electric circuit analysis.pdf

PDF generated using the open source mwlib toolkit. See http://code.pediapress.com/ for more information.PDF generated at: Tue, 23 Nov 2010 07:36:26 UTC

ELECTRIC CIRCUITANALYSIS

ContentsPassive sign convention 1Simple Resistive Circuits 7Resistors in Series 12Resistors in Parallel 18Circuit Analysis Quiz 1 23Kirchhoff's Voltage Law 25Kirchhoff's Current Law 30Nodal analysis 35Mesh Analysis 41Circuit Analysis Quiz 2 47

ReferencesArticle Sources and Contributors 49Image Sources, Licenses and Contributors 50

Article LicensesLicense 51

Passive sign convention 1

Passive sign convention

Wikiversity Electrical Engineering SchoolThe Lessonsin

ELECTRIC CIRCUITS ANALYSIS COURSE

http://en.wikiversity.org/w/index.php?title=File:Crystal_Clear_app_kfm_home.png


http://en.wikiversity.org/w/index.php?title=Topic:Electrical_Engineering


Lessons in Electric Circuit Analysis

Lesson #1: Passive signconvention← You arehere

Lesson #2: Simple ResistiveCircuits

Lesson #3:Resistors in Series

Lesson #4:Resistors in Parallel

Quiz Test:Circuit Analysis Quiz1

Lesson #5:Kirchhoff's VoltageLaw

Lesson #6:Kirchhoff's CurrentLaw

Lesson #7:Nodal analysis

Lesson #8:Mesh Analysis

Quiz Test:Circuit Analysis Quiz2

HomeLaboratory: Circuit Analysis -

Lab1

http://en.wikiversity.org/w/index.php?title=File:Start_hand.svg

http://en.wikiversity.org/w/index.php?title=Electric_Circuit_Analysis

http://en.wikiversity.org/w/index.php?title=File:1_2_1.svg

http://en.wikiversity.org/w/index.php?title=File:Isimple_system_icons_app_edit.png




http://en.wikiversity.org/w/index.php?title=File:Crystal_Clear_app_help_index.svg






http://en.wikiversity.org/w/index.php?title=Circuit_Analysis_-_Lab1


http://en.wikiversity.org/w/index.php?title=File:Gartoon-Gnome-desktop-config.png


IntroductionThis is the first of eight lessons in Electric Circuit Analysis. This course is a pre-requisite courseto most Level 2 courses in this school. As such it is imperative that a student gains insight intothe methods and theory introduced and explained in this course.

There are plenty of worked examples and an exercises at the end of the lesson. Work through theexercise on your own, and only then you can compare your results with the solutions given on alinked Sub-page.

Lesson Preview

This Lesson is about Passive sign convention. This Lesson introduces a student to CircuitComponents which will be encountered in Electric Circuit Analysis. The student/User isexpected to understand the following at the end of the lesson

• Active Components• Passive Components• Passive Sign Convention• Guidelines for Passive sign Convention

Remember that Open Learning is all about you. You can set your own pace in this course andyou will be helped to evaluate your self along the way.


Part 1: Electric CircuitAn electric circuit is a connection ofcomponents (Voltage/Current sources,Resistors, Inductors and Capacitors) suchthat there is some power supplied anddissipated. This means that if you connect aresistor to a battery using conductor wires,then you have created an electrical circuit.

Figure 1.1: Active components

Active Components:

All components that Supply electric powerare called Active components. the followingpicture shows circuit symbols used to depicta Voltage Source and a Current Source.Notice that the components show a generalorientation of where the direction ofconventional current.

Figure 1.2 and 1.3: Passive components

Passive Components:

All components that Absorb or Dissipateelectric power are called Passivecomponents. the following picture showscircuit symbols used to depict a Resistor.Figure 1.2 is generally the preferred symbolof a resistor and will be used throughout thiscourse.Please note that capacitors and inductors arebeyond the scope of this course as theyintroduce complex resistance where real andreactive power complexities come in.

Part 2: Passive Sign ConventionThe concept of passive sign convention comes directly from the definition of voltage.

Voltage is a difference of charge between two places in space. Not an absolute quality. You couldthink of it in terms of depth and height.

Something has an elevation or height only with respect to something else such as sea level. Likewisedepth, something is only deep compared to some level, again such as sea level.

There is one difference between depth and height. We consider height to be positive and depth to benegative. One of the reasons why we do this is because we usually deal more with height then depth,and we wish to minimize the amount of subtraction that we perform.

The passive sign convention is the same concept. It is an algorithm to decide what is addingpotential energy to the system and what is taking it away.

Here are some basic ground rules:

• All resistors are either positive or negative uniformly. Which means that if you consider oneresistor to be positive (which is the common case) then all the resistors are positive.

• At least one source is the opposite sign of the resistors. If only one is present then that is the one.

• Always start by making your loop.

Why do we use this Passive sign convention?

One of the most important ideas of an electric circuit is that there is a source of power and adissipator of power. As circuit connections become more intricate this basic idea becomes moreblurred. In some cases there are more than one power supply at different circuit locations, such thatsimple addition of their power magnitudes is not possible. We need to know which direction powersupply and consumption is. The next examples will illustrate this.

http://en.wikiversity.org/w/index.php?title=File:EE-102-L01-Fig1.1.png

http://en.wikiversity.org/w/index.php?title=File:EE-102-L01-Fig1.2%261.3.png


Part 3

Figure 1.4: Passive Sign Convention scenario 1

Here is what we can deduce from figure 1.4. PointsA and B are physical end points of Resistor R. A ismore positive than B thus electrical charge at pointA is higher than the electrical charge at point B.This creates electric potential.

Explanation of Part 3This simply means that an electrical charge Q at point A will easily move to point B if a pathis set up (i.e Points A and B connected by a conductor.) Thus the resistor loses electricpotential and the electric charge is evenly spread. If electric charge is forced to point A frompoint B, then point A gains electric potential.

Thus for -V ; -I and +V ; +I cases The Electrical charge will lose electric potential byeffectively moving from high electric potential to low electric potential.The resistor haseffectively absorbed power from the electric charge to enable it to move to a low potentialpoint. Hence, the resistor, a passive component, absorbs power in this case.

Thus for -V ; +I and -V ; +I cases the electrical charge will gain electric potential byeffectively moving from low electric potential to high electric potential. The resistor haseffectively powered the electric charge to a high potential point. Hence, the resistor, a passivecomponent, supplies power in this case.

It is therefore important to understand the flow and direction of conventional current in orderto correctly apply passive sign convention. This becomes important later on in the coursewhen we treat Mesh and Nodal Analysis.

The following examples are related to the lesson. The answers to the exercise questions aregiven as a link to a sub page. Attempt the problems before viewing the answers.

Example 1.1

Figure 1.5: Example 1.1

Figure 1.5 shows a simpleresistor with the followingparameters.

, Find and Determine ifthis resistor is supplyingpower or dissipating it.Solution:

.

Since power is positive thisresistor is absorbing power.

Example 1.2


Figure 1.6 shows a simpleresistor with the followingparameters:

,

Find and Determine ifthis resistor is supplyingpower or dissipating it.Solution:

.

Since power is negative thisresistor is supplying power.





Example 1.3


Figure 1.7 shows a simpleresistor with the followingparameters.

,

Find and Determine ifthis Resistor is Supplyingpower or Dissipating it.Solution:

.

Since Power is Positive thisResistor is Dissipating power.Surprised?

Well let's look at figure 1.7again. If Voltage is Given as-6V it means that despite thegiven sign convention of theresistor, Point A is MorePositive Compared to point B.

The Current is shown enteringPoint A but by the fact thatcurrent is -3A it means that thecurrent is in fact leaving atpoint A. Thus The resistor iseffectively Dissipating power.Refer to part 3 & 4 of thislesson.

Try the exercises.

Exercises1. From Figure 1.5 given current is 4 Amps and the Voltage across the resistor is 4 Volts how much power is

being produced or consumed?2. From Figure 1.5 given current is 1.5 milli-Amps and the Voltage across the resistor is -1.5 Volts how much

power is being produced or consumed?3. From Figure 1.7 given current is 15 Amps and the Voltage across the resistor is 15 Volts how much power is

being produced or consumed?4. From Figure 1.5 given current is -20 milli-Amps and the Voltage across the resistor is -1.5 Volts how much

power is being produced or consumed?

• Answers to Exercise 1

Completion listOnce you finish your Exercises you can post your score here! To post your score just e-mail your course

co-ordinator your name and score *Click Here [1] .1. Ozzimotosan -- 75% & Corrected2. Doldham -- 75% & Corrected3. … 4. …

previous lesson previous page next page lesson intro next lesson course menu


http://en.wikiversity.org/w/index.php?title=Passive_sign_convention/Answers

http://en.wikiversity.org/w/index.php?title=File:Nuvola_apps_edu_miscellaneous.svg

http://en.wikipedia.org/wiki/Special%3Aemailuser%2Fthuvack

http://en.wikiversity.org/w/index.php?title=File:Crystal_Clear_app_xfmail.png

http://en.wikiversity.org/w/index.php?title=User:Ozzimotosan

http://en.wikiversity.org/w/index.php?title=File:Crystal_Clear_action_1uparrow.png


http://en.wikiversity.org/w/index.php?title=File:Crystal_Clear_action_1leftarrow.png

http://en.wikiversity.org/w/index.php?title=File:Blue_Glass_Arrow.svg

http://en.wikiversity.org/w/index.php?title=File:Crystal_Clear_action_reload.png

http://en.wikiversity.org/w/index.php?title=File:Crystal_Clear_action_1downarrow.png




Resource type: this resource contains a lecture or lecture notes.

