electric circuit analysis.pdf

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ELECTRIC CIRCUIT

ANALYSIS

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Passive sign convention

1 Simple Resistive Circuits

7 Resistors in Series

12 Resistors in Parallel

18 Circuit Analysis Quiz 1

23 Kirchhoff's Voltage Law

25 Kirchhoff's Current Law

30 Nodal analysis

35 Mesh Analysis

41 Circuit Analysis Quiz 2

47 References

Article Sources and Contributors

49 Image Sources, Licenses and Contributors

50 Article Licenses

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Passive sign convention

1 Passive sign convention

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Lessons in Electric Circuit Analysis

Lesson #1:

Passive sign convention← You are here Lesson #2: Simple Resistive Circuits Lesson #3: Resistors in Series Lesson #4: Resistors in Parallel Quiz Test:

Circuit Analysis Quiz 1 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

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Passive sign convention

3 Introduction

This is the first of eight lessons in Electric Circuit Analysis. This course is a pre-requisite course to most Level 2 courses in this school. As such it is imperative that a student gains insight into the 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 the exercise on your own, and only then you can compare your results with the solutions given on a linked Sub-page.

Lesson Preview

This Lesson is about Passive sign convention. This Lesson introduces a student to Circuit Components which will be encountered in Electric Circuit Analysis. The student/User is expected 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 and you will be helped to evaluate your self along the way.

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Part 1: Electric Circuit

An electric circuit is a connection of components (Voltage/Current sources, Resistors, Inductors and Capacitors) such that there is some power supplied and dissipated. This means that if you connect a resistor 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 power are called Active components. the following picture shows circuit symbols used to depict a Voltage Source and a Current Source. Notice that the components show a general orientation of where the direction of conventional current.

Figure 1.2 and 1.3: Passive components

Passive Components:

All components that Absorb or Dissipate electric power are called Passive components. the following picture shows circuit symbols used to depict a Resistor. Figure 1.2 is generally the preferred symbol of a resistor and will be used throughout this course.

Please note that capacitors and inductors are beyond the scope of this course as they introduce complex resistance where real and reactive power complexities come in.

Part 2: Passive Sign Convention

The 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 could think of it in terms of depth and height.

Something has an elevation or height only with respect to something else such as sea level. Likewise depth, 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 be negative. 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 adding potential 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 one resistor 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 a dissipator of power. As circuit connections become more intricate this basic idea becomes more blurred. In some cases there are more than one power supply at different circuit locations, such that simple addition of their power magnitudes is not possible. We need to know which direction power supply and consumption is. The next examples will illustrate this.

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Passive sign convention

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Part 3

Figure 1.4: Passive Sign Convention scenario 1 Here is what we can deduce from figure 1.4. Points A and B are physical end points of Resistor R. A is more positive than B thus electrical charge at point A is higher than the electrical charge at point B. This creates electric potential.

Explanation of Part 3

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

Thus for -V ; -I and +V ; +I cases The Electrical charge will lose electric potential by effectively moving from high electric potential to low electric potential.The resistor has effectively absorbed power from the electric charge to enable it to move to a low potential point. 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 by effectively moving from low electric potential to high electric potential. The resistor has effectively powered the electric charge to a high potential point. Hence, the resistor, a passive component, supplies power in this case.

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

The following examples are related to the lesson. The answers to the exercise questions are given 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 simple resistor with the following parameters.

,

Find and Determine if this resistor is supplying power or dissipating it. Solution:

.

Since power is positive this resistor is absorbing power.

Example 1.2

Figure 1.6: Example 1.2 Figure 1.6 shows a simple resistor with the following parameters:

,

Find and Determine if this resistor is supplying power or dissipating it. Solution:

.

Since power is negative this resistor is supplying power.

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Example 1.3

Figure 1.7: Example 1.3 Figure 1.7 shows a simple resistor with the following parameters.

,

Find and Determine if this Resistor is Supplying power or Dissipating it. Solution:

.

Since Power is Positive this Resistor is Dissipating power. Surprised?

