Circuit Analysis – DC Circuits

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

http://creativecommons.org/licenses/by/1.0/

Table of Contents

Click the lecture name to jump onto the first slide of the lecture.

1 1. lecture2 2. lecture3 3. lecture4 4. lecture5 5. lecture6 6. lecture

7 7. lecture8 8. lecture9 9. lecture10 10. lecture11 11. lecture12 12. lecture


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.


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 :-).


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.


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.


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

Electronics 1, Automotive Electrical Engineering Labs, . . .

Important!

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.


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.


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”.


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.


1. lecture

Now, Let’s Get into Business

Any questions on the practical arrangements of the course?


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

-


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!


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


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

?


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


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-


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).


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.


1. lecture

A simple DC circuit

A light bulb is wired to a battery. The resistance of the filament is10Ω.

+

−

12V @@

I =?-


1. lecture

A simple DC circuit


+

−

12V 10Ω

I =?-


1. lecture

A simple DC circuit


+

−

12V 10Ω

I =?-

12V

?


1. lecture

A simple DC circuit


+

−

12V 10Ω

I = 1,2A-

12V

?

U = RI

I = U

R= 12V

10Ω = 1,2A


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?


2. lecture

Homework 1 - Model solution

Homework 1


+

−

1,5V

+

−

1,5V

R = 20Ω

I?


2. lecture


Homework 1


+

−

1,5V

+

−

1,5V

R = 20Ω

I?U1

?

U2

?UR

?

U1 + U2 − UR = 0 ⇔ UR = U1 + U2


2. lecture


Homework 1


+

−

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


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-


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.


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.


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.


2. lecture

Series and Parallel Circuits

Series circuit

I-

I-

Parallel Circuit

U -

U -


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.


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.


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


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


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.


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


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-


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


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-


2. lecture

Homework 2 (released 3rd Sep, to be returned 7th Sep)

Homework 2


+

−

E

R1 R3 R5

R2 R4 R6

I-

R1 = R2 = R3 = R4 = R5 = R6 = 1Ω E = 9V


3. lecture


Homework 2


+

−

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.


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.


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


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


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


+

−

E1 +

−

E2R3

R1 R2

I?


3. lecture

Example


+

−

E1 +

−

E2R3

R1 R2

I?

I1-

I2


3. lecture

Example


+

−

E1 +

−

E2R3

R1 R2

I?

I1-

I2

U3

?


3. lecture

Example


+

−

E1 +

−

E2R3

R1 R2

I?

I1-

I2

U3

?

I = I1 + I2


3. lecture

Example


+

−

E1 +

−

E2R3

R1 R2

I?

I1-

I2

U3

?

I = I1 + I2

E1 − U3- E2 − U3


3. lecture

Example


+

−

E1 +

−

E2R3

R1 R2

I?

I1-

I2

U3

?

I = I1 + I2

E1 − U3- E2 − U3

U3

R3=

E1 − U3

R1+

E2 − U3

R2


3. lecture

Example


+

−

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


3. lecture

Example


+

−

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


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!


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).


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


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


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)


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.


4. lecture

Example 1

Find I and U.

+

−

E1 +

−

E2

− +E3

R1

R2

J6I?

U

?


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


4. lecture

Example 2

Find U2 and I1.

+

−

E1 +

−

E2 +

−

E3

J1

-

J2

RU2

I1


4. lecture

Example 2

Find U2 and I1.

+

−

E1 +

−

E2 +

−

E3

J1

-

J2

RU2

I1

I1 = (E1 − E3)G + J1

E2 + U2 = E3


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


http://users.tkk.fi/~ksilvone/Lisamateriaali/lisamateriaali.htm

http://users.tkk.fi/~ksilvone/Lisamateriaali/teht100.pdf

http://www.linear.com/designtools/software/ltspice.jsp

4. lecture

Example 3

Find U4.

+

−

E R2 R4

R1 R3

U4

?


4. lecture

Example 3

Find U4.

+

−

E R2 R4

R1 R3

U4

?

U2

?

(E − U2)G1 = U2G2 + (U2 − U4)G3

(U2 − U4)G3 = G4U4


4. lecture


Homework 4

Find the voltage U1. All resistors have the value 10Ω, E = 10V jaJ = 1A.

J6

R1 R3

R2

R4

+

−

EU1

?


5. lecture


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


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


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.


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.


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.


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!


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


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.


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.


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


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!


5. lecture


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.


6. lecture


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.


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).


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.


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.


6. lecture


+

−

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


6. lecture


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!


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


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


6. lecture

Example 1

Calculate the Thevenin equivalent. All component values = 1.

J16

R1 R2

− +E

b

b


6. lecture

Example 2

Calculate the Thevenin equivalent. All component values = 1.

+

−

E R2

J6R3

b

bR1


6. lecture


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


7. lecture


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

?


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).


7. lecture

The Final Circuit

+

−

ET = 13 V

RT = 23 Ω

b

b


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.


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.


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


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


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.


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.


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


8. lecture


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


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.


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.


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.


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.


8. lecture

Example 1

R4

R1 R2 R3

I1? I2? I3?

J6


8. lecture

Example 1

R4

R1 R2 R3

I1? I2? I3?

J6

I1 = J G1G1+G2+G3


8. lecture

Example 1

R4

R1 R2 R3

I1? I2? I3?

J6

I1 = J G1G1+G2+G3

I2 = J G2G1+G2+G3


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


8. lecture

Example 2

R1 R2 R3

R4+

−

E

U1

W U2

W U3

W

U4

9


8. lecture

Example 2

R1 R2 R3

R4+

−

E

U1

W U2

W U3

W

U4

9

U1 = E R1R1+R2+R3+R4


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


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


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


8. lecture


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


9. lecture


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.


9. lecture

Inductors and Capacitors

Ri

- u

W Li

- u

W

Ci

- u

W

u = Ri u = L di

dti = C du

dt


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.


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.


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.


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.


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


10. lecture


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


10. lecture


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


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.


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.


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).


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).


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.


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.


11. lecture


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.


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


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


11. lecture

Recapitulation

On this lesson, we solve some refresher assignments. If you have solved allthe circuits, solve the home assignment.


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.


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.


11. lecture

Recap assignment 3

Recap assignment 3

Find U3.

G1 = 1S G2 = 2S G3 = 3S G4 = 4S G5 = 5S g = 6S J = 3A

J6

G1 G2 G3

G4 G5

gU1

U1

?

U3

?

r

U3 = −48115

V ≈ −417mV


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-


12. lecture


Homework 11


R1 = 5Ω R2 = 4Ω R3 = 2Ω R4 = 5Ω R5 = 6Ω E2 = 30V

+

−

E1 +

−

E2

R1 R3 R2

R4 R5

I3-


12. lecture


Homework 11


R1 = 5Ω R2 = 4Ω R3 = 2Ω R4 = 5Ω R5 = 6Ω E2 = 30V

+

−

E1 +

−

E2

R1 R3 R2

R4 R5

I3-

U4

?

U5

?


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.


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.


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


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


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

http://creativecommons.org/licenses/by/1.0/

Circuit Analysis – DC Circuits

Education

Transcript of Circuit Analysis – DC Circuits