Simple Resistive Circuits



http://en.wikiversity.org/w/index.php?title=File:Nuvola_apps_edu_languages.svg

http://en.wikiversity.org/w/index.php?title=Help:Resources_by_type

http://en.wikiversity.org/w/index.php?title=Category:Lectures




Simple Resistive Circuits 8

Lesson Review: Lesson 1The first Lesson was about Passive sign convention. The Lesson introduced CircuitComponents which will be encountered in Electric Circuit Analysis.

• Active Components• Passive Components• Passive Sign Convention• Guidelines for Passive sign Convention

Lesson PreviewThis Lesson is about Simple resistive Circuits. The student/User is expected tounderstand the following at the end of the lesson.

• Voltage: (V or v - Volts)The electrical potential between two points in a circuit.• Current: (I or i - Amperes)The amount of charge flowing through a part of a circuit.• Power: (W - Watts)Simply P = IV. It is the current times the voltage.• Source: A voltage or current source is the supplier for the circuit.• Resistor: (R measured in Ω - Ohms)A circuit element that "constricts" current flow.


Lesson #1: Passive sign convention

Lesson #2: Simple Resistive Circuits←You are here



Quiz Test:Circuit Analysis Quiz 1

Lesson #5:Kirchhoff's Voltage Law

Lesson #6:Kirchhoff's Current Law



Quiz Test:Circuit Analysis Quiz 2

HomeLaboratory: Circuit Analysis - Lab1
















Part 1

Voltage Source

This is possibly the simplest circuit. The voltage source supplies a voltage to the circuit. When thisvoltage is applied over a resistor, R, there is a current.

Equation 2.1

This equation explains the relation to all three elements in the circuit. In this case the voltage source hasthe same magnitude as the voltage drop across the resistor. We know that it is V. The resistor has acertain amount of Ohms depending on its rating. We now know R. With algebra I = V/R. So as long asyou know two of the variables then you can find the third.

Now comes the power part of the circuit analysis.

Equation 2.2

Once Equation 2.1 is solved then this equation should follow quickly. The I and V are the same variablesso insert them into the equation and solve for P (Watts). With these two equations, 1.1 and 1.2, and alittle bit of algebra you get:

Equation 2.3

Equation 2.4

Part 2As an explanation the power runningthrough is the voltage times the current.This is instantaneous power rather thanpower used over time. Power has to besupplied and consumed. In a perfectworld without heat-loss both are equal.The source supplies the required powerthat is consumed in this case by theresistor.

Example 2.1

Figure 2.1: Voltage Source

Given': Find: I, the current in Amps. The powerproduced by the source. The powerconsumed by the resistor.

Solution: Using the equations:

Remember the power supplied equalsthe power consumed.

http://en.wikiversity.org/w/index.php?title=File:SimpleResistiveCircuitsSimple1.png


Part 3

Current Source

All that will happen here is that the givens will change. Rather than knowing what the voltage is acrossthe resistor we now know what the current is flowing through the resistor.

Don't forget in the description of resistors that a similar model in fluid physics is a smaller pipe thatconstricts the amount of flow. Well, current is flow of charge. With the fluid the side of the smallersection being supplied with fluid will have a greater pressure than the out flowing side. The differencebetween these is potential. In circuits this potential is known as voltage, but then again this is all review,right?

So now we use equation 2.1 again. The current source gives us the current through the resistor. Given theresistor value it should be just a matter of multiplication.

Part 4

Example: 2.1

Figure 2.2: Current Source

Given': Find: V, the voltage. The powerproduced by the source. The powerconsumed by the resistor.


Of course, power consumed equalspower supplied in this perfect universe,this course.

Part 5: More Examples

Example 2.2

Given': Find: I, the current in Amps. The power produced by the source. Thepower consumed by the resistor.


Part 6

Example 2.3

Given': Find: V, the voltage. The power produced by the source. The powerconsumed by the resistor.


Part 7

Example 2.4

Given': Find: V, the voltage. R, the resistance


Part 8

Example 2.5

Given': Find: I, the current in Amps. R, the resistance.


http://en.wikiversity.org/w/index.php?title=File:SimpleResistiveCircuitsSimple2.png


Part 9

Example 2.6

Given':

Find: V, the voltage. I, thecurrent.

Solution: Using theequations:

OR

OR

Part 10: Exercise 21. If the given current is 300 Amps and the resistance is 2 Ohms, what is the Voltage across the resistor and how

much power is being produced or consumed?2. An engineer measures the resistivity of a resistor before putting it into a simple circuit. It is 50 Ohms. After

putting the resistor into place the engineer measures 2 Volts across the resistor. How many Amps are goingthrough the resistor?

3. A 60 Watt bulb is found to have 300 Ohms of resistance. What is the necessary voltage and current to have thebulb run optimally?

4. A large voltage supply has 10,000V. The company wants to know how big a resistance can be put on the voltagesupply along with how much power will be consumed. Is this a solvable problem?

5. Practice drawing the elements of a simple resistive circuit. Draw a resistor and Ω 5 times. Draw 5 current sourcesand 5 voltage sources.


Completion listOnce you finish your Exercises you can post your score here! To post your score just e-mail your course co-ordinator

your name and score *Click Here [1] .1. Ozzimotosan -- 100%2. Doldham -- 75% & Corrected3. … 4. …



http://en.wikiversity.org/w/index.php?title=Simple_Resistive_Circuits/Answers















Resistors in Series 12

Resistors in Series







Lesson 2 : ReviewWhat you need to remember from Simple resistive Circuits. If you ever feel lost, do not be shy togo back to the previous lesson & go through it again. You can learn by repetition.

• Voltage: (V or v - Volts)The electrical potential between two points in a circuit.• Current: (I or i - Amperes)The amount of charge flowing through a part of a circuit.• Power: (W - Watts)Simply P = IV. It is the current times the voltage.• Source: A voltage or current source is the supplier for the circuit.• Resistor: (R measured in Ω - Ohms)A circuit element that "constricts" current flow.


Lesson #1: Passive signconvention


Lesson #3:Resistors in Series←You are here


Quiz Test:Circuit AnalysisQuiz 1







Lab1

Lesson 3: PreviewThis Lesson is about Resistors in Series. The student/User is expected to understand the followingat the end of the lesson.

• Total Series Resistance: ( )The total Resistance of Resistors in series is the sum of allreistors in series.

















Part 1

Resistors in Series

Series of resistors means resistors connected end to end in a line.

This means that the resistance for the circuit is different than any oneresistor. Take two resistors in series in a circuit with a voltage supply.

To find the overall resistance of the circuit, add the resistances of theresistors.

Part 2Equation 3.1

So what if there were 10 resistors in series? Just add up all of theresistances and you have the equivalent over all resistance. In generalthis can be expressed:

Equation 3.2

Where R equivalent is the sum of all N of the resistors in series. So itreally doesn't matter how many resistors there are. If they are inseries they can be added up into an equivalent resistance.

Part 3

Voltage Divider

There comes a time when the boss or the project demands that you know whatthe voltage is between these millions of resistors in series. No need to panicthough because it isn't too much harder.

Lets take the two resistor problem first. There is a voltage source with tworesistors in series. We know that the overall voltage drop across the tworesistors is the same as the voltage the source is supplying in our exampleworld. So the voltage drop across one resistor would be a portion of theoverall drop. What proportion would we use to figure out the answer? Oneresistor over the two added together times the over all voltage drop:

Equation 3.3

Remember, this is the voltage drop across the first resistor. If you want theactual voltage there you still need to do some adding or subtracting to get it.Say that you have a 12V source and a drop over the first resistor of 3V. Thenyou actually need to subtract 3V from 12V to get the actual voltage betweenthe resistors.

At this point it seems that everything isn't quite as simple as it started. Withour example and equation for two resistors in series something else canhappen. What if the second resistor was set in the first resistor's place in theequation? Well, simply we would get the other side of the proportion:

Part 4Equation 3.4

This is the drop over the second resistor. But if it is dropping tozero, ground, or the negative side of the source then adding it tozero would give us the same answer as above.

For more than two resistors in series it is just a matter of keepingtrack of which resistor is on which side and summingappropriately.

Equation 3.5

Where is the voltage drop over N resistors out of a total of Mresistors. Remember that the resistors where the voltage drop isbeing calculated should be continuous. If they aren't all that can besaid about the answer derived from the equation is that it is part ofthe whole voltage drop and somewhat worthless otherwise.If the resistors are in the middle of the series then it will benecessary to calculate the voltage drop on one of the sides to beable to calculate the voltage.

http://en.wikiversity.org/w/index.php?title=File:Series_resitors.PNG

http://en.wikiversity.org/w/index.php?title=File:Series_resitors.PNG

http://en.wikiversity.org/w/index.php?title=File:Series_resistors_with_supply2.PNG


Part 5It becomes clear then that, two equal resistors will divide the source voltage intotwo equal voltages (half of the source's voltage is dropped across each resistor). Ifthe ratio of the resistance values is 3 to 1, there will be 3/4 of the source voltagedropped across the higher resistance, and of the source voltage dropped across

the lower resistance.

