
?
j = gu6
The current j of VCCS is dependent of some voltage u.
The multiplier g is called transconductance.
A real-world example: a field-effect transistor (JFET or MOSFET).
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 107 / 125
10. lecture
Current Controlled Current Source (CCCS)
r
r
i?
j = βi6
The current j of CCCS is dependent of some current i .
The multiplier β is called current gain.
A real-world example: a (bipolar junction) transistor.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 108 / 125
10. lecture
Homework 10 (released 1st Oct, to be returned 5th Oct)
Homework 10
Find the voltage U.
E1 = 10V R1 = 10Ω R2 = 20Ω R3 = 30Ω
R4 = 40Ω r = 2Ω
+
−
E1
R1 R2
R3
i-
+
−
e2 = ri
R4
U
9
Note that the source on the right is a controlled source.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 109 / 125
11. lecture
Homework 10 - Model Solution
Homework 10
Find the voltage U.
E1 = 10V R1 = 10Ω R2 = 20Ω R3 = 30Ω
R4 = 40Ω r = 2Ω
+
−
E1
R1 R2
R3
i-
+
−
e2 = ri
R4
U
9
Note that the source on the right is a controlled source.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 110 / 125
11. lecture
+
−
E1
R1 R2
R3
i-
+
−
e2 = ri
R4
U
9
Let’s denote the total resistance of R1:n ja R2 with symbol R12 and write anodal equation:
UG3 = (E1 − U)G12 + (ri − U)G4
There are two unknowns in the circuit and therefore we need anotherequation with the same unknowns:
i = (E1 − U)G12
Then we substitute i to the first equation:
E1G12 − UG12 + rG4G12E1 − rG4G12U − UG4 = UG3
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 111 / 125
11. lecture
E1G12 − UG12 + rG4G12E1 − rG4G12U − UG4 = UG3
from which we get
G12E1(1 + rG4) = U(G3 + G12 + G4 + rG4G12).
Then we substitute the component values and solve U:
U =1030 (1 +
240)
130 + 1
30 + 140 + 2
40·30
= 3,75V
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 112 / 125
11. lecture
Recapitulation
On this lesson, we solve some refresher assignments. If you have solved allthe circuits, solve the home assignment.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 113 / 125
11. lecture
Recap assignment 1
Recap assignment 1
Find U and I first by using the method of superposition and then by someother method of your choice.
R1 = 1Ω R2 = 2Ω J = 1A E = 3V
+
−
E
I-
R1
R2
J
?U
?
Vastaus: I = 4A ja U = 1V.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 114 / 125
11. lecture
Recap assignment 2
Recap assignment 2
Form a Thevenin equivalent of the circuit on the left. Then, compute thecurrent IX, when the switches are closed and RX is a) 0Ω, b) 8Ω ja c)12Ω.
R1 = 5Ω R2 = 3Ω R3 = 8Ω R4 = 4Ω E = 16V
R1
R2
R3
R4
b b
b b+
−
E
RX
IX?
: Vastaus: RT = 8Ω, ET = 8V. a) 1A b) 0,5A c)0,4A.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 115 / 125
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 116 / 125
11. lecture
Homework 11 (released 5th Oct, to be returned 8th Oct)
Homework 11
We are given a fact that the current I3 = 0A. Find E1.
R1 = 5Ω R2 = 4Ω R3 = 2Ω R4 = 5Ω R5 = 6Ω E2 = 30V
+
−
E1 +
−
E2
R1 R3 R2
R4 R5
I3-
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 117 / 125
12. lecture
Homework 11 - Model Solution
Homework 11
We are given a fact that the current I3 = 0A. Find E1.
R1 = 5Ω R2 = 4Ω R3 = 2Ω R4 = 5Ω R5 = 6Ω E2 = 30V
+
−
E1 +
−
E2
R1 R3 R2
R4 R5
I3-
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 118 / 125
12. lecture
Homework 11 - Model Solution
Homework 11
We are given a fact that the current I3 = 0A. Find E1.
R1 = 5Ω R2 = 4Ω R3 = 2Ω R4 = 5Ω R5 = 6Ω E2 = 30V
+
−
E1 +
−
E2
R1 R3 R2
R4 R5
I3-
U4
?
U5
?
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 118 / 125
12. lecture
Because I3 = 0A, the current through R1 equals the current through R4
and the current through R2 equals the current through R5. Therefore, theresistors are in series2 and we may use the voltage divider formula to findvoltages over R4 and R5. The voltage over R5 is U5 = E2
R5R2+R5
= 18V.Therefore the voltage over R4 is 18V too. Now, by the voltage dividerrule:
U4 = E1R4
R1 + R4⇒ 18V = E1
5Ω
5Ω + 5Ω
from which we can solve E1 = 36V. Note: it is completely correct to writenodal equations for the circuit and solve E1 from them, too.
2Because and only because we know that I3 is zero.Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 119 / 125
12. lecture
Recapitulation
On this lesson, we solve some refresher assignments. I can alsodemonstrate some examples on the blackboard or to your booklets, too.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 120 / 125
12. lecture
Recap assignment 4
R2 = 5Ω E1 = 3V E2 = 2V
+
−
E1
+
−
E2
R2
R1
I1-
I2
a) How should we choose R1, if we want I2 to be 0A?b) How large is I1 then?a) 10 Ω ja b) 0,2A.
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 121 / 125
12. lecture
Recap assignment 5
R1 = 100Ω R2 = 500Ω R3 = 1,5 kΩ R4 = 1 kΩ E1 = 5V
J1 = 100mA J2 = 150mA
J16
R1 +
−
E1 R4
R2 R3
J2
-
r
U4
?
Find U4.U4 = 92V
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 122 / 125
12. lecture
Recap assignment 6
R1 = 12Ω R2 = 25Ω J = 1A E1 = 1V E2 = 27V
+
−
E1
J6
− +E2
R1
R2U
?
Find voltage U.Solution: 1
37V ≈ 27mV
Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 123 / 125
12. lecture
Homework 12 (released 8th Oct, to be returned 12th Oct)
Write a short essay on following subjects:
What did you learn on the course?
Did the course suck or was it worthwhile?
What could the lecturer do better?
How should this course be improved?
The essay will not affect the grading of the exam — please give honestfeedback3.How to return this homework: Write the essay as a plain text email (noattachments) and send it to me no later than the exam day at 18:00. Thesubject of the email message must be ’DC Circuits course feedback
2009 Firstname Surname’.
3I am really interested in how I could make the course better.Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 124 / 125
12. lecture
Final Notices on these Slides
The slides are licensed with CC By 1.0 4. In short: you can use andmodify the slides freely as long as you mention my name (= VesaLinja-aho) somewhere.
Single examples and circuits can be of course used without any namementioning, because they are not an object of copyright (legal term:”Threshold of originality”).
The origin of these slides is the DC Circuits course in Metropoliapolytechnic in Helsinki, Finland.
If you find typos, misspellings or errors in facts, please give mefeedback.
4http://creativecommons.org/licenses/by/1.0/Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 125 / 125