Well let's look at figure 1.7 again. If Voltage is Given as -6V it means that despite the given sign convention of the resistor, Point A is More Positive Compared to point B. The Current is shown entering Point A but by the fact that current is -3A it means that the current is in fact leaving at point A. Thus The resistor is effectively Dissipating power. Refer to part 3 & 4 of this lesson.

Try the exercises.

Exercises

1. 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 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. Ozzimotosan -- 75% & Corrected 2. Doldham -- 75% & Corrected 3. …

4. …

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Passive sign convention

7

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

Simple Resistive Circuits

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Lesson Review: Lesson 1

The first Lesson was about Passive sign convention. The Lesson introduced Circuit Components which will be encountered in Electric Circuit Analysis.

• Active Components • Passive Components • Passive Sign Convention

• Guidelines for Passive sign Convention

Lesson Preview

This Lesson is about Simple resistive Circuits. The student/User is expected to understand 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

Lesson #3:

Resistors in Series

Lesson #4:

Resistors in Parallel

Quiz Test:

Lesson #5:

Kirchhoff's Voltage Law

Lesson #6:

Kirchhoff's Current Law

Lesson #7:

Nodal analysis

Lesson #8:

Mesh Analysis

Quiz Test:

Home

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Simple Resistive Circuits

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Part 1 Voltage Source

This is possibly the simplest circuit. The voltage source supplies a voltage to the circuit. When this voltage 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 has the same magnitude as the voltage drop across the resistor. We know that it is V. The resistor has a certain amount of Ohms depending on its rating. We now know R. With algebra I = V/R. So as long as you 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 variables so insert them into the equation and solve for P (Watts). With these two equations, 1.1 and 1.2, and a little bit of algebra you get:

Equation 2.3

Equation 2.4

Part 2

As an explanation the power running through is the voltage times the current. This is instantaneous power rather than power used over time. Power has to be supplied and consumed. In a perfect world without heat-loss both are equal. The source supplies the required power that is consumed in this case by the resistor.

Example 2.1

Figure 2.1: Voltage Source Given':

Find: I, the current in Amps. The power produced by the source. The power consumed by the resistor. Solution: Using the equations:

Remember the power supplied equals the power consumed.

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Part 3 Current Source

All that will happen here is that the givens will change. Rather than knowing what the voltage is across the 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 that constricts the amount of flow. Well, current is flow of charge. With the fluid the side of the smaller section being supplied with fluid will have a greater pressure than the out flowing side. The difference between 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 the resistor 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 power produced by the source. The power consumed by the resistor. Solution: Using the equations:

Of course, power consumed equals power 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. The power consumed by the resistor.

Solution: Using the equations:

Part 6 Example 2.3

Given':

Find: V, the voltage. The power produced by the source. The power consumed by the resistor.

Solution: Using the equations:

Part 7 Example 2.4

Given':

Find: V, the voltage. R, the resistance Solution: Using the equations:

Part 8 Example 2.5

Given':

Find: I, the current in Amps. R, the resistance. Solution: Using the equations:

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Simple Resistive Circuits

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Part 9 Example 2.6

Given':

Find: V, the voltage. I, the current.

Solution: Using the equations:

OR

Part 10: Exercise 2

1. 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 going through the resistor?

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

4. A large voltage supply has 10,000V. The company wants to know how big a resistance can be put on the voltage supply 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 sources and 5 voltage sources.

Completion list

1. Ozzimotosan -- 100% 2. Doldham -- 75% & Corrected 3. …

4. …

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

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Resistors in Series

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Resistors in Series

13 Lesson 2 : Review

What you need to remember from Simple resistive Circuits. If you ever feel lost, do not be shy to go 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 sign convention Lesson #2: Simple Resistive Circuits Lesson #3: Resistors in Series← You are here

Lesson #4: Resistors in Parallel Quiz Test: Circuit Analysis Quiz 1 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

Lesson 3: Preview

This Lesson is about Resistors in Series. The student/User is expected to understand the following at the end of the lesson.