Three equal resistances in a series circuit with a single voltage source would drop1/3 of the source voltage across each resistor. If the three had 1-2-3proportionality (100,200,300 ohms for instance) they would drop , and

of the source voltage each. That is:

× VTotal

, × VTotal

,and × VTotal

.

Part 6

Current

Where does current come into any of this? Current, in this case,plays a similar role to that of the current in the Simple ResistiveCircuits. Once the equivalent resistance of all the resistors in aseries is found, effectively making a simple circuit again, thenthe current can be found with:

Equation 3.6

Just as a reminder. But the interesting thing is that the currentthrough all resistors in series is the same. If the resistor is 30Ω ithas the same current flow as the resistor with 500Ω, so long as itis in series. Thinking about everything above we are adding upall of the resistors to make a single equivalent resistor. Socurrent isn't different from 30Ω to 500Ω because together theresistance is 530Ω. That resistance is used then to calculate thecurrent.


Part 7

More Examples

Figure 3.1: Series resistors

Figure 3.1 shows a Series resistive circuit with thefollowing parameters. Vs=100Volts ; R1=15;R2=30; Find V1 and V2.

Solution: from Equation 3.3 we see that.

.

Similarily:

.

Thus it can be said that The Supply Voltage hasbeen divided between R1 and R2 by and

respectively.

Related Topic(s) in Wikiversity

Please visit the following page to supplementmaterial covered in this lesson.

• The role of resistors in electrical circuits

Part 8: Exercise 3Here are some questions to test yourself with.

1. Given 2 Resistors: R1 = R2 = 5 and Supply Volatage is 20 Volts find The V1; V2 andCurrent drawn by the Resistors.

2. Given 3 Resistors: R1 = 2 ; R2 = 3 and R3 = 7 and Supply Voltage is 15 Volts.Find V1; V2 & V3 and Current drawn by these Resistors.

3. Given 3 Resistors: R1 = 2 ; R2 = 3 and R3 = 7 and 3 Batteries with negligibleinternal resistances connected in series as follows: Vs1 = 3V ; Vs2 = 5V and Vs3 = 1.5V,Find V1; V2 & V3 dropped by individual Resistors.


http://en.wikiversity.org/w/index.php?title=File:EE-102-L02-fig1.png

http://en.wikiversity.org/w/index.php?title=The_role_of_resistors_in_electrical_circuits

http://en.wikiversity.org/w/index.php?title=Resistors_in_Series/Answers



Completion listOnce you finish your Exercises you can post your score here! To post your score just e-mail

your course co-ordinator, your name, and score Click Here [1] .1. Ozzimotosan -- 100%2. Doldham -- 100% & Corrected3. Sonu rockin -- 100% and corrected4. … 5. … 6. … 7. … 8. … 9. … 10. … 11. … 12. … 13. … 14. … 15. … 16. …


fr:Résistance et impédance/Résistance












http://en.wikiversity.org/w/index.php?title=Fr:R%C3%A9sistance_et_imp%C3%A9dance/R%C3%A9sistance




Resistors in Parallel 18

Resistors in Parallel







Lesson 3 : ReviewWhat you need to remember from Resistors in Series. If you ever feel lost, do not be shy to goback to the previous lesson & go through it again. You can learn by repitition.

• Total Series Resistance: ( )The total Resistance of Resistors in series is the sum of allresistors in series.

• Voltage Divider Equation 2.3:

• Current through Resistors connected in Series is the same for all resistors.





Lesson #4:Resistors inParallel← You arehere








Lab1

Lesson 4: PreviewThis Lesson is about Resistors in Parallel. The student/User is expected to understand thefollowing at the end of the lesson.

• two resistors connected in Parallel:

• Current Divider Principle:

















Part 1

Introduction

The best way to understand Parallel circuits is tostart with the definition. A circuit is parallel toanother circuit or several circuits if and only if itshare common terminals. That is if both of thebranches touch each other endpoints they are inparallel. Here is an example:

Figure 4.1: A Parallel circuit

R1, R2, and the voltage source are all in parallel.To prove this fact consider the top and bottomparts of the circuit.

Figure 4.2: Components in parallel share acommon nodes

The areas in yellow all are connected together, aswell as the areas in blue. So all the branches havethe same terminals, which means that R1, R2, andthe source are all in parallel.If we take this discussion of the water flowanalogy. Electric current can be seen as water andthe conductors as water pipes. Somethinginteresting happens as the current reaches thecommon node of Resistors that are connected inparallel, The total current is divided into theparallel branches.

Part 2

Voltage Rule

If two or more branches are parallel then the voltage across them is equal. So based on this wecan conclude that VR1=VR2=5volts. However unlike series resistors, the current across thebranches is not necessarily equal.

Equivalent resistance

For series resistors to find the total resistance we simply add them together. For parallelresistors its a little more complicated. Instead we use the following equation:

However for the case of only two resistors and only two resistors we can use this simplifiedform

Equation 4.2: Total Parallel Resistance

It is well to note at this point that The total Resistance of parallel connected Resistors willalways be Less than the smallest of the individual Resistors.

Current Rule

In Series Connection we deduced that Voltage is divide amongst resistors. For Parallelconnected Resistors, Current is divided. So here is a mathematical formula as we did withvoltage division principle.

Equation 4.3: Current Divider Formula

Using this formula you can workout the currents flowing through individual Resistors.

http://en.wikiversity.org/w/index.php?title=File:Circuit1.JPG

http://en.wikiversity.org/w/index.php?title=File:Circuit2.PNG


Part 3

Application

We have spent three lectures hacking on about What & Why Resistors & resistive circuitsin two connection schemes are used, (i.e Series and parallel connections). The questionnow is, where & how in Real life do these connections happen?

One simple application of these connection schemes is the Shunt application. In ElectricMeasurement industry, most often enough, we wish to measure Currents and Voltages ofVery High Magnitudes ( e.g some ranges of 500kV and upwards or 1000kA and upwards). The problem is that metering devices have delicate electronic components and usuallyhave small Voltage and Ampere operating ratings.

Solution to the above is to have a metering device connected in parallel to a resistor, thisresistor is thus called a "shunt" resistor since it is there to protect the metering device asshown in the next figure in part 4.

Part 4

Figure 4.3: Application of Parallel Resistive circuits.Shunt connection

If we know what the ampere rating of a device andwhat the total current is then we can work out the shuntcurrent and thus the Shunt Resistor.

Part 5: Examples


Figure 3.4 shows a Parallel resistive circuitwith the following parameters.

; ; ; Find ; Solution: from Equation 4.2 we see that.

Here are the solutions to the above problem:

.. .

Thus it can be said that The Supply Currenthas been divided between R1 and R2 .

We know that when solving these problems,we look at the Data given and thus we cansee how we need to manipulate ourequations in order to achieve ourobjective.The Following ExampleHighlights this point, see that you can followthe Method used and the reasoning behind.

Part 6: Examples


Figure 4.5 shows a Parallel resistive circuitwith the following parameters.

; ; ;Find: and .Solution: from Equation 3.2 we see that.

Here are the solutions to the above problem:

First Find: :

.Then;

.

..





Part 7

Do you Remember?

Let's take some time to Reflect on Material covered thus far. We havelearned a great deal about simple resistive circuits and the possibleconnections they afford us. Here I think you'll want to remember:

• Voltage: (V or v - Volts)The electrical potential between two points ina circuit.

• Current: (I or i - Amperes)The amount of charge flowing through apart of a circuit.

• Power: (W - Watts)Simply P = IV. It is the current times the voltage.• Source: A voltage or current source is the supplier for the circuit.• Resistor: (R measured in Ω - Ohms)A circuit element that "constricts"

current flow.

• Total Series Resistance: ( )

• Voltage Divider :

• Current through Resistors connected in Series is the same for allresistors.

• Two resistors connected in Parallel:

• Current Divider Principle:

Do Exercise 4 in part 8. After being completely satisfied of your work, youcan go on and try The next Page which is a quick quiz test. Good luck :-) !

Related Topic(s) in Wikiversity

Please visit the following page to supplement material covered in thislesson.

1. The role of resistors in electrical circuits2. Resistor Reduction

Part 8: Exercise 4Here are some questions to test yourself with.

1. Given 2 Resistors: in parallel find The .2. Given 3 Resistors: ; R2 = 3 and R3 = 7 in

parallel and Supply Current is 15Amps. Find: ; ; &and Supply Voltage across these Resistors.

3. Given 4 Resistors: R1 = 2 is connected in series to a parallelbranch consisting of R2 = 3 ; R3 = 7 and R4 = 4 Find:Total Resistance as seen by the Voltage source.

