Chapter Seven circuit theorems. linearity. linearity is the property of an element describing a linear relationship between the cause and the effect. A linear circuit is one whose output is linearly related or directly proportional to its input. One of the properties of an electric circuit is that it is homogeneous. We sometimes refer to it as the homeowner genetic D or scaling property of an electric circuit.

For example, we can take a circuit described by ohms law V equals IR, and we can scale it by multiplying it by any scaling number. In this case, we just call it K. We can move Supply key times the voltage. And if we long as we multiply the right hand side of the equation by K, it is said to be homogeneous or we can scale it. There's also the additive property. And we'll use this as we go through some of our complex calculations and other lessons. And that is if we have two values for a circuit with one voltage is equal to one current times a resistor, and another voltage is equal to another current types of resistor, we can add those two things together as long as we add the left side of the equation and the right side of the equation.

The property still holds true. In other words, v one plus v two is equal to i, one r times r plus i two times R, or you can collect like terms On the right hand side which is equal to the quantity of I one plus i two times R. I am going to look at power sources now and more specifically voltage sources in current sources. First of all, let's have a look at voltage sources. And we've already used voltage sources in in some of our previous work in the way of a battery and we all assume that the output of a battery is constant. And those batteries you can buy them off the shelf or double A's or triple A's or 1.5 volts output. You can buy a nine volt battery or you can get a 12 volt battery for a car or various assortments of little disc batteries from 1.5 all the way up to 32 volts.

The constant thing or the common thing about a voltage source is that the voltage output is assumed to be constant. And indeed, most of the time it is. So when we add a load to this voltage source, we can use ohms law to determine the current that's going to flow to the load, because we know the voltage and it's constant. And depending on the resistor we put in there, if it's a particular value, we can divide the voltage by the resistance to find out what the current is. And the voltage as I said at the risk of repeating will remain constant. Now if we start to increase the load by reducing the resistor, in other words, current then we'll start to go up because the voltage doesn't change.

If the resistor becomes very small The current will become very large. And that's given to us by ohms law. And in fact, if you go to the extreme, where you have zero resistance or a short circuit, then ultimately ohms law tells us that the current will go to infinity. However, I think we've all experienced this because one time or another, we might have shorted a battery out. And certainly, it might spark an arc but it the current doesn't go to infinity. So what is the limiting factor when you start to get a short circuit?

Well, as you start to reduce the resistance in the load, then the actual resistance of the wire becomes significant. And that will indeed be the limiting factor for your current. So when you short out a battery, the wire to the battery, the resistance, even though it's very small, becomes significant. And, indeed, voltage sources have what they call internal resistance that has to be taken into consideration. Now, you don't have to do this all the time, because 99% of the time, a battery is not short circuited, so it is safe to assume that battery is a constant voltage output. And it's not just a battery, because we have all various assortment of DC power sources or voltage sources.

If you go in a computer or a cell phone, you have to plug those in for charging. That is a DC power supply. And instead of putting a battery in our circuit, we represent a DC power source or a voltage source in this manner, a circle with a polarity attached to it. So it tells us where the positive is in a circuit. So we'll know which direction the currents going to flow, but we know that the voltage will remain constant. Now, a current source is a relatively rare animal, there are some out there and they are there are calculations that you have to make considering a current source.

So we are going to look at it very quickly now, but most of the sources of energy in a DC Circuit are voltage sources. However, there are current sources out there and it's represented by something like this circle with an arrow depicting the direction of the current flow. And the takeaway from this particular device is that it produces a constant current, regardless of the load, and we've shown the circuit has closed here and in fact it is actually short circuited. However, it's not a problem. Because the current source maintains a constant current, regardless of the load. In this type of circuit, now the current will be constant.

So the thing that's going to change as we change the resistance in a load is the voltage drop across the load, because we will use ohms law again for calculating the voltage drop across the resistor. So if we start to increase the resistance of the load, then the voltage is going to go up as well. As we can see here, till ultimately you would open circuit or if you did open circuit, the resistance would go to infinity. And indeed, you would have a problem because then the voltage is going to go to infinity, which of course it never does, because it's going to be limited by several factors including the breakdown of air, but it could be very dangerous to open circuit a constant current source. However, just like the voltage sources 99 times out of 100 year who are going to be using closed circuits on a constant current source.

So you just have to be aware of the fact that a current source will maintain a constant current when you're doing your calculations. So as we move through analyzing circuits, you'll run into two types of sources one's a voltage source and the other is a current source and they are or they are represented by circles. And in the case of the voltage source, you got a polarity plus and minus. And in the case of a current source, you have an arrow and you have to remember that the voltage is a constant voltage Put the current in a current source is a constant current output. This example will demonstrate the use of a current source as well as demonstrate what we mean by linearity. In this particular circuit, the current generator at one particular time is pumping out five amps that is is equal to five amps if you go through the calculation, you will find or we will find that I naught which is the current through the five ohm resistor is one amp.

Now, if we increase the source current to 15 amps, we've essentially scaled it up by three. So, we because of the linearity we can scale up our answer, which used to be one amp by a factor of three, so if we increase is by a factor of three and is 15 amps that I not by virtue of linearity will be three apps. The handy thing here for linearity is, we don't have to go through all of the calculation using ohms law and circuitry to come up with a new answer for AI not because we just have to scale it up by three. That brings us to the first of the theorems that we're going to look at that is the superposition theorem, which states that in a linear circuit with several sources, and by a linear circuit, we mean it has a constant current source and or a constant voltage source.