• Total Series Resistance: ( )

The total Resistance of Resistors in series is the sum of all reistors in series.

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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 one resistor. Take two resistors in series in a circuit with a voltage supply.

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

Part 2

Equation 3.1

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

Equation 3.2

Where R equivalent is the sum of all N of the resistors in series. So it really doesn't matter how many resistors there are. If they are in series 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 what the voltage is between these millions of resistors in series. No need to panic though because it isn't too much harder.

Lets take the two resistor problem first. There is a voltage source with two resistors in series. We know that the overall voltage drop across the two resistors is the same as the voltage the source is supplying in our example world. So the voltage drop across one resistor would be a portion of the overall drop. What proportion would we use to figure out the answer? One resistor 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 the actual 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. Then you actually need to subtract 3V from 12V to get the actual voltage between the resistors.

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

Part 4

Equation 3.4

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

For more than two resistors in series it is just a matter of keeping track of which resistor is on which side and summing

appropriately. Equation 3.5

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

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Resistors in Series

15

Part 5

It becomes clear then that, two equal resistors will divide the source voltage into two equal voltages (half of the source's voltage is dropped across each resistor). If the ratio of the resistance values is 3 to 1, there will be 3/4 of the source voltage dropped 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 drop 1/3 of the source voltage across each resistor. If the three had 1-2-3

proportionality (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 Resistive Circuits. Once the equivalent resistance of all the resistors in a series is found, effectively making a simple circuit again, then the current can be found with:

Equation 3.6

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

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Part 7 More Examples

Figure 3.1: Series resistors

Figure 3.1 shows a Series resistive circuit with the following 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 has been divided between R1 and R2 by and respectively.

Part 8: Exercise 3

Here are some questions to test yourself with.

1. Given 2 Resistors: R1 = R2 = 5 and Supply Volatage is 20 Volts find The V1; V2 and Current 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 negligible internal resistances connected in series as follows: Vs1 = 3V ; Vs2 = 5V and Vs3 = 1.5V, Find V1; V2 & V3 dropped by individual Resistors.

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Resistors in Series

17 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. Ozzimotosan -- 100% 2. Doldham -- 100% & Corrected 3. Sonu rockin -- 100% and corrected 4. … 5. … 6. … 7. … 8. … 9. … 10. … 11. … 12. … 13. … 14. … 15. … 16. …

fr:Résistance et impédance/Résistance

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Resistors in Parallel

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Resistors in Parallel

19 Lesson 3 : Review

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

The total Resistance of Resistors in series is the sum of all resistors in series.

• Voltage Divider Equation 2.3:

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

Lesson #1: Passive sign convention Lesson #2: Simple Resistive Circuits Lesson #3: Resistors in Series Lesson #4: Resistors in Parallel← You are here Quiz Test: Circuit Analysis Quiz 1 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

Lesson 4: Preview

This Lesson is about Resistors in Parallel. The student/User is expected to understand the following at the end of the lesson.

• two resistors connected in Parallel:

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Part 1 Introduction

The best way to understand Parallel circuits is to start with the definition. A circuit is parallel to another circuit or several circuits if and only if it share common terminals. That is if both of the branches touch each other endpoints they are in parallel. 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 bottom parts of the circuit.

Figure 4.2: Components in parallel share a common nodes

The areas in yellow all are connected together, as well as the areas in blue. So all the branches have the same terminals, which means that R1, R2, and the source are all in parallel.

If we take this discussion of the water flow analogy. Electric current can be seen as water and the conductors as water pipes. Something interesting happens as the current reaches the common node of Resistors that are connected in parallel, The total current is divided into the parallel branches.

Part 2 Voltage Rule

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

Equivalent resistance

For series resistors to find the total resistance we simply add them together. For parallel resistors 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 simplified form

Equation 4.2: Total Parallel Resistance

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

Current Rule

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

Equation 4.3: Current Divider Formula

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Resistors in Parallel

21

Part 3 Application

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

One simple application of these connection schemes is the Shunt application. In Electric Measurement industry, most often enough, we wish to measure Currents and Voltages of Very 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 usually have small Voltage and Ampere operating ratings.