4. Is it possible to connect Voltage sources in Parallel, If so whatconditions must be met?


Completion listOnce you finish your Exercises you can post your score here! To postyour score just e-mail your course co-ordinator your name and score

*Click Here [1] .1. Ozzimotosan -- 100%2. Doldham -- 75% & Corrected3. … 4. … 5. … 6. … 7. … 8. … 9. … 10. … 11. …


http://en.wikiversity.org/w/index.php?title=The_role_of_resistors_in_electrical_circuits

http://en.wikiversity.org/w/index.php?title=Resistor_Reduction

http://en.wikiversity.org/w/index.php?title=Resistors_in_Parallel/Answers









http://en.wikiversity.org/w/index.php?title=Kirchhoff%2527s_Voltage_Law






Circuit Analysis Quiz 1

WARNING: Article could not be rendered - ouputting plain text.Potential causes of the problem are: (a) a bug in the pdf-writer software (b) problematic Mediawiki markup (c) tableis too wide

You have done well to get to this point, this your chance to test just how well you are doing. Remember that you set your pace, in your Open-Learning. You are advised to go through Lectures 1 ; 2 & 3 and do Exercises 1; 2 & 3 thoughroughly before attempting this quiz. Here are some pointers to answering this Quiz Test. Please read them carefully before attempting the questions. Be honest to your self, After attempting all Questions click on the Submit button, to View your score and Model Answers. Due to the foregoing please attempt this quiz test Once. This Quiz test is on Material covered thus far and as follows: Single resistor voltage problems. Single resistor resistance problems. Single resistor power problems. Series resistor problems. Series resistor Voltage problems. Parallel resistor problems. Parallel resistor current problems. Select the most correct answer of the four possible answers to each question. A calculator is allowed. Feel free to do work on a piece of paper. Can't understand a specific Question? Click Here to ask for help. Electric Circuit AnalysisLessons in Electric Circuit Analysis Lesson #1: Passive sign convention Lesson #2: Simple Resistive Circuits Lesson #3: Resistors in Series Lesson #4: Resistors in Parallel Quiz Test: Circuit Analysis Quiz 1← You are here Lesson #5: Kirchhoff's Voltage Law Lesson #6: Kirchhoff's Current Law Lesson #7: Nodal analysis Lesson #8: Mesh Analysis Quiz Test: Circuit Analysis Quiz 2 Home Laboratory: Circuit Analysis - Lab1 <quiz display=simple> { 3 amps flow through a 1 Ohm resistor. What is the voltage? type="()" } - (A) 1V. - (B) \frac{1}{3}V + (C) 3V. - (D) None of the above. { Why do we say the "voltage across" or "the voltage with respect to?" Why can't we just say voltage? type="()" } + (A) Voltage is a measure of Electric Potential difference between two electrical points. - (B) It's an Electrical Cliche'. - (C) The other point could be Negative or positive. - (D) None of the above. { A resistor consumes 5 watts, and its current is 10 amps. What is its voltage? type="()" } - (A) 10V. + (B) 0.5V. - (C) 2V. - (D) 15V. { A resistor has 10 volts across it and 4 amps going through it. What is its resistance? type="()" } + (A) 2.5\Omega. - (B) 3.5\Omega. - (C) 4.5\Omega. - (D) None of the above. { If you plot voltage vs. current in a circuit, and you get a linear line, what is the significance of the slope? type="()" } - (A) Power. - (B) Discriminant. + (C) Resistance. - (D) None of the above. { A resistor has 3 volts across it. Its resistance is 1.5 ohms. What is the current? type="()" } - (A) 3A - (B) 12A + (C) 2A - (D) 1.5A { A resistor has 8 volts across it and 3 Amps going through it. What is the power consumed? type="()" } + (A) 24W - (B) 3W - (C) 8W - (D) 2.2W { A resistor has a voltage of 5 volts and a resistance of 15 ohms. What is the power consumed? type="()" } + (A) 1.67 Watts - (B) 11.67 Joules - (C) 2.5 Watts - (D) None of the above { A resistor is on for 5 seconds. It consumes power at a rate of 5 watts. How many joules are used? type="()" } - (A) 5 Joules + (B) 25 Joules - (C) 3 Joules - (D) None of the above { A 1 ohm resistor has 5 volts DC across its terminals. What is the current (I) and the power consumed? type="()" } + (A) I = 5A & P = 25W. - (B) I = 25A & P = 5W. - (C) I = 3A & P = 9W - (D) I = 9A & P = 3W. { The voltage across two resistors in series is 10 volts. One resistor is twice as large as the other. What is the voltage across the larger resistor? What is the voltage across the smaller one? type="()" } - (A) V_{small-Resistor} = 5V and V_{Big-Resistor} = 5V. - (B) V_{Big-Resistor} = 3.33V andV_{small-Resistor} = 6.67V. + (C) V_{Big-Resistor} = 6.67V and V_{small-Resistor} = 3.33V. - (D) None of the above. { A 1 ohm, 2 ohm, and 3 ohm resistor are connected in series.




Circuit Analysis Quiz 1 24

What is the total resistance? type="()" } - (A) R_{Total} = 2\Omega. + (B) R_{Total} = 6\Omega. - (C) R_{Total}= 3\Omega. - (D) None of the above. { Two identical resistors are connected in series. The voltage across both ofthem is 250 volts. What is the voltage across each one? type="()" } + (A) R_1 = 125V and R_2 = 125V. - (B) R_1 =200V and R_2 = 200V. - (C) R_1 = 150V and R_2 = 200V. - (D) None of the above. { A 1 ohm, 2 ohm, and 3 ohmresistor are connected in parallel. What is the total resistance?type="()" } - (A) \frac{6}{3}\Omega. - (B)\frac{3}{6}\Omega. - (C) \frac{11}{6}\Omega. + (D) \frac{6}{11}\Omega. { A 5 ohm and a 2 ohm resistor areconnected in parallel. What is the total resistance? type="()" } + (A) \frac{10}{7}\Omega. - (B)\frac{7}{10}\Omega. - (C) \frac{10}{6}\Omega. - (D) \frac{6}{10}\Omega. { A 7 ohm and a 3 ohm resistor areconnected in parallel. What is the total resistance? type="()" } - (A) \frac{10}{21}\Omega. + (B)\frac{21}{10}\Omega. - (C) \frac{11}{7}\Omega. - (D) \frac{7}{11}\Omega. { Three 1 ohm resistors are connectedin parallel. What is the total resistance? type="()" } - (A) \frac{3}{2}\Omega. - (B) \frac{2}{3}\Omega. + (C)\frac{1}{3}\Omega. - (D) 3\Omega. { If you put an infinite number of resistors in parallel, what would the totalresistance be? type="()" } + (A) R_{total} would approach Zero as The No. of Resistors In parallel ApproachesInfinity. - (B) R_{total} would approach 1 as The No. of Resistors In parallel Approaches Infinity - (C) It is notpossible to connect that Number of Resistors in parallel. - (D) None of the above. { What is the current through R1and R2 in Diagram 1? type="()" } - (A) I_1 = 15A and I_2 = 25A. - (B) I_1 = 0.1A and I_2 = 0.1667A. + (C) I_1 =1A and I_2 = 1.667A. - (D) I_1 = 10A and I_2 = 16.67A. { What is the current through R1, R2, R3, and R4 inDiagram 2? type="()" } - (A) I_1 = 10A; I_2 = 50A; I_3 = 33A; I_4 = 25A.. - (B) I_1 = 1A; I_2 = 5A; I_3 = 3.3A;I_4 = 2.5A. - (C) I_1 = 0.25A; I_2 = 0.33A; I_3 = 0.5A; I_4 = 0.1A. + (D) I_1 = 1A; I_2 = 0.5A; I_3 = 0.33A; I_4 =0.25A. { Two resistors are in parallel with a voltage source. How do their voltages compare? type="()" } + (A) Theyboth have the same voltage as the source. - (B) They both have half the voltage of the source. - (C) One has fullvoltage, the other has none. - (D) None of the above. </quiz> Take some time off, you've done well. If you're aworkaholic then you can go to the next page. Resistors in Parallelprevious lesson Resistors in Parallelprevious pageKirchhoff%27s Voltage Lawnext page Passive sign conventionlesson intro Kirchhoff%27s Voltage Lawnext lessonElectric Circuit Analysiscourse menu Resource type: this resource is a quiz.

Kirchhoff's Voltage Law 25

Kirchhoff's Voltage Law







Circuit Analysis Part II (Laws & Theorems)This part is an introduction to some useful Electric Circuit Laws and theorems. You areencouraged to master the theorems and laws that will be discussed herein as they form a basis uponwhich most Circuit analysis methods are built.

Lesson ReviewWhat you need to remember from Previous Lessons.

• Review all lessons thus far ( i,e Read and be sure you understand lesson reviews done thus far )

Lesson 5: PreviewThis Lesson is about Kirchhoff's Voltage Law. The student/User is expected to understand thefollowing at the end of the lesson.

• Remember what was learned in Passive sign convention, You can go back and revise Lesson 1.

• Define Kirchhoff's Voltage Law ( word-by-word ).

• Kirchhoff's Voltage Law:







Lesson #5:Kirchhoff's VoltageLaw← You are here






Lab1


















Part 1: Kirchhoff's Voltage LawKirchhoff's Voltage Law states:

The sum of the voltages around a closed circuit path must be zero.

Notice that a closed circuit path insists that if one circuit element is chosen as a starting point, then onemust be able to traverse the circuit elements in that loop and return to the element in the beginning.

Mathematically, The Kirchhoff's Voltage Law is given by

For reference, this law is sometimes called Kirchhoff's Second Law, Kirchhoff's Loop Rule, andKirchhoff's Second Rule.

Part 2:Kirchhoff'sVoltage Law (Cont...)

Figure 5.1:

We observe five voltages in Figure 5.1:v4 across a voltage source, and the fourvoltages v1, v2, v3 and v5 across theresistors R1, R2, R3 and R5, respectively.The voltage source and resistors R1, R2and R3 comprise a closed circuit path,thus the sum of the voltages v4, v1, v2and v3 must be zero:

The resistor R5 is outside the closed pathin question, and thus plays no role in thecalculation of Kirchhoff's Voltage Lawfor this path. (Note that alternate closedpaths can be defined which include theresistor R5. In these cases, the voltage v5across R5 must be considered incalculating Kirchhoff's Voltage Law.)

http://en.wikiversity.org/w/index.php?title=W:Gustav_Robert_Kirchhoff

http://en.wikiversity.org/w/index.php?title=Voltage

http://en.wikiversity.org/w/index.php?title=File:KVL.png


Part 3Now, if we take the point d in the image as our reference point and arbitrarily set its voltage tozero, we can observe how the voltage changes as we traverse the circuit clockwise.