The current through and the voltage drop for any element in the circuit is the sum of the country. currents and voltages produced by each source acting in dependently. In other words, you can do a calculation of each source with the other sources removed, then remove that source and replace it with the second source, then remove that one, replace it with the third and do that for all the sources that you have in a circuit. Then you just arithmetically add up the results to come up with the total values. The first thing that pops into mind though, is how do you replace or remove sources? What does that look like?

Well, there's going to be two types of sources in our circuit, a voltage source and a current source. When we want to do the calculations for a voltage source and remove all the other voltage and current sources As the voltage sources that we are not going to leave in a circuit are replaced by a short in other words replaced by a wire. And in the case of current sources, if we are going to remove a current source and do our calculations with either a current or a voltage source a single one, then we have to replace all the other current sources by an open circuit. I think the best way to understand this is go through an example because it is pretty simple, and it's very intuitive. But let's look at our first example. Okay, the best way to have a look at the superposition theorem is demonstrating it with an example here we have a circuit made up of a three or a five ohm resistor and a toggle resistor.

We have a current source of eight amps and a voltage source of 20 volts. all connected as you see in the diagram here, we would like to calculate what the voltage drop across the two ohm resistor is. One of the ways we can do it is calculate what each source provides in the way of current through the two ohm resistor and then linearly add the two of them together and find out what the final effect is. So at the risk of being repetitive, here are the steps to apply the superposition principle. Turn off all independent sources except one and replace with an open or a short depending on if it is a current or voltage source. Step to solve the voltage or current Solve for voltage and current due to that one active source.

Repeat step one for each of the independent sources and then find the total contribution by adding algebraically all the contributions due to the independent sources. Okay, the first thing we do is replace the eight amp current generator and replace it with an open circuit of course, and that leaves us with a 20 volt power supply feeding a five three and a two ohm resistor in series. The next step will be to replace that voltage generator with a short circuit and that will leave a current generator feeding to parallel circuits of five ohms to the right and three plus a tool to the left. Going back to our first slide Get, we can calculate the current flowing in that circuit by simply using ohms law which is, is equal to V over R and V being 20 volts, and the ARB being a sum of five, three and two ohms, which is 10 ohms, which would leave us with two amps.

So two amps flowing through a two ohm resistor would give us a voltage drop of four volts. In the bottom circuit, the current generator is split between current going to the left and current going to the right. And in these circuits, it just so happens that to the right is five ohms. And to the left is three plus two which is also five ohms. So you get an equal split of the current half going one direction, half going the other and gives us four ads Going through the two ohm resistor, and that would provide a voltage drop of eight volts. So we then arithmetically add the two answers that were gained independently four volts plus eight volts to give us 12 volts.

In this example, we have a six volt voltage source and a three amp current source feeding an eight Ohm four ohm resistor. And connected as you see here, the first thing we do is we can replace a three amp generator with a with an open circuit that leaves the six volt power supply feeding eight ohms forums in series, which gives us by Holmes law. a current of point five amps which point five amps flowing through the four ohm resistor will give us a voltage drop for v1 to volts. Next we have the current source is feeding the Edelman the forum in parallel. What you would do to calculate the current is you want to know what the voltage drop across each resistor is and the voltage drop across each resistor is the voltage drop across eight ohms and four ohms in parallel, so you find the equivalent resistance value for the eight Ohm forum in parallel and calculate the voltage.

Then you'll use the voltage drop across each resistor to calculate the current split through that resistor and ultimately come up with the current flow through the two which would be two amps. The current asset essentially splits in one third two thirds that's what you'll find out, but the current through the forum resistor is two amps, and that provides a voltage drop of eight volts. And the answer then would be simply adding v one and v two together, or two plus eight, which would give us 10 volts. The next theorem that we want to look at is called Thevenin theorem. And this is a very useful a theorem not just for simple DC circuits. It is used in AC circuits and especially when you get into three phase fault calculations of asymmetrical short circuits in a three phase system.

However, we're not going to go there today we're only going to talk about the Thevenin theorem as it applies to DC circuits. And what the theorem says that a linear terminal circuit can be replaced by an equivalent circuit consisting of a voltage source, V subscript th, which stands for the Thevenin and voltage in series with a resistor, r subscript th for the Sabin equivalent of the resistance. So that square box could have been made up of several generators of different types and different loads. But what we're what davidon says is you can replace that circuit and figure out how it affects the load say in this diagram, by replacing everything that was inside the box by one voltage source and one series resistor. The trick is to find out how to calculate the Thevenin and our Damon well, V davidon is simply the open circuit voltage at those terminals.

And our thievin is the input or equivalent resistance at that at those terminals. So it's pretty simple, but we need to go through a few calculations to demonstrate how easily it works. Thevenin theorem is especially useful in analyzing power systems and other circuits where one particular resistor in the circuit called the load resistor is subject to change and recalculation of the circuit is necessary with each trial value of load resistance to determine the voltage across it and the current through it. Let's take another look at an example. Let's suppose that we decide to designate our two as the load resistor in this circuit Davidson's theorem Makes it easy by temporarily removing the load resistance from the original circuit and reducing what's left to an equivalent circuit composed of a single voltage source and a series resistance. The load resistance can then be reconnected to this even an equivalent circuit and calculations carried out as if the whole work of whole network was nothing but a simple series circuit after the Thevenin conversion.

Calculating the equivalent Thevenin source voltage in series resistance is actually quite easy. First, the chosen load resistor is removed from the original circuit and replaced with a break or an open circuit. Next, the voltage between the two points where the load resistor used to be attached is determined. You can use whatever method that's at your disposal. In this case, you could pick The superposition theorem and calculate the current contributed from the 28 volt battery than the current contributed by the seven volt battery arithmetically, add the two, and you would end up with as our table indicates a circuit value for the current of 4.2 amps, there is only one path for the current to flow, and it's 4.2 amps in all the components in the circuit. So that would give us a voltage drop across r1 of 16.8 voltage drop across our three of 4.2.