Solution to the above is to have a metering device connected in parallel to a resistor, this resistor is thus called a "shunt" resistor since it is there to protect the metering device as shown 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 and what the total current is then we can work out the shunt current and thus the Shunt Resistor.

Part 5: Examples

Figure 4.4: Example 3.1 Figure 3.4 shows a Parallel resistive circuit with the following parameters.

; ;

; Find ;

Here are the solutions to the above problem:

.

. .

Thus it can be said that The Supply Current has been divided between R1 and R2 . We know that when solving these problems, we look at the Data given and thus we can see how we need to manipulate our equations in order to achieve our objective.The Following Example

Highlights this point, see that you can follow the Method used and the reasoning behind.

Part 6: Examples

Figure 4.5: Example 3.1 Figure 4.5 shows a Parallel resistive circuit with the following parameters.

; ; ;

Find: and .

Here are the solutions to the above problem: First Find: : . Then; . . .

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Part 7 Do you Remember?

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

• Voltage Divider :

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

• Two resistors connected in Parallel:

• Current Divider Principle:

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

Part 8: Exercise 4

Here 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 parallel branch 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 what conditions must be met?

Completion list

1. Ozzimotosan -- 100% 2. Doldham -- 75% & Corrected 3. … 4. … 5. … 6. … 7. … 8. … 9. … 10. … 11. …

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Resistors in Parallel

23 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) table

is 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.

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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 of

them 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 ohm

resistor 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 are

connected 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 are

connected 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 connected

in 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 total

resistance be? type="()" } + (A) R_{total} would approach Zero as The No. of Resistors In parallel Approaches

Infinity. - (B) R_{total} would approach 1 as The No. of Resistors In parallel Approaches Infinity - (C) It is not

possible to connect that Number of Resistors in parallel. - (D) None of the above. { What is the current through R1

and 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 in

Diagram 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) They

both have the same voltage as the source. - (B) They both have half the voltage of the source. - (C) One has full

voltage, the other has none. - (D) None of the above. </quiz> Take some time off, you've done well. If you're a

workaholic then you can go to the next page. Resistors in Parallelprevious lesson Resistors in Parallelprevious page

Kirchhoff%27s Voltage Lawnext page Passive sign conventionlesson intro Kirchhoff%27s Voltage Lawnext lesson

Electric Circuit Analysiscourse menu Resource type: this resource is a quiz.

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Kirchhoff's Voltage Law

25 Kirchhoff's Voltage Law

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Circuit Analysis Part II (Laws & Theorems)

This part is an introduction to some useful Electric Circuit Laws and theorems. You are

encouraged to master the theorems and laws that will be discussed herein as they form a basis upon which most Circuit analysis methods are built.

Lesson Review

What 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: Preview

This Lesson is about Kirchhoff's Voltage Law. The student/User is expected to understand the following 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 #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's Voltage Law← You are here

Lesson #6: Kirchhoff's Current Law Lesson #7: Nodal analysis Lesson #8: Mesh Analysis Quiz Test: Circuit Analysis Quiz 2 Home

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Kirchhoff's Voltage Law

27 Part 1: Kirchhoff's Voltage Law

Kirchhoff'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 one must 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, and Kirchhoff's Second Rule.

Part 2:Kirchhoff's

Voltage Law (Cont...)

Figure 5.1:

We observe five voltages in Figure 5.1: v4 across a voltage source, and the four

voltages v1, v2, v3 and v5 across the

resistors R1, R2, R3 and R5, respectively.