Going from point d to point a across the voltage source, we experience a voltage increase of v4volts (as the symbol for the voltage source in the image indicates that point a is at a positivevoltage with respect to point d).

On traveling from point a to point b, we cross a resistor. We see clearly from the diagram that,since there is only a single voltage source, current must flow from it's positive terminal to itsnegative terminal--clockwise around the circuit path. Thus from Ohm's Law, we observe thatthe voltage drops from point a to point b across resistor R1.

Likewise, the voltage drops across resistors R2 and R3. Having crossed R2 and R3, we arriveback at point d, where our voltage is zero (just as we defined). So we experienced one increasein voltage and three decreases in voltages as we traversed the circuit.

The implication from Kirchhoff's Voltage Law is that, in a simple circuit with only one voltagesource and any number of resistors, the voltage drop across the resistors is equal to the voltageapplied by the voltage source:

Kirchhoff's Voltage Law can easily be extended to circuitry that contains capacitors.

Part 4 : Example

Figure 5.2: Example 1

Consider Figure 5.2 with the followingParameters:

Find current through using Kirchhoff'sVoltage Law.Solution:

Figure 5.3: Example 1 loops

We can see that there are two closed paths (loops)where we can apply KVL in, Loop 1 and 2 asshown in figure 5.3

From Loop 1 we get:

From Loop 2 we get:

A bit confused? well look at the explanation inPart 3 of this lesson and Review Passive signconvention.

http://en.wikiversity.org/w/index.php?title=Ohm%27s_Law

http://en.wikiversity.org/w/index.php?title=File:EE-102-L05-Fig2.jpg



Part 5 : Example (Continued)The above results can further be simplified as follows:

..... (1)

and

..... (2)

By equating above (1) and (2) we can eliminate and hence get the following:

..... (3)

We end up with the above three equations and now substitute the Values given in theabove equations and solve the variables.If you feel lost up to this point do go back to the beginning of the example. Think ofthis as just another mathematical problem requiring solving by use of simultaneousequations with two unknowns!

Notice that we work with Variables only and try to solve the equation to its simplestform. Only after we have arrived at a simplified equation then that we can substitute invalues of Resistors, Voltages and current. This can save you a lot of trouble because, ifyou go wrong you can easily trace your work to the problem.

Part 6 : Example (Continued)It is clear that: from (3)

.

Substitute the Above Result into (2)

.

The Positive sign for only tells us that Current flows in the same direction to our initial assumeddirection. Thus now we can calculate Current through as follows:

.

The Negative sign for only tells us that Current flows in the same direction to direction.If you are lost repeat this example and try to follow thelogic, otherwise just send a message to the courseinstructor as outlined in Part 8.

Part 7:The method used to solve the above problem is very tedious, when the complexity of the circuit isincreased the method becomes very cumbersome and almost impossible to use in solving circuitequations. This method is used just to illustrate KVL in this lesson. In Lesson 08 a more efficientmethod of solving these kinds of Circuit problems using KVL is introduced in a form of MeshAnalysis.

Try the following exercise on your own and compare your answers with the given possible solution.

Part 8:Further Reading Links:

• Kirchhoff's circuit laws

References:

• Nilsson, James W. and Riedel, Susan A.Electric Circuits (5th ed.).Addison-Wesley. (1996).ISBN020155707X

http://en.wikiversity.org/w/index.php?title=W:Kirchhoff%27s_circuit_laws

http://en.wikiversity.org/w/index.php?title=W:James_W._Nilsson

http://en.wikiversity.org/w/index.php?title=W:Susan_A._Riedel



Exercise 5

Figure 5.4: Exercise 5

Consider Figure 5.4 with the following Parameters:

Find current through using Kirchhoff's Voltage Law.


Completion listOnce you finish your Exercises you can postyour score here! To post your score juste-mail your course co-ordinator your name

and score *Click Here [1] .1. … 2. … 3. … 4. … 5. … 6. … 7. …


es:Ley de Voltaje de Kirchhoff fr:Loi de Kirchhoff/Loi des mailles

Kirchhoff's Current Law




http://en.wikiversity.org/w/index.php?title=Kirchhoff%27s_Voltage_Law/Answers





http://en.wikiversity.org/w/index.php?title=Kirchhoff%2527s_Current_Law







http://en.wikiversity.org/w/index.php?title=Es:Ley_de_Voltaje_de_Kirchhoff

http://en.wikiversity.org/w/index.php?title=Fr:Loi_de_Kirchhoff/Loi_des_mailles




Kirchhoff's Current Law 31

Lesson Review 5:What you need to remember from Kirchhoff's Voltage Law. If you ever feel lost, do not be shy to goback to the previous lesson & go through it again. You can learn by repitition.




Lesson 6: PreviewThis Lesson is about Kirchhoff's Current Law. The student/User is expected to understand thefollowing at the end of the lesson.


• Define Kirchhoff's Current Law ( word-by-word ).

• Kirchhoff's Current Law:




Lesson #3:Resistors inSeries

Lesson #4:Resistors inParallel


Lesson #5:Kirchhoff'sVoltage Law

Lesson #6:Kirchhoff'sCurrent Law




HomeLaboratory: Circuit Analysis

- Lab1
















Part 1: Kirchhoff's Current LawKirchhoff's Current Law states:

The sum of the currents entering a particular point must be zero.

We can now define the electrical point physically connecting two or more electric circuit components, asa NODE. Note that a positive current leaving a point is considered to be a negative current entering thatpoint.

Mathematically, Kirchhoff's Current Law is given by

For reference, this law is sometimes called Kirchhoff's first law, Kirchhoff's point rule, Kirchhoff'sjunction rule, and Kirchhoff's first rule.

Part 2:Kirchhoff'sCurrent Law(Cont...)

Figure 6.1:

We observe four currents "entering" thejunction depicted as the bold black dotin Figure 6.1. Of course, two currentsare actually exiting the junction , but forthe purposes of circuit analysis it isgenerally less restrictive to considerwhat are in actuality positive currentsflowing out of a junction to be negativecurrents flowing into that junction(mathematically the same thing). Doingso allows us to write Kirchhoff's law forthis example as:

http://en.wikiversity.org/w/index.php?title=W:Gustav_Robert_Kirchhoff

http://en.wikiversity.org/w/index.php?title=Current

http://en.wikiversity.org/w/index.php?title=File:KCL.png


Part 3It may not be clear at this point why we insist on thinking of negative currents flowing into ajunction instead of positive currents flowing out. But note that Figure 6.1 provides us with moreinformation that we generally can expect to get when analysing circuits, namely the helpfularrows indicating the direction of current flow. If we don't have such assistance, we generallyshould not pass judgment on the direction of current flow (i.e., placing a negative sign beforeour current variable) until we calculate it, lest we confuse ourselves and make mistakes.

Nevertheless in this case we have the extra information of directional arrows in Figure 6.1, sowe should take advantage of it. We know that currents i2 and i3 flow into the junction and thecurrents i1 and i4 flow out. Thus we can write

Kirchhoff's Current Law as written is only applicable to steady-state current flow (i.e., noalternating current, no signal transmission). It can be extended to include time-dependentcurrent flow, but that is beyond the scope of this section.

Kirchhoff's Current Law is used in a method of circuit analysis referred to as nodal analysis tobe discussed in lecture 8. A node is a section of a circuit where there is no change in voltage(where there are no components, wire is often assumed to be perfectly conductive).

Each node is used to form an equation, and the equations are then solved simultaneously,giving the voltages at each node.

Part 4 : Example



Find current through using Kirchhoff'sCurrent Law.Solution:

Figure 6.3: Voltages at nodes

This is the same example we solved in Exercise 5.

Figure 6.3 shows Voltages at Nodes a, b, c and d.

We use node a as common node ( ground if youlike ). thus .

Part 5 : Example (Continued)From Node b we get:

From Node d we get:

It is clear that we must solve V_c, in order to complete Voltage definitions at allnodes. V_c will be found by applying KCL at Node c and solving resulting equationsFollows:

and

We can group like terms to get the following equation:

Part 6 : Example (Continued)Substitute values into previous equations you get:

thus

Thus now we can calculate Current through as follows:

.

Just as we expected! Note that current here is simplifiedbecause of following Voltage definitions and current pathsin Figure 6.3.

This method becomes tedious as the complexity of circuitsis increased.





• Kirchhoff's circuit laws

Refferences:

• Nilsson, James W. and Riedel, Susan A.Electric Circuits (5th ed.). Addison-Wesley.(1996).ISBN 020155707X

Exercise 6

Figure 6.4: Exercise 6


Find current through using Kirchhoff'sCurrent Law.