So we have the voltage drops as well as the voltage as indicated by the power supplies so we can arithmetically calculate what the voltage between the two points where our load resistor was connected and that works out too 11.2 volts, the voltage between the two load connection points is now our davidon equivalent voltage. Now to find the Thevenin series resistance for our equivalent circuit we need to take the original circuit with the load resistor still removed. Remove the power sources in the same style as we did with a superposition theorem voltage sources replaced with wires and current sources replaced with open circuits or brakes and figure the resistance from one load terminal to the other. With the removal of the two batteries, the total resistance measured at this location is equal to r1 and our three in parallel or point eight ohms This is our Thevenin resistance for the equivalent circuit.

With the load resistor two ohms attached between the connection points, we can determine the voltage across it and the current through it as though the whole network were nothing more than a simple series circuit. So we can fill out our table here right now we got part of the table filled out because we can put in the total voltage, which is that they've been an equivalent 11.2. And we know that they've been in impedance our system Thevenin resistance is point eight ohms and our load resistance is two ohms. So we can calculate the series resistance just by adding the two of them together and come up with 2.8 ohms. And use ohms law to calculate where what the current would be flowing in the circuit and that is four amps. With four amps flowing through our load, we can then calculate the load voltage drop, which is eight volts is what we were trying to do in the first place.

And of course, we could also calculate the voltage drop across that even in resistance if we needed that 3.2 volts. So we can fill out our whole table now with using the oven in serum. And if we wanted to replace the load resistor with a tomb for old age old panel, we can do it very quickly and the calculations are very easy and very quick. Norton's theorem is similar to saving in theorem with the difference that Norton's theorem is dealing with current sources you want to end up with a current source and a parallel resistor. And in the case of feminine It was a voltage source in series with a resistor Norton serum state That is possible to simplify any linear circuit no matter how complex to an equivalent circuit with just a single current source and a parallel resistor connected across the load. Norton's theorem states that a linear two terminal circuit can be replaced by an equivalent circuit consisting of a current source I subscript Norton which will be the Norton current source in parallel with a resistor R subscript Norton, which is the Norton resistance are Norton is the input or equivalent resistance resistance of at the terminals and I Norton is the current flowing if those terminals were shorted.

I Norton would be the current flowing through that short And we're going to use the same example that we started out with with the Thevenin equivalent. This time we're going to calculate the Norton theorem equivalent for this circuit. Again, we're choosing our two the two ohm load resistance as our load. So we're going to now replace everything else in the circuit with the Norton equivalent, which is a current source in parallel with its internal internal resistance are Norton. Remember, that a current source is a component whose job it is to provide a constant amount of current out putting as much or as little energy necessary to maintain a constant current flow. That means the voltage may rise and and lower but the current will remain constant.

As with David Allen's theorem, everything in the original circuit except the load resistance has been reduced to an equivalent circuit that is simpler to analyze. Also similar to Stephens theorem are the steps used in Norton's theorem. To calculate the Norton source current i Norton and the Norton resistance are Norton. As before, the first step is to identify the load resistance and remove it from the original circuit. Then to find the Norton current. For the current source of the Norton equivalent circuit, place a direct wire or a short connection between the load points and determine the result and current.

Note that this step is exactly opposite. The respective step in Thevenin theorem where we replace load resistor with a breaker or an open circuit with zero voltage drop between the load resistor connection points the current through r1 is strictly a function of B one's voltage and RS resistance seven amps and AI is equal to V over R. And likewise the current through R three is no strictly a function of B twos voltage and our threes resistance seven amps and we can use again those law i equals e all over R. The total current through the short between the load connection points is the sum of these two currents seven amps plus seven amps is equal to 14 amps therefore, the Norton source current i Norton in our equivalent circuit is 14 amps. To calculate the Norton resistance, we do exactly the same thing as we did for calculating the Thevenin resistance.

Take the original circuit with the load resistors still removed, remove the power sources in the same style as we did with a superposition theorem voltages are voltage sources are replaced with shorts and current sources are replaced with brakes and figure out the total resistance from one load connection point to the other. And our Norton is point eight ohms. Now our Norton equivalent circuit looks like this. If we reconnect our original load resistance of two ohms we can now analyze the Norton circuit as a simple parallel arrangement. We know that the reason distances in this circuit are point eight ohms for the Norton equivalent and two ohms. For our load resistance, we know that the current flowing through the Norton current generator is 14 amps.

We'd like to know what the voltage is across our parallel loads. In order to do that, we have to find the equivalent parallel impedance sort of the parallel resistance in this circuit. So it is our Norton in parallel with two ohms, which works out 2.57143 ohms or you can call it 571.43 million ohms. Once we have that, resistance calculated we can then calculate our voltage drop because we know the current flowing through that resistance and that will give us a voltage drop of eight volts across the load resistor and the Norton resistor. Once we have the voltage drop we can calculate using ohms law what the current is in the load resistor, which is four amps. And with we want to calculate the current through the Mr. Norton, it would be 10 amps.

So, this makes calculation of the circuit very simple. The last thing we want to talk about in this chapter is source transformation and by source transformation when you mean the transformation of a power source from a voltage source to a current source or from a current source to a voltage source. Sometimes this becomes a matter of convenience or a matter of simplifying further calculations downstream. Regardless of why we want to do it, we can do it and we can do it on paper, as long as Know how to convert one to the other. So let's start with a voltage source transformation, a voltage source would look like this voltage source being the volts. So, we know what the voltage would be, and it would have an internal resistance of R subscript p again a known quantity, what we want to do is we want to have a current source now, that would be equivalent to the voltage source if we were to use it in a circuit.