The voltage source and resistors R1, R2

and R3 comprise a closed circuit path,

thus the sum of the voltages v4, v1, v2

and v3 must be zero:

The resistor R5 is outside the closed path

in question, and thus plays no role in the calculation of Kirchhoff's Voltage Law for this path. (Note that alternate closed paths can be defined which include the resistor R5. In these cases, the voltage v5

across R5 must be considered in

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Part 3

Now, if we take the point d in the image as our reference point and arbitrarily set its voltage to zero, 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 v4

volts (as the symbol for the voltage source in the image indicates that point a is at a positive voltage 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 its negative terminal--clockwise around the circuit path. Thus from Ohm's Law, we observe that the 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 arrive back at point d, where our voltage is zero (just as we defined). So we experienced one increase in 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 voltage source and any number of resistors, the voltage drop across the resistors is equal to the voltage applied 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 following Parameters:

Find current through using Kirchhoff's Voltage 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 as shown in figure 5.3

From Loop 1 we get:

From Loop 2 we get:

A bit confused? well look at the explanation in Part 3 of this lesson and Review Passive sign convention.

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Kirchhoff's Voltage Law

29 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 the above equations and solve the variables.

If you feel lost up to this point do go back to the beginning of the example. Think of this as just another mathematical problem requiring solving by use of simultaneous equations with two unknowns!

Notice that we work with Variables only and try to solve the equation to its simplest form. Only after we have arrived at a simplified equation then that we can substitute in values of Resistors, Voltages and current. This can save you a lot of trouble because, if you 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 assumed direction. 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 the logic, otherwise just send a message to the course instructor as outlined in Part 8.

Part 7:

The method used to solve the above problem is very tedious, when the complexity of the circuit is increased the method becomes very cumbersome and almost impossible to use in solving circuit equations. This method is used just to illustrate KVL in this lesson. In Lesson 08 a more efficient method of solving these kinds of Circuit problems using KVL is introduced in a form of Mesh Analysis.

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).ISBN 020155707X

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Exercise 5

Figure 5.4: Exercise 5

Consider Figure 5.4 with the following Parameters:

Find current through using Kirchhoff's Voltage Law.

Completion list

1. … 2. … 3. … 4. … 5. … 6. … 7. …

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

Kirchhoff's Current Law

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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 go back to the previous lesson & go through it again. You can learn by repitition.

Lesson 6: Preview

This Lesson is about Kirchhoff's Current Law. The student/User is expected to understand the following 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 Current Law ( word-by-word ).

• Kirchhoff's Current Law:

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'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

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Part 1: Kirchhoff's Current Law

Kirchhoff'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, as a NODE. Note that a positive current leaving a point is considered to be a negative current entering that point.

Mathematically, Kirchhoff's Current Law is given by

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

Part 2:Kirchhoff's

Current Law

(Cont...)

Figure 6.1:

We observe four currents "entering" the junction depicted as the bold black dot in Figure 6.1. Of course, two currents are actually exiting the junction , but for the purposes of circuit analysis it is generally less restrictive to consider what are in actuality positive currents flowing out of a junction to be negative currents flowing into that junction (mathematically the same thing). Doing so allows us to write Kirchhoff's law for this example as:

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Kirchhoff's Current Law

33

Part 3

It may not be clear at this point why we insist on thinking of negative currents flowing into a junction instead of positive currents flowing out. But note that Figure 6.1 provides us with more information that we generally can expect to get when analysing circuits, namely the helpful arrows indicating the direction of current flow. If we don't have such assistance, we generally should not pass judgment on the direction of current flow (i.e., placing a negative sign before our 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, so we should take advantage of it. We know that currents i2 and i3 flow into the junction and the currents 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., no alternating current, no signal transmission). It can be extended to include time-dependent current 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 to be 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's Current 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 you like ). 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 all nodes. V_c will be found by applying KCL at Node c and solving resulting equations Follows:

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 simplified because of following Voltage definitions and current paths in Figure 6.3.

This method becomes tedious as the complexity of circuits is increased.

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Part 7:

Further Reading Links: • 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 Consider Figure 6.4 with the following Parameters:

Find current through using Kirchhoff's Current Law.

Part 8: Completion List

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

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Nodal analysis

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Lesson Review 5 & 6:

What you need to remember from Kirchhoff's Voltage & Current Law . If you ever feel lost, do not be shy 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 an end. Links to relevant Mathematical theories will be supplied to assist the student.