Part 8: Completion List

Once you finish your Exercises you can post your score here! To post your score just e-mail

your course co-ordinator your name and score *Click Here [1] .1. … 2. … 3. … 4. … 5. … 6. … 7. … 8. … 9. … 10. … 11. … 12. … 13. … 14. … 15. …


es:Ley de Corriente de Kirchhoff fr:Loi de Kirchhoff/Loi des nœuds

http://en.wikiversity.org/w/index.php?title=W:Kirchhoff%27s_circuit_laws

http://en.wikiversity.org/w/index.php?title=W:James_W._Nilsson

http://en.wikiversity.org/w/index.php?title=W:Susan_A._Riedel


http://en.wikiversity.org/w/index.php?title=Kirchhoff%2527s_Current_Law/Answers













http://en.wikiversity.org/w/index.php?title=Es:Ley_de_Corriente_de_Kirchhoff

http://en.wikiversity.org/w/index.php?title=Fr:Loi_de_Kirchhoff/Loi_des_n%C5%93uds

Nodal analysis 35

Nodal analysis






Nodal analysis 36

Lesson Review 5 & 6:What you need to remember from Kirchhoff's Voltage & Current Law . If you ever feel lost, do not beshy to go back to the previous lesson & go through it again. You can learn by repitition.




• Define Kirchhoff's Current Law ( word-by-word ).

• Kirchhoff's Current Law:

This part of the course onwards will collaborate with the Mathematics Department extensively.Mathematical Theory will be kept minimal as mathematical tools are only used here as a means to anend. Links to relevant Mathematical theories will be supplied to assist the student.




Lesson #3:Resistors inSeries





Lesson #7:Nodal analysis←You are here



HomeLaboratory: Circuit Analysis

- Lab1
















Nodal analysis 37

Lesson 7: PreviewThis Lesson is about Kirchhoff's Current Law. Thestudent/User is expected to understand the following at theend of the lesson.

• Use KCL at super nodes to formulate circuit equations.• Create matrix from circuit equations.• Solve for Unknown Node Voltages using Kramers Rule.

Part 1: Pre-reading MaterialThe student is advised to read the following resources fromthe Mathematics department:

• College Algebra• Linear algebra

The following external link has an excellent summary onusing Kramer's rule to solve linear equations:

* Solutions/kramer [1]

After you have satified yourself of the above resources, youcan go to Part 2.

Part 2: Nodal analysisLet's start off with some useful definitions:

• Node:

A point in a circuit where terminals of atleast two electriccomponents meet. This point can be on any wire, it is infinitely smalland dimensionless.

• Major Node:

This point is a node. A set of these nodes is used to create constraintequations.

• Reference Node:

The node to which Voltages of other nodes is read with regard to.This can be seen as ground ( V = 0).

• Branch:

This is a circuit element(s) that connect two nodes.

Basic rule: The sum of the currents entering any point (Node) must equal the sum ofthe currents leaving.( From KCL in Lecture 6).

http://en.wikiversity.org/w/index.php?title=College_Algebra

http://en.wikiversity.org/w/index.php?title=Linear_algebra

http://www.ece.umd.edu/class/enee204-1.S2000/solutions/kramer.pdf

Nodal analysis 38

Part 3The following is a general procedure for using Nodal Analysis method to solve electric circuitproblems. The aim of this algorithm is to develop a matrix system from equations found byapplying KCL at the major nodes in an electric circuit. Kramer's rule is then used to solve theunkown major node voltages.

Once the Node voltages are solved, normal circuit analysis methods ( Ohm's law; Voltage andCurrent Divider principles etc... ) can then be used to find whatever circuit entity. Remember toconsult previous lessons if you are not confident in using normal circuit analysis techniques thatwill be used in this lesson.

Manual Nodal Analysis Algorithm:

1.) Choose a reference node. ( Rule of thumb: take Node with most branches connecting to it )

2.) Identify and Number major nodes. ( Usually 2 or 3 major Nodes )

3.) Apply KCL to identified major nodes and formulate circuit equations.

4.) Create Matrix system from KCL equations obtained.

5.) Solve Matrix for unknown node voltages by using Kramer's rule ( It is simpler although youcan still use gaussian method as well )

6.) Use solved Node voltages to solve for the desired circuit entity.

The above algorithm is very basic and usefull for 2 x 2 and 3 x 3 size matrices. Generally as thenumber of major node voltages increase and the size of the matrix exceeds 3 x 3, numericalmethods ( Beyond scope of this course ) are employed with the aid of computers to solve suchcircuit networks.

Let's try an example to illustrate the above nodal analysis algorithm.

Part 4 : Example



Find current through using Nodal Analysismethod.Solution:

Figure 7.2: Voltages at nodes

This is the same example we solved in Exercise 6,except that in this case we have added extraResistors to increase the complexity of the circuit.

Figure 7.2 shows Voltages at Nodes a, b, c and d.

We use node a as common node ( ground if youlike ). thus as we did previously.

Follow Carefully theconstruction of the Nodalanalysis algorith explained inpart 3 of this lesson as follows inPart 5 and 6.

http://en.wikiversity.org/w/index.php?title=File:EE-102-L07-Fig1a.jpg


Nodal analysis 39

Part 5 : Example (Continued)Now that we have labelled the currents flowing in this circuit using passive signconvention, and have identified Nodes b; c and d as major nodes, we proceed asfollows:

KCL @ Node b:

Thus by applying Ohms law to above equation we get.

Therefore

............... (1)

KCL @ Node c:


Part 6 : Example (Continued)

Therefore

............... (2)

KCL @ Node d:


Therefore <<<This part is wrong!!! You have forgotten tochange all of your signs when moving across the equals sign>>>

............... (3)

Part 7:The next step in this algorithm is to construct a matrix. Inorder to do that easily we substitute all resistances in above equations 1; 2 & 3 with theirequivalent Admittances as follows:

etc thus equations 1; 2 & 3 will be re-written as follows:

Now we can create a matrix with the above equations as follows:

The following matrix is the above with values substituted:

→

Now that we have arranged equations 1; 2 & 3 into a matrix we need to get Determinants of the General matrix, and Determinants of alterations ofthe general matrix as follows:

http://en.wikiversity.org/w/index.php?title=W:Determinants


Nodal analysis 40

Part 8:Solving determinants of:

• Matrix A : General matrix Afrom KCL equations

• Matrix A1 : Genral Matrix Awith Column 1 substituted by .



As follows:

Part 9:

Now we can use the solved determinants to arrive at solutions for Node voltages as follows:

1.

2.

3.

Now we can apply Ohm's law to solve for the current through as follows:

As we have seen previously, the positive sign in the above current tells us that the effective current flowingthrough is in fact in the direction we chose when drawing up the circuit in figure 7.2.

Please use the provided link for details on working out the determinant of a 3 x 3 Matrix.

To apreciate the algorithm we have just used, try solving the above problem using either KVL or KCL as wedid in lessons 5 & 6 and see just how cumbersome the process would be.

As usual the following part is an Exercise to test your self on content discussed in this lesson. Please look atPart 11 for further reading material and interesting related External links.

Part 10:Exercise 7

Figure 7.3: Exercise


Find current through using NodalAnalysis method.


Part 11:

Further Reading & Other Interesting Links:

• w:Nodal analysis

• Free Nodal Analysis EBook by Dr. Yaz Li [2]

• Node voltage Analysis Web Application [3]

• MIT Circuits and Electronics (Video) Lecture - Basic Circuit Analysis Method [4]

References:

• Add reference here!

Completion List

Once you finish your Exercises you can post your score here! To post your score just e-mail your

course co-ordinator your name and score *Click Here [1] .1. … 2. … 3. …




http://en.wikiversity.org/w/index.php?title=File:EE-102-L08-Fig3-Final.jpg

http://en.wikiversity.org/w/index.php?title=Nodal_analysis/Answers


http://en.wikiversity.org/w/index.php?title=W:Nodal_analysis

http://circuits.solved-problems.com/nodal-analysis/

http://mazlumi.webng.com/node.asp

http://www.youtube.com/watch?v=2vHGYdepKLw&eurl=http%3A%2F%2Focw%2Emit%2Eedu%2FOcwWeb%2FElectrical%2DEngineering%2Dand%2DComputer%2DScience%2F6%2D002Spring%2D2007%2FVideoLectures%2Fdetail%2Fembed02%2Ehtm&feature=player_embedded#t=2014















Nodal analysis 41

Mesh Analysis






Mesh Analysis 42

Lesson Review 5 & 6:What you need to remember from Kirchhoff's Voltage & Current Law .


• Kirchhoff's Voltage Law

• Kirchhoff's Current Law

This part of the course onwards will collaborate with the Mathematics Department extensively.Mathematical Theory will be kept minimal as mathematical tools are only used here as a means to anend. Links to relevant Mathematical theories will be supplied to assit the student.










Lesson #8:Mesh Analysis←You are here



Lab1

Lesson Review 7:What you need to remember from Nodal analysis. If you ever feel lost, do not be shy to go back tothe previous lesson & go through it again. You can learn by repitition.

• Use KCL at super nodes to formulate circuit equations.• Create matrix from circuit equations.• Solve for Unknown Node Voltages using Kramers Rule.
















Mesh Analysis 43

Lesson 8: PreviewThis Lesson is about Mesh Analysis. The student/User is expected tounderstand the following at the end of the lesson.

• Use KVL at meshes or loops to formulate circuit equations.• Create matrix from circuit equations.• Solve for Unknown Mesh Currents using Kramers Rule.

Part 1: Pre-reading MaterialThe student is advised to read the following resources from theMathematics department:

• College Algebra• Linear algebra

Part 2: Mesh AnalysisLet's start off with some useful definitions:

• Branch:

This is a circuit element(s) that connect two nodes.