So, we would replace the voltage source with a current source in parallel with a new internal current resistor. Now, they will be happy to determine is what that AI is for the current source and what the R subscript is for the internal resistance of the current source. And we do this using Norton serum and Norton serum states that if you want to find out what the equivalent Resistance is in this case, we want to find out what r subscript is, we remove the voltage source and replace it with a short. So we would then measure the the impedance or sorry, the resistance of the circuit, and that would be simply RV. So are the would equal our I. We're partway there now in determining what the current is we again use Norton's theorem and to find out what the current is in a Norton equivalent circuit is shorting out the terminals and measuring the current that would flow in this case, it would be simply determined by ohms law of the amount of current being drawn by RV when the terminals are shorted when driven by the voltage V. So the current of our current source would be the all over subscript or R subscript P. Now We have a way of determining what i n r is an R in our current source.

So we now have a way of converting a voltage source into a current source. If we want to convert a current source into a voltage source, we take the same steps only this time we're going to be using the Thevenin equivalent circuit. Our current source would look like this, of course, and we want to end up with a voltage source that would look like this. So we have to be able to calculate what our V and V is in terms of I and R subscript is. So we start out by using Thevenin equivalent and they say replace the sources with their equivalent shorts or, or opens and in this case, it's a current source. So we replace it with an open and measure the resistance and the resistance is r subscript i.

So no big surprise here. R subscript is going to be equal to r subscript v. So now we want to term and determine what V is. And it's simply what would be the voltage across the terminals, if we let the, if the we let the source, in this case source current run through the internal resistance, and in this case, it is simply again ohms law, so the V would be i times r subscript i. So, we have a way of converting a current source to a voltage source and these are the formulas that you'd use. As you may know or have guessed by now, there's more than one way to analyze an electric circuit. We are going to look at one now that is very powerful, but a very simple method of analyzing an electric circuit It uses ohms law, catch offs, Voltage Law and the superposition theorem all at once to convert what is an electrical circuit to a mathematical equation that if you follow simple rules, you come to the actual answer.

I'm going to go through the theory first very quickly, and then I'll analyze some circuits. So you'll get a feel for this mesh current or loop analysis. mesh current, or loop analysis is by virtue of its name, analyzes a circuit that looks essentially like a mesh. And a mesh has several loops in it. In this case, this particular example has three loops in it. And we're going to call them loop number one, loop number two and loop number three It goes without saying that there is going to be current flowing in this circuit because of the voltage supplies or the batteries that are on the left hand side of the loops.

And if we consider current flowing in each of the loops, I'm going to make an assumption right now that the current flowing in loop one looks like this, the current flowing in loop two looks like this, and the current flowing in loop three looks like this. Now the beauty of doing mesh current or loop analysis is once we've made the assumption of currents flowing in the direction the currents are flowing in, we very quickly come to a set of mathematical equations which we can analyze and come up with the values for the currents. As I said, the beauty of this is if we assume the currents are flowing in the wrong direction, then the answer we get for those currents will come out negative. And we'll see that through the mathematical analysis. So I'm going to assume my current is flowing in a clockwise direction and each one of the loops and it you can make the assumption that they're flowing in other directions or opposing directions if you want.

It doesn't matter because the mathematics will look after it. For now. I'm assuming I one is flowing clockwise and I two is flowing clockwise and three is flowing clockwise. The part that is the superposition theorem is that we can analyze each one of the currents independently of the other. That's what the superposition theorem tells us. So we will look at each one of the currents independently and Do our analysis.

If current flowing in I one looks like that, then we're going to get a voltage drop across the bottom resistor there as I've indicated plus to minus and the resistor in the top of the three meshes. Because of the current AI two flowing in that direction we'll set up a voltage drop from positive to negative as indicated in the diagram. And in loop three, the outermost resistor because of the current i three will have a voltage drop as indicated plus two minus in the diagram. Now looking at the common resistors, the common resistor between I one and I two, the current flowing in that resistor will be I one minus i two, because I two is flowing in the opposite direction of AI one. And again, it doesn't matter whether AI one is greater than or less than AI two, because if it turns out that AI one is smaller than I two, that our answer will come up such that I one minus i two is negative and I'll know the direction is wrong.

But for now, because we're assuming I one flowing clockwise and I two is flowing clockwise, the direction of the current i one minus i two is left to right. And the current flowing in the common resistor between I two and i three is similarly I to minus i three, and they read the current flowing in the common resistor between I one and i three is going to be I will One minus i three. Once we have assumed the loop currents I one I two and i three, we then write the kerchief Voltage Law equations for each of the loops. Now remember current shops voltage law states that the voltage drops around a circuit must add to zero. So, we will go around each loop using the assumed current flows of I one, I two and i three and write down the voltage drops for that particular circuit using that particular current.

Remember now we're going to analyze each current independently and then add them together. That is what the superposition theorem dictates. So, the rules for voltage drops are the voltage drops are positive and across each resistor and are given by ohms law. So, the voltage drops on the outermost resistors will have a positive to negative voltage drops as indicated because the currents flowing through them are in the direction as indicated in our diagram. When we come to the power supplies which are voltage supplies, the voltage will be rising and voltage rises are negative and are equal to the voltage source ratings. In other words, if they're 1.5 volts, they will be 1.5 volt volt drives if they are nine volts it'll be nine volt voltage rise or 12 volts or whatever the battery voltage is, if the batteries are in the reverse direction to the current, such as that the voltage is no longer a voltage rise but it will be a voltage drop from positive to negative and therefore, the voltage drop will be positive.