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's Voltage Law Lesson #6: Kirchhoff's Current Law Lesson #7: Nodal analysis← You are here

Lesson #8: Mesh Analysis Quiz Test: Circuit Analysis Quiz 2 Home

Laboratory: Circuit Analysis - Lab1

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Nodal analysis

37 Lesson 7: Preview

This Lesson is about Kirchhoff's Current Law. The student/User is expected to understand the following at the end 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 Material

The student is advised to read the following resources from the Mathematics department:

• College Algebra • Linear algebra

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

* Solutions/kramer [1]

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

Part 2: Nodal analysis

Let's start off with some useful definitions: • Node:

A point in a circuit where terminals of atleast two electric

components meet. This point can be on any wire, it is infinitely small and dimensionless.

• Major Node:

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

• 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 of the currents leaving.( From KCL in Lecture 6).

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Part 3

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

Once the Node voltages are solved, normal circuit analysis methods ( Ohm's law; Voltage and Current 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 analysis techniques that will 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 you can 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 the number of major node voltages increase and the size of the matrix exceeds 3 x 3, numerical methods ( Beyond scope of this course ) are employed with the aid of computers to solve such circuit networks.

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

Part 4 : Example

Find current through using Nodal Analysis method.

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 extra Resistors 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 you like ). thus as we did previously.

Follow Carefully the construction of the Nodal analysis algorith explained in part 3 of this lesson as follows in Part 5 and 6.

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Nodal analysis

39 Part 5 : Example (Continued)

Now that we have labelled the currents flowing in this circuit using passive sign convention, and have identified Nodes b; c and d as major nodes, we proceed as follows:

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 to change 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 their equivalent 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 of the general matrix as follows:

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Part 8:

Solving determinants of: • Matrix A : General matrix A

from KCL 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 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 flowing through 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 we did 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 at Part 11 for further reading material and interesting related External links.

Part 10:Exercise 7

Figure 7.3: Exercise Consider Figure 7.3 with the following Parameters:

Find current through using Nodal Analysis 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

1. … 2. … 3. …

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Nodal analysis

41 Mesh Analysis

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Lesson Review 5 & 6:

What you need to remember from Kirchhoff's Voltage & Current Law .

• Remember what was learned in Passive sign convention, You can go back and revise Lesson 1. • 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 an end. Links to relevant Mathematical theories will be supplied to assit the student.

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's Voltage Law Lesson #6: Kirchhoff's Current Law Lesson #7: Nodal analysis Lesson #8: Mesh Analysis← You are here

Quiz Test:

Home

Lesson Review 7:

What you need to remember from Nodal analysis. If you ever feel lost, do not be shy to go back to the 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.

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Mesh Analysis

43 Lesson 8: Preview

This Lesson is about Mesh Analysis. The student/User is expected to understand 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 Material

The student is advised to read the following resources from the Mathematics department:

• College Algebra • Linear algebra

Part 2: Mesh Analysis

Let'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 are used to create constraint equations.

• Mesh:

A loop passing though atleast one branch. Basic rule: The sum of Voltages arround any loop must be Zero.( From KVL in Lecture 5).

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Part 3

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

Once the Mesh Currents are solved, normal circuit analysis methods ( Ohm's law; Voltage and Current 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 analysis techniques 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 although you 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 the number of loops or meshes increase and the size of the matrix exceeds 3 x 3, numerical methods ( Beyond scope of this course ) are employed with the aid of computers to solve such circuit networks.

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

Part 4 : Example

Find current through using Mesh Analysis method.

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, 2 and 3 as shown in figure 8.2

We can now apply KVL arround the loops remembering Passive Convention when defining Currents and voltages.

KVL arround abca loop:

Therefore

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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 above equations 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 the General 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:

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Part 7: Exercise 8

Figure 8.3: Exercise Consider Figure 8.3 with the following Parameters:

Find current through using Mesh Analysis method.

Part 8:

Further Reading Links: • w:Mesh analysis

• Mesh-Currents Analysis Web Application [1] Refferences:

• Add refference here!

Completion List

1. … 2. … 3. …