• Loop:

This a closed path in a circuit. A set of these loops areused to create constraint equations.

• Mesh:

A loop passing though atleast one branch.

Basic rule: The sum of Voltages arround any loopmust be Zero.( From KVL in Lecture 5).

http://en.wikiversity.org/w/index.php?title=College_Algebra

http://en.wikiversity.org/w/index.php?title=Linear_algebra

Mesh Analysis 44

Part 3The following is a general procedure for using Mesh or Loop Analysis method to solve electriccircuit problems. The aim of this algorithm is to develop a matrix system from equations foundby applying KVL arround Loops or Meshes in an electric circuit. Kramer's rule is then used tosolve the unkown Mesh Currents.

Once the Mesh Currents are solved, normal circuit analysis methods ( Ohm's law; Voltage andCurrent Divider principles etc... ) can then be used to find whatever circuit entity.

Remember to consult previous lessons if you are not confident in using normal circuit analysistechniques that will be used in this lesson.

Manual Mesh/Loop Analysis Algorithm:

1.) Choose a conventional current flow.

2.) Identify and Number loops or meshes. ( Usually 2 or 3 meshes/ loops )

3.) Apply KVL to identified meshes/loops and formulate ciruit equations.

4.) Create Matrix system from KVL equations obtained.

5.) Solve Matrix for unknown Mesh Currents by using Kramer's rule ( It is simpler althoughyou can still use gaussian method as well )

6.) Used solved Mesh Currents to solve for the desired circuit entity.

The above algorithm is very basic and usefull for 2 x 2 and 3 x 3 size matrices. Generally as thenumber of loops or meshes increase and the size of the matrix exceeds 3 x 3, numericalmethods ( Beyond scope of this course ) are employed with the aid of computers to solve suchcircuit networks.

Generally speaking, Mesh Analysis is shorter than Nodal Analysis although not alwayspreffered.Let's try an example to illustrate the above Mesh/Loop analysis algorithm.

Part 4 : Example



Find current through using Mesh Analysismethod.Solution:

Figure 8.2: Currents in loops

This is the same example we solved in Exercise 7.

We can see that there are three closed paths(loops) where we can apply KVL in, Loop 1, 2and 3 as shown in figure 8.2

We can now apply KVL arround the loopsremembering Passive Convention when definingCurrents and voltages.

KVL arround abca loop:

Therefore

............(1)



Mesh Analysis 45

Part 5 : Example(Continued)KVL arround acda loop:

Therefore

............... (2)

KVL arround bdcb loop:

Therefore

............... (3)

Now we can create a matrix with the aboveequations as follows:

Part 6 : Example (Continued)

The following matrix is the above with values substituted:

→

Now that we have arranged equations 1; 2 & 3 into a matrix we need to get Determinants of theGeneral matrix, and Determinants of alterations of the general matrix as follows:

Solving determinants of:

• Matrix A : General matrix A from KVL equations

• Matrix A1 : Genral Matrix A with Column 1 substituted by .

• Matrix A2 : Genral Matrix A with Column 2 substituted by .

• Matrix A3 : Genral Matrix A with Column 3 substituted by .As follows:

Part 7: Exercise 8 Part 8:

Now we can use the solved determinants to arrive at solutions for Mesh Currents as follows:

1.

2.

3.

Now we can solve for the current through as follows:

The answer is as we expected it to be, flows in the direction of .



Mesh Analysis 46

Part 7: Exercise 8

Figure 8.3: Exercise


Find current through using MeshAnalysis method.



• w:Mesh analysis

• Mesh-Currents Analysis Web Application [1]

Refferences:

• Add refference here!

Completion ListOnce you finish your Exercises you can post your score here! To post your score just e-mail your

course co-ordinator your name and score *Click Here [1] .1. … 2. … 3. …



el:Μέθοδος βροχικών ρευμάτων

http://en.wikiversity.org/w/index.php?title=File:EE-102-L08-Fig3-Final.jpg

http://en.wikiversity.org/w/index.php?title=Mesh_Analysis/Answers


http://en.wikiversity.org/w/index.php?title=W:Mesh_analysis

http://mazlumi.webng.com/mesh.asp














http://en.wikiversity.org/w/index.php?title=El:%CE%9C%CE%AD%CE%B8%CE%BF%CE%B4%CE%BF%CF%82_%CE%B2%CF%81%CE%BF%CF%87%CE%B9%CE%BA%CF%8E%CE%BD_%CF%81%CE%B5%CF%85%CE%BC%CE%AC%CF%84%CF%89%CE%BD


Circuit Analysis Quiz 2

WARNING: Article could not be rendered - ouputting plain text.Potential causes of the problem are: (a) a bug in the pdf-writer software (b) problematic Mediawiki markup (c) tableis too wide

You have done well to get to this point, The final Quiz Test for this course.This your chance to test just how wellyou are doing. Remember that you set your pace, in your Open-Learning. You are advised to go through AllLectures and to have done All Exercises thoughroughly before attempting this quiz. Be honest to your self, Afterattempting all Questions click on the Submit button, to View your score and Model Answers. Due to the foregoingplease attempt this quiz test Once. This Quiz test is on Material covered thus far and as follows: Passive signconvention. Single resistor problems. Series resistor problems. Parallel resistor problems. Kirchhoff's Voltage Lawproblems. Kirchhoff's Current Law problems. Nodal Analysis problems. Mesh Analysis problems. Select the mostcorrect answer of the four possible answers to each question. A calculator is allowed. Feel free to do work on a pieceof paper. Can't understand a specific Question? Click Here to ask for help. Electric Circuit AnalysisLessons inElectric Circuit Analysis Lesson #1: Passive sign convention Lesson #2: Simple Resistive Circuits Lesson #3:Resistors in Series Lesson #4: Resistors in Parallel Quiz Test: Circuit Analysis Quiz 1 Lesson #5: Kirchhoff'sVoltage Law Lesson #6: Kirchhoff's Current Law Lesson #7: Nodal analysis Lesson #8: Mesh Analysis Quiz Test:Circuit Analysis Quiz 2← You are here Home Laboratory: Circuit Analysis - Lab1 <quiz display=simple> { What isthe significance of a negative ( - ) sign from a calculation when solving circuit problems? type="()" } - (A) None -(B) It means you did something wrong on your calculation + (C) Real resulting current or Voltage is in the oppositedirection to one assumed - (D) You probably used a smaller scalling factor { What will be the reading on thevoltmeter for the following circuit? type="()" } - (A) 7.5V - (B) Zero - (C) Infinity + (D) 9V { Which Resistor in theParallel brach of the following circuit will always have most current flowing through irrespective of the amount ofthe supply voltage? type="()" } + (A) R4 - (B) R2 - (C) R3 - (D) None { Which one of the following statements ismore accurate? type="()" } + (A) KVL is concerned with Voltage Drops - (B) KVL is calculated at Super Nodesonly - (C) Both (A) and (B) - (D None { KCL is used when solving circuits with ... type="()" } - (A) Closed Loops +(B) Sufficient Nodes / Junctions - (C) Capacitors - (D None { Which method would be simpler to implement in thefollowing circuit? type="()" } + (A) KCL - (B) Millman's theorem - (C) KVL - (D) Norton { Nodal Analysis appliesthe following principles … type="()" } - (A) KVL & Ohm's Law + (B) KCL & Ohm's Law - (C) KVL &Superposition - (D) KCL & Superposition { Which of the following statements is true? type="()" } - (A) MeshAnalysis is easiest when a circuit has more than two nodes - (B) Mesh Analysis is more difficult than Nodal Analysis+ (C) Mesh Analysis employs KVL to solve for Loop currents - (D) All of the above { Given the following problem.which method would yield simpler equations for solving circuit elements? type="()" } + (A) Mesh Analysis - (B)Nodal Analysis - (C) KVL - (D) Thervenin { Which of the following statements is False? type="()" } + (A) Nodal &Mesh Analysis are only used for Linear circuits - (B) Kramers rule is used in solving Nodal Analysis Problems - (C)Gaussian Method can be used to solve Nodal analysis problems - (D) Nodal Analysis can be used to verify any Meshanalysis problem </quiz> This is it, The last Quiz Test for this course.The next page, is an Introduction to HomeLaboratory Exercises. Mesh Analysisprevious lesson Mesh Analysisprevious page Circuit Analysis - Lab1next pagePassive sign conventionlesson intro Circuit Analysis - Lab1next lesson Electric Circuit Analysiscourse menuResource type: this resource is a quiz.