However, in our case, the battery is in that direction. So, it will be considered a voltage rise and it will be negative. Now, the voltage drops across the common resistors that is the resistors with two currents flowing in them are also given by ohms law the equals times are and are positive, the polarity is according to the loop current being followed. In other words, if we're following I one, the polarity will be positive to negative left to right. If we're following AI to the common resistor we'll have voltage drop positive to negative right to left. The current in the ohms law equation is the arithmetic sum of the two currents flowing through it.

The Loop current being followed is positive, the other current is positive or negative depending on its direction of flow with respect to the loop current being followed. Now that's a mouthful. What that means in our diagram is that you will see that in a common resistor between I one and I two, that I two opposes I one because they're both flowing in each of the respective loops in a clockwise direction. So, I two flows in the opposite direction of I one in the common resistor. If one or the other current was flowing in the other direction, then it would be positive, not negative. Let's have a closer look at that in our example.

And we can see that if we are following current i one in the current loop the resistor in the bottom of that loop is pretty simple we see that the current flowing through it will set up a a positive voltage drop from right to left. The voltage drop across the common resistor is also going to be positive to negative because we are following I one and it is also going to be given by ohms law. However, the current in our ohms law is going to be made up of AI one which is positive and because the current the other current flowing in that resistor is due to eye to eye two is flowing in the opposite direction to AI one, so that the current is given by I one minus i To now remember we have arbitrarily assumed that the current flowing in the direction in each of those two loops are clockwise.

If, when we're finished doing our analysis, we have chosen the wrong direction for the current, the current values will be negative and we'll show you that in an example. However, in the logic that we're setting out here, the current flowing in the common resistor between I one and I two, because we are following the current in AI one, the current in the common resistor will be I one minus i two and voltage drop is considered positive. So the last section In that paragraph the loop current being followed is positive, the other current is positive or negative depending on the direction of its flow. If I two was flowing in the other direction, it would be positive because the current flowing in the common resistor would be made up of I one plus itu, but they're flowing in opposite directions. So it's going to be I one minus i two.

So let's apply some values to our example here and work through the mesh analysis for this circuit. Now, remember, we have assumed three circuits, or three currents flowing in each of the three loops. So ultimately, in analyzing this circuit, we want to know what the three currents are I one, I two and i three will Once we know what these loop currents are, we can calculate everything else that's associated with this circuit. In other words, we can know at all the currents flowing in all of the branches of the circuit, and all of the voltage drops across cross each one of the resistors is going to be. So we're going to, as I said earlier, convert this electrical circuit to a set of mathematical equations that will allow us to solve for the unknowns which are the three currents in this case. So we're going to set up three equations in three unknowns.

And once we've got three equations and three unknowns, we will have the capability of analyzing for the unknowns and how we come across these equations by writing Kershaw's Voltage Law around each loop for each one of the currents. Starting with eye one, eye one will be flowing in loop one, and it's going to set up voltage drops across the resistors as indicated in the diagram. Now remember, we're falling back on the superposition theorem which allows us to analyze each current separately. So now I'm looking at the current shops voltage law or equation for current number one. The current that's flowing in the common resistor, the five ohm resistor is going to be made up of I one and I two, and because we are following I one, it's going to be I one minus i two and the voltage drop Again, according to following the loop of I, one is from left to right positive to negative in the common resistor, the forum common resistor between a loop one and three, the current flowing in that resistor is going to be made up of I one and i three, but because we're following the loop or the current source voltage law around the loop for I one, the current or the voltage drop across the four ohm resistor is going to be due to the current flowing I one minus i three.

I'm now going to set up the equation for Current shops Voltage Law in loop one, according to the current i one, and we will start we can start anywhere in the loop. But the first thing I'm going to start with is the power supply the 18 volt power supply and remember our rules, it is a voltage dries so it will be minus, so we have minus 18. We will then come to the common resistor, the five or the five ohm resistor. And remember, the voltage drop across that resistor is going to be V is equal to i times R. The eye that we're looking at is I one minus i two, so it's going to be five ohms times I one minus i two is the voltage drop across the five Ohm common resistor. The voltage drop across the forum common resistor is going Could be again given by ohms law.

And it is going to be the current in the forum resistor times the resistor, which is four times the quantity I one minus i three. And the voltage drop across the one ohm resistor is only due to the current i one. And it again is going to be given by ohms law, but it is only due to one current. So it's one Ohm times I one. And of course, kerchief states that the voltage drops around that loop must add to zero. So that has given us our first equation.

Now we can manipulate this equation now we're dealing with mathematical terms, we can almost forget the fact that we're dealing And electrical values, because we just have numbers and unknown quantities. And the there are three unknown quantities in that equation I one, I two and three. So I want to arrange the values in that equation. So I'm going to expand the brackets, and the equation will look like this minus 18 plus five times I one minus five times I two, plus four times I one minus four times i three plus one times AI one is equal to zero. Now I want to collect the like terms in terms of our unknown starting with AI one and going through I two i three. We will set up an equation that has 10 I one minus five.

I two minus four, i three is equal to eight teen, I've taken the the definite value of the power supply which was minus 18 on the left hand side of the equation just moved it to the right hand side of the equation. And in doing so the sign changes of course, so we have three unknowns in this equation. We are now going to look at the current i two in loop number two, and the voltage drops across the components of that loop are indicated in the diagram. Now remember, according to the superposition theorem, we can look at only I two in terms of how it is going to react to the circuit. And then once we're finished with it Two and Three, we can add them together superposition wise and come up with the answer. But right now we want to look at only I two, and I too will set up voltage drops in that loop.

As you see in the circuit here. The voltage drop in the common resistor the five Ohm common resistor with loop one, the current in that resistor is going to be made up of I two of course flowing through it, but we're going to have to subtract off I one and I one is flowing in the opposite direction to AI two. So the current according to the I two value is going to be I to minus i one because we are evaluating The itu circuit. So the current flowing according to itu in the common resistor, five Ohm common resistor is given by I to minus i one. Now looking at the three old common resistor, that current flowing in it is going to be made up of I two, but we have to subtract off the contribution from current i three. So the current flowing in the three ohm resistor is going to be I two, minus pi three.

Now, looking at the mathematical equation for AI to the same way as we did for AI one, we're going to start with With a power supply and it's a voltage rise so it's going to be minus 12 and the current flowing in the two ohm resistor is only going to be I two. So, the voltage drop on a two ohm resistor is two times i to the voltage drop in the three ohm resistor is going to be due to I to minus i three. So the voltage drop is three times the quantity I to minus three and the voltage drop across the five ohm resistor common resistor with I two and I one is going to be made up of i to the current that is going to be made up of I to minus i one and the voltage drop is five times I to minus i want and of course current shops voltage law states that it has to equal zero will not Expand the quantities outside the brackets or get remove the brackets and we get the equation minus 12 plus two times I two plus three times I two, minus three times three plus five times I to minus five times I one.

And of course, purchase voltage law states that the sum of those around the loop have to equal zero. We will now collect like terms of I one, I two and i three and that will leave us with the equation minus five I one plus 10 I two minus three i three equals 12 volts. We are now going to look at loop three and more specifically the current item Three flowing in that loop and the voltage drops around the loop in loop three are only due to the voltage drops of the resistors there is no power supply in that loop. So, looking at the common resistor, the three own common resistor between I two and i three, the voltage drop across it will be i three times that resistance but we have to subtract off the contribution of I two. So, the actual current flowing in the three ohm resistor is made up of i three minus i two and looking at the common resistor the forum common resistor between i three and I one the current flowing in it will be due to three minus i one because I want us flowing against i three.

So the current flowing in the forum resistor will be i three minus i one. So the current jobs voltage ally equation going around the loop starting with the three ohm resistor we see that the voltage drop across the three ohm resistor will be three times the current i three minus i two. The voltage drop across the two ohm resistor will be just two ohms times i three and the voltage drop across the four ohm resistor will be four times the quantity i three minus one and that will all equals zero according to current shops. Voltage Law. Now we can expand The brackets again as we did before, and we will be left with three, three minus three is two plus two is three, plus four is three minus four is one, all equaling zero. collecting like terms, we will end up with minus four, I one minus three is two, plus nine is three is equal to zero.

So we have three equations, and we've set them up in a matrix form. If you don't recognize it, it is a matrix form, where the unknowns are from left to right I one I two and i three, and we can solve for that are those equations to get our unknown quantities of current. As I said, we have now three equations in three unknowns. So we can solve for the three currents that are in those three equations. And you can use any method you want for solving it. And I'm not going to go through the mechanics of actually solving it in factors.

There's apps on the internet that you can download, or you can actually do it online. We have a matrix C of three unknowns, and you can just plug in the values in that matrix C to come up with the solutions or you can go through the process of subtracting and adding the the equations themselves until you end up with a solution. Regardless of how you find the solution. You will come out with the fact that I one is 7.02 amps. And that is the current in the one ohm resistor in the loop one, I two is six point to eight amps, the current in the two ohm resistor in Luke two, and I three is 5.21 amps, which is the current in a two amp reset or two ohm resistor in loop three. And looking at these solutions, we now configure out just about anything that has to do with this circuit.

But we'll take it a little bit further in the analysis process. Because of the solution, we see that I one is greater than I two and I two is greater than i three. So if we want to find out the I one minus i two, we just subtract I two from my one and we get a positive answer which is 0.74 amps which we assumed was flowing left to right and because it's a positive answer, we have assumed the right direction for I one minus i too, and I two minus i three, which is the current flowing into three ohm resistor is I to minus i three or 6.28 amps minus 5.21 amps, which is 1.07 amps which is a positive number. So again we have chosen the right direction for the current i to minus i three. Finally, the current in the forum resistor is given by I one minus i three and that again is a positive number because I one is greater than I two is greater than 93.

The answer is 1.81 amps, which is the current and the four ohm resistor and again it's a positive answer. So, we have again chosen the right direction. Now, if any one of those turned out to be negative any one of those equations, I one through one minus side three, if any one of them were a negative answer that would mean that the assumed direction of the current at any stage of our calculation was wrong and we just reverse the direction to come up with the right answer. Let's look at another example very quickly just to get a feel for mesh analysis. Now the mesh circuit that you see in front of you only has one power supply which is 100 volts with 12345 resistors that are connected to it. And we can start the process out as we did the last time by assuming currents that are flowing in all three of the meshes that are in this circuit.

I one will be the current through the power supply and through the five ohm and the 40 ohm resistor. I too will be The current through the 20 Ohm the 10 ohm and the five ohm resistor and i three will be the current through the 40 ohm, the 10 ohm and 15 ohm resistor. Once again, we're assuming that the current is flowing in a clockwise rotation. It may not be when we finish and we will see that from the from the analysis. However, it's a pretty good guess that that's the direction currents are going to flow, seeing how the power supply is connected the way it is. We will now use purchase Voltage Law to sum up the voltage drops around the loop.