Article Sources and Contributors 49

Article Sources and ContributorsPassive sign convention Source: http://en.wikiversity.org/w/index.php?oldid=613787 Contributors: Bgorges, Dhamal, Linelor, McCormack, Mortense, Remi, Seanusmc, The Isiah, Thuvack, 5anonymous edits

Simple Resistive Circuits Source: http://en.wikiversity.org/w/index.php?oldid=534885 Contributors: Bgorges, McCormack, Remi, Rsbhatia16, Thuvack, 2 anonymous edits

Resistors in Series Source: http://en.wikiversity.org/w/index.php?oldid=618097 Contributors: Bgorges, Dhamal, Landreu, McCormack, Redemption, Remi, The Isiah, Thuvack, 5 anonymousedits

Resistors in Parallel Source: http://en.wikiversity.org/w/index.php?oldid=618118 Contributors: Bsheehan5013, Dhamal, JWS, McCormack, The Isiah, Thuvack, 5 anonymous edits

Circuit Analysis Quiz 1 Source: http://en.wikiversity.org/w/index.php?oldid=515348 Contributors: Driscoll, Erkan Yilmaz, Hr1354, Mario51t, McCormack, Redemption, Remi, The Isiah,Thuvack, 8 anonymous edits

Kirchhoff's Voltage Law Source: http://en.wikiversity.org/w/index.php?oldid=462195 Contributors: Dhamal, Electroguy, HenkvD, Javier Carro, Lerdsuwa, McCormack, Remi0o, RobertLutz,Thuvack, 7 anonymous edits

Kirchhoff's Current Law Source: http://en.wikiversity.org/w/index.php?oldid=462196 Contributors: 24cell, Electroguy, HenkvD, Javier Carro, Lerdsuwa, McCormack, Remi0o, RobertLutz,Sebmol, Stinky sock, Thuvack, 6 anonymous edits

Nodal analysis Source: http://en.wikiversity.org/w/index.php?oldid=626186 Contributors: Jtneill, McCormack, Mu301, PeteX, The Isiah, Thuvack, 8 anonymous edits

Mesh Analysis Source: http://en.wikiversity.org/w/index.php?oldid=535539 Contributors: Assassingr, McCormack, Thuvack, 2 anonymous edits

Circuit Analysis Quiz 2 Source: http://en.wikiversity.org/w/index.php?oldid=622225 Contributors: McCormack, Sakatti, Thuvack, 2 anonymous edits

Image Sources, Licenses and Contributors 50

Image Sources, Licenses and ContributorsImage:Crystal Clear app kfm home.png Source: http://en.wikiversity.org/w/index.php?title=File:Crystal_Clear_app_kfm_home.png License: unknown Contributors: CyberSkull, J.delanoy,Loyna, Mysid, Pseudomoi, Rocket000, Tcrow777, 1 anonymous editsImage:Start_hand.svg Source: http://en.wikiversity.org/w/index.php?title=File:Start_hand.svg License: Creative Commons Attribution-Sharealike 2.5 Contributors: Peter SchwarzImage:1 2 1.svg Source: http://en.wikiversity.org/w/index.php?title=File:1_2_1.svg License: Public Domain Contributors: HejkompisImage:Isimple system icons app edit.png Source: http://en.wikiversity.org/w/index.php?title=File:Isimple_system_icons_app_edit.png License: GNU General Public License Contributors:http://www.kde-look.org/usermanager/search.php?username=dirceuFile:Crystal Clear app help index.svg Source: http://en.wikiversity.org/w/index.php?title=File:Crystal_Clear_app_help_index.svg License: GNU Lesser General Public License Contributors:User:쿠도군

Image:Gartoon-Gnome-desktop-config.png Source: http://en.wikiversity.org/w/index.php?title=File:Gartoon-Gnome-desktop-config.png License: GNU General Public License Contributors:Berto, Cwbm (commons), Cycn, Denniss, Javierme, Mikm, Trần Nguyễn Minh HuyImage:EE-102-L01-Fig1.1.png Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L01-Fig1.1.png License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:EE-102-L01-Fig1.2&1.3.png Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L01-Fig1.2&1.3.png License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:EE-102-L01-Fig1.4.png Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L01-Fig1.4.png License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:EE-102-L01-Fig1.6.png Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L01-Fig1.6.png License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:EE-102-L01-Fig1.7.png Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L01-Fig1.7.png License: GNU Free Documentation License Contributors: ThuvackImage:Nuvola apps edu miscellaneous.svg Source: http://en.wikiversity.org/w/index.php?title=File:Nuvola_apps_edu_miscellaneous.svg License: GNU Lesser General Public License Contributors: User:StanneredImage:Crystal Clear app xfmail.png Source: http://en.wikiversity.org/w/index.php?title=File:Crystal_Clear_app_xfmail.png License: unknown Contributors: Color probe, CyberSkull,Rocket000, 3 anonymous editsImage:Crystal Clear action 1uparrow.png Source: http://en.wikiversity.org/w/index.php?title=File:Crystal_Clear_action_1uparrow.png License: unknown Contributors: Abu badali,Augiasstallputzer, CyberSkull, Rocket000Image:Crystal Clear action 1leftarrow.png Source: http://en.wikiversity.org/w/index.php?title=File:Crystal_Clear_action_1leftarrow.png License: unknown Contributors: Augiasstallputzer,CyberSkull, Rocket000Image:Blue Glass Arrow.svg Source: http://en.wikiversity.org/w/index.php?title=File:Blue_Glass_Arrow.svg License: GNU Lesser General Public License Contributors: User:EveraldoCoelhoImage:Crystal Clear action reload.png Source: http://en.wikiversity.org/w/index.php?title=File:Crystal_Clear_action_reload.png License: unknown Contributors: CyberSkull, Freddo, Loyna,Ms2ger, Rocket000, 3 anonymous editsImage:Crystal Clear action 1downarrow.png Source: http://en.wikiversity.org/w/index.php?title=File:Crystal_Clear_action_1downarrow.png License: unknown Contributors: Abu badali,Augiasstallputzer, CyberSkull, Foroa, Juliancolton, Rocket000, 6 anonymous editsImage:Crystal Clear action 2uparrow.png Source: http://en.wikiversity.org/w/index.php?title=File:Crystal_Clear_action_2uparrow.png License: unknown Contributors: Augiasstallputzer,CyberSkull, Rocket000, 1 anonymous editsImage:Nuvola apps edu languages.svg Source: http://en.wikiversity.org/w/index.php?title=File:Nuvola_apps_edu_languages.svg License: GNU Lesser General Public License Contributors:User:StanneredImage:SimpleResistiveCircuitsSimple1.png Source: http://en.wikiversity.org/w/index.php?title=File:SimpleResistiveCircuitsSimple1.png License: unknown Contributors: BgorgesImage:SimpleResistiveCircuitsSimple2.png Source: http://en.wikiversity.org/w/index.php?title=File:SimpleResistiveCircuitsSimple2.png License: unknown Contributors: BgorgesImage:Series resitors.PNG Source: http://en.wikiversity.org/w/index.php?title=File:Series_resitors.PNG License: Public Domain Contributors: The IsiahImage:series_resistors_with_supply2.PNG Source: http://en.wikiversity.org/w/index.php?title=File:Series_resistors_with_supply2.PNG License: Public Domain Contributors: The IsiahImage:EE-102-L02-fig1.png Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L02-fig1.png License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:Circuit1.JPG Source: http://en.wikiversity.org/w/index.php?title=File:Circuit1.JPG License: Public Domain Contributors: The IsiahImage:Circuit2.PNG Source: http://en.wikiversity.org/w/index.php?title=File:Circuit2.PNG License: Public Domain Contributors: The IsiahImage:EE-102-L03-Fig3.3.png Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L03-Fig3.3.png License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:EE-102-L03-Fig3.4.png Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L03-Fig3.4.png License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:Circuit3.PNG Source: http://en.wikiversity.org/w/index.php?title=File:Circuit3.PNG License: Public Domain Contributors: The IsiahImage:Crystal Clear app kblackbox.png Source: http://en.wikiversity.org/w/index.php?title=File:Crystal_Clear_app_kblackbox.png License: unknown Contributors: AVRS, CyberSkull,Überraschungsbilder, 1 anonymous editsImage:KVL.png Source: http://en.wikiversity.org/w/index.php?title=File:KVL.png License: GNU Free Documentation License Contributors: Glenn, It Is Me Here, Kwinkunks, Pieter KuiperImage:EE-102-L05-Fig2.jpg Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L05-Fig2.jpg License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:EE-102-L05-Fig3.jpg Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L05-Fig3.jpg License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:EE-102-L05-Fig4.jpg Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L05-Fig4.jpg License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:KCL.png Source: http://en.wikiversity.org/w/index.php?title=File:KCL.png License: GNU Free Documentation License Contributors: Glenn, Pieter Kuiper, SteveZodiacImage:EE-102-L06-Fig3.jpg Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L06-Fig3.jpg License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:EE-102-L07-Fig1a.jpg Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L07-Fig1a.jpg License: GNU Free Documentation License Contributors: User:Thuvack,User:Thuvack/Mind-LogsImage:EE-102-L07-Fig2.jpg Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L07-Fig2.jpg License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:EE-102-L08-Fig3-Final.jpg Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L08-Fig3-Final.jpg License: Creative Commons Attribution-Sharealike 3.0 Contributors:ThuvackImage:EE-102-L08-Fig1.jpg Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L08-Fig1.jpg License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:EE-102-L08-Fig2.jpg Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L08-Fig2.jpg License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:EE-102-L09-Fig1.jpg Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L09-Fig1.jpg License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:EE-102-L09-Fig2.jpg Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L09-Fig2.jpg License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:EE-102-L09-Fig3.jpg Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L09-Fig3.jpg License: Creative Commons Attribution-Sharealike 3.0 Contributors: ThuvackImage:EE-102-L09-Fig4.jpg Source: http://en.wikiversity.org/w/index.php?title=File:EE-102-L09-Fig4.jpg License: Creative Commons Attribution-Sharealike 3.0 Contributors: Thuvack

License 51

LicenseCreative Commons Attribution-Share Alike 3.0 Unportedhttp:/ / creativecommons. org/ licenses/ by-sa/ 3. 0/

http://creativecommons.org/licenses/by-sa/3.0/

electric circuit analysis.pdf

Documents

Transcript of electric circuit analysis.pdf