In fact, we're going to do it for each loop but we're going to start with loop number one which is current number one, the KVL for AI one or the current shows voltage law for AI one start With the power supply of 100 volts, the voltage drop of the power supply is minus 100 because it is a voltage rise actually. So in our equation it is minus 100 and the voltage drop across the five ohm resistor is five times the current which is I one minus i two, because I two is going against I one, it has to be subtracted and the voltage drop across the 40 ohm resistor is 40 times I one minus i three and again, I three is opposing I one so it shows up as a minus sign when going around the loop number one, and all of those voltage drops have to add to zero.

Now, we just proceed with my mathematical manipulation. And the first step is to open up the brackets which will give us minus 100 plus five is one minus five is Two plus 41 minus 43 is all equal to zero, we can now collect like terms, which are 45. I one minus five to minus 43 is equal to 100, giving us our first equation. Looking at loop number two, and current i to the KVL for I two is going to be starting with the resistor, the 20 ohm resistor, which gives us a voltage drop of 20 times I two going around the circuit, we come to resistor, the 10 ohm resistor and the voltage drop across it is going to be given by 10 times I to minus i three and we finally come to The five ohm resistor again, but this time we're following it too.

So the current in the five ohm resistor is going to be I to minus i one because I one now opposes I two, and we're going to get the voltage drop by multiplying the resistance times that current, which will be five times the quantity to minus one all equal to zero. Now we will expand the brackets again this time we'll get 20 I two plus 10 is two minus 10 is three plus five is two, minus five by one all equal to zero. collecting like terms, we're left with the equation minus five I one plus 35 I two minus 10. i three is equal to zero, which gives us our second equation. Looking at the third loop of our mash More specifically i three, the KVL for AI three, starting out with the 10 ohm resistor. The voltage drop across the channel resistor is given by 10 times i three minus i two, because we're following i three s plus three, and I two opposes i three in the 10 ohm resistor.

So the voltage drop is 10 times three minus i two plus the voltage drop across the 15 ohm resistor which only has three flowing in it. So it's 15 times i three. And finally the 40 amp or 40 ohm resistor gives us a voltage drop of 40 times i three minus one because I one is flowing in the opposite direction died three and because we're following it three, it's AI three minus AI one all of those voltage drops add up to zero. Next step, open up the brackets which gives us 10. i three minus 10 is two plus 15 is three plus 40 is three minus 40 is equal to zero. collecting like terms, we're left with the equation, minus 40 is one minus 10 is two plus 65. i three all equaling zero. That leaves us with the matrix equations that you see there in highlighted in red.

And now we can solve those three equations for three unknowns and we can use whatever method we want to use. But after we do that solution, we will find that it generates I one equal to six amps. I two is equal to two amps and i three is equal to four amps. We now have all of the information we need to analyze this circuit. However we'll take it another step further and because the current flowing in the five ohm resistor was given by one minus i two and I, one is six amps and I two is two amps so I one minus i two is four amps. The current flowing in the 40 ohm resistor is given by I one minus three, which is six minus four, which gives us two amps which is the current flowing in the 40 old resistor.

And the current flowing in the 10 ohm resistor is given by i three minus i two, which is two amps because i three is four and I To is to i three minus i two or four minus two is equal to two. In the beginning, I said we could choose the direction of the current flows in each one of the meshes arbitrarily In other words, we can choose clockwise or counterclockwise direction we arbitrarily chose all of the currents and be flowing in the clockwise direction. However, you can choose whatever you like and let's work through the same example this time. I'm going to assume it to flowing in a counterclockwise direction leaving I one and i three flowing in clockwise direction. We have now look at the equations for kerkhof voltage block going around the loop, we get minus 100 for the power supply, and the current flowing into five ohm resistor is going to be I one plus i two Which, if we want to find the voltage drop is five times that quantity.

And the voltage drop across the 40 ohm resistor is still going to be 40 times the quantity I one minus i three because i three opposes the curve of I one, all of those equals zero. Expanding the equation gives us minus 100 plus five by one plus five by two plus 41 minus 43. All equal to zero, which generates the equation 45 I one plus five I to minus 40 i three all equal to 100. Looking at a loop to and specifically current i to the KVL going around that loop starting with a 20 ohm resistor the voltage drop is 20 times I too because the only current flowing in the 20th resistor is I two and the voltage drop across the five ohm resistor with respect to the current i two is going to be given by five times the current i two plus i one because I want an eye to are flowing in the same direction through the five ohm resistor.

Similarly, for the 10 ohm resistor, the voltage drop across it is going to be given by 10 times the quantity I two plus three, because again i three is flowing in the same direction as I two when it flows through the 10 ohm resistor. All of those voltage drops when added together have to add to zero. Opening up the brackets will give us 20 I two plus five I two plus five I one plus 10 I two plus 10 i three all equal to zero collecting like terms. We get plus five I one plus 35 I two plus 10 i three all equal to zero. And finally we'll look at the current flowing in loop three or AI three, and calculating or writing down the KVL for i three, we get 10 times the quantity i three plus i two because again I two and i three are flowing in the same direction through that resistor plus 15. i three because i three is the only current flowing through the 15 ohm resistor plus 40 times i three minus i one because I one is moving opposing i three, the voltage drop is given by 40 times i three minus i one, all of those voltage drops equal to zero.

Expanding the brackets give us 10 i three plus 10 Two plus 53 plus 43 minus 41, all equal to zero. and collecting like terms leaves us with the equation minus 40 is one plus 10 is two plus 65 is three, all equal to zero. I'm highlighting again our final equations in red and they are the matrix of three equations and three unknowns which you again can use whatever process you need to solve three equations and three unknowns. But what you will get is I one, again equal to six amps. I two though, will work out to minus two amps. This loop current then is flowing in the opposite direction.

We assumed it was flowing counterclockwise, but we can came up with a minus number. So we know that we assume the wrong direction, which is not a problem because now we know which direction that current is flowing in. It's still two amps. And I three works out to four amps. So, we have chosen the right direction for i three I one plus i two is four amps. I one minus i three is two amps.

And I three plus i two is two amps. So you can see it did not matter which direction we chose the current to flow in. We still arithmetically ended up with the right answers. As I said before, there is more than one way to analyze a circuit and by analyze we mean find the currents and the voltage drops in the circuit as a whole We just went through a very powerful and simplistic way of analyzing an electric circuit that is using mesh analysis or loop analysis. However, there is still a basic way of analyzing a circuit. And that is reducing all the passive elements to single components.

In other words, we have one power supply here, it'd be nice if we could make the equivalent circuit of all those resistors to one lump circuit, and then work backwards from there to find the individual currents in the various branches. And I'm going to do that right now. But the first thing you're going to run into when you're trying to reduce this resistor network is trying to come up with what are power parallel and what are series circuits and trying to reduce them. First thing you run into is that there is no clear duplication or perish. Have resistors that we can form a series circuit or a pair that could form a parallel circuit so that we could reduce it. We keep running into a branch flowing out through the middle of it.

So you might be at a quandary, how can we analyze this circuit? Well, there is one trick that I want to show you here. And it's something that you can use for more than just this one circuit analysis, but it will demonstrate the point. Let's give quantity to the various resistors here so that we can follow them through the analysis process. We're going to start with 100 volt power supply 100 volt, voltage source, feeding into resistors are a RB RC and r one and r two. One of the tricks that We can use is converting what we call a Delta arrangement to a y arrangement or a y arrangement to a Delta arrangement.

This will, at times, ease the process of analysis. And we can convert and there are proofs for this, I'm not going to go through the proof, it's the end result that we're more more interested in. If we are going to convert the circuit which is in a Delta format, but there are three points of contact A, B and C, then we can replace that delta arrangement of our AC RBC and our A B with the wire arrangement of our a RC and RB because if we looked at the circuit from the point of view of the three points A, B and C, the resistance would be the same So how do we do that, but just before proceeding and giving you the formulas, there is another name you might run across with these types of arrangements rather than delta why sometimes a Delta circuit is called a PI circuit, the, you can still see that there are common points A, B, and C in the pipe network.

So they're just arranged in, up and down left and right version rather than in a, in a Delta version. And the Y configuration can be a T configuration, which again, is left and right and up and down. Bottom line is we still have three points that we're looking at that circuit A, B, and C, and we're going to replace one with the other. We just have to come up with a formula for the conversion and the conversion formula would love Like this to convert from a delta to a Wye circuit, the values of our a, rb and RC are going to be given by in the case of our a, it's going to be that delta resistors are a B, B times r a c, all over the some of the three resistors. In the case of the R b, it's going to be r a b times R, BC all over the some other Delta resistors.

And in the case of the RC, it's our AC times r BC, all over the some of the resistors. In the case of converting from a y to a delta, the R A, B of the Delta is now going to be given by r A times RB plus r A times RC plus RB times RC all oversee the R BC is going to be given by the same numerator as our A B, this time divided by our a and our a C again is going to have the same numerator, but this time dividing by our B, these are the conversions that you could use for changing the configuration of a y to a Delta or in the previous instance a delta to a y. So let's have a look at how we do this in this practical example. This is the example or the circuit that we have started with and the internal resistance are in a T formation or a y formation.

Our a our be at RC This would more closely resemble a Wye configuration, which is a T configuration. But I've labeled the points A, B and C. And I'm going to replace that y configuration with its equivalent Delta configuration. The y configuration can be replaced with this delta configuration at points A, B and C. With resistors are a CR BC and r a b, we know what those values are, because they can be calculated from the Y configuration of resistors. And we can very quickly see that we now have at least two resistors in parallel R one and r a c. So we can replace those two resistors with one resistor, that resistor being the parallel equivalent of rec parallel to r1. Now in order to further reduce this configuration, I'm going to redraw it so it, it's a little bit easier to recognize the parallel and series circuits.

And this circuit can be redrawn to look like this. We now see that we have two further resistors in parallel R two and R BC. So they can be replaced by the parallel combination, which is our bc parallel to r two. And very quickly, we can also see that we now have two resistors in series so we can replace those two resistors with one resistor equivalent To the series connection of the two, and that resistor would be the quantity r AC in parallel with r one plus r bc in parallel with r two. And very easily, we can see that we now have two resistors again in parallel and they can be replaced with one resistor. And that resistor would have a value of r a b in parallel with a quantity that I've put in square brackets there and I colored the square brackets red so we can see what we're dealing with.

And it was the two resistors that were out there before in series. They are our a C in parallel with r one plus r bc in parallel with r two That, once calculated will give us the one resistor that will draw the same current from the hundred Ohm voltage power supply that all the network of resistors that we started with, and we can calculate that current now because we know what the voltage is and we know what the resistance is, and we can just use ohms law, which puts the V over R to give us the current and that is 100. All over that big long value that we have for the one resistor. Now that we have the current being drawn for the hundred from the hundred Ohm power supply, we can now start working backwards to calculate the individual currents that are flowing in the branch circuits and the voltage drops in the branch circuits.

We can now work backwards and get the currents in All of the circuits and the voltage drops across everyone at the resistors. This ends chapter seven