Understanding Thevenin and Norton Circuits

Okay, so starting with the first question I left for you all to do on Monday. You have to find the 7-in-equivalent circuit for the circuit below. The first thing you would notice about this circuit is it has two supplies, meaning you can't perform any circuit analysis on it or you can't get the desired parameters using basic laws. It is, by definition, a complex circuit. It is... However, the simplest of complex circuits. Yeah. For your example. So what we need to do is find a VTH and RTH because we want to reduce this complex circuit to its 7N equivalent. Its 7N equivalent is an equivalent voltage source of a value VTH and a series resistance, equivalent resistance of a value RTH. Yeah. so according to the rules the first step of solving this problem of the reducing this circuit to its equivalent uh counterpart is removing the load resistor we have the load resistor here i labeled it rl it's definitely the load resistor because it's between the terminals of this particular circuit the output terminals yeah So we have to remove it before we go forward with the analysis. So when we remove that, we have the circuit on which we have to perform circuit analysis using one of the complex methods of analysis. Yeah. So what is VTH? Thevenin's theorem states that that voltage source as equivalent voltage source is the open circuit voltage. across the terminals. So after we remove this load, the voltage VAB is the voltage we need to find. And that voltage is the value of the Thevenin's voltage source. And then after we have to find RTH, the Thevenin's resistance. And the Thevenin's resistance is simply the equivalent resistance looking in from these terminals here. When I say looking, I mean, At these two points, what is the equivalent resistance, or between these two points, sorry, what is the equivalent resistance of the circuit? And to do that, we have to turn off the sources. Yeah? How do we turn off sources? We need to short-circuit voltage sources so that the terminals of the voltage source no longer give us a potential difference of 32 volts. It gives us a potential difference of zero. because a short circuit is the same point you are bringing these points together when you short circuit thereby applying a zero volt at this point instead of 32 volts yeah and to take off the current source you know usually this current source will send it two amps to this branch and to the rest of the circuit taking it off means sending zero amps to the rest of the circuit so open circuit you have to open circuit the current source so that it would send zero arms through this branch yeah so let's go forward and i want to do so is there no stepping through here okay i take off the animation all right so let's find the evidence equivalent circuit for the circuit below so we have the original circuit we want to find bth the first thing we need to identify is the 7 and equivalent voltage okay so the first thing we're doing is the uh we're removing the load resistor yeah and we need to find out what voltage the open circuit voltage is so the 7 and equivalent voltage is the voltage v a b the voltage across this two points these the potential difference across a and b Yep, and what we can see here when we remove this load is that we open-circuited this loop. Do you all see that? If we remove RL, we open-circuited the loop, meaning no current is flowing through the 1-ohm resistor because we pull out that load resistor. So the 1-ohm resistor will be inactive in the circuit when the load is removed. because current could only flow in a full loop a closed loop where branches make a full a closed path for current to flow so the seven ends equivalent voltage if no current is flowing through the one ohm is also equal to the ab is also equal to the voltage across the current source do you all agree because if no current is flowing through one the one ohm resistor there is no voltage drop across it and therefore the voltage across AB is equal to the voltage across the 2-amp current source. And since the current source is in parallel with the 12-ohm resistor, the Thevenin's voltage is also equal to the voltage across the 12-ohm resistor. So what we are looking for is the voltage at this point here and this is one node which is the voltage across the 12 ohm or the voltage across the current source this point here you're following me so far this one ohm resistor left the circuit why because no current is flowing through it and you could listen to this and and always apply it whenever current isn't flowing through our branch or a number of particular components a particular number of components you could ignore those components because if no current is flowing through them they're inactive in the circuit and hence doesn't need uh you to include them for the circuit analysis no current flowing to them no voltage drops across them they're not contributing to the circuit that's why this part came out and this is what we have here as the equivalent circuit when we remove the just the source in series with four ohm in parallel with the 12 ohm and then that 12 ohm is in parallel with the current source yeah everybody gets over so now that we know so one end the biggest things is to find what we need to find yeah and what we need to find is this voltage here to find the seven ends equivalent voltage which is the open circuit voltage across the terminal taking all the load did something to the circuit so all this is included in your skills of circuit analysis it's not just taking a circuit and doing it you know performing methods of analysis you have to make changes and you have to roll with those changes you have to in in uh infer what those changes mean yeah so we need this voltage here using nodal analysis we could find this voltage it's still a complex circuit because the two sources there we have to use complex methods of analysis we have three options i'll show you the three on this suit yeah To find this voltage here using nodal analysis, we need to analyze this node. We need to analyze the node, and to analyze that node, we need to choose the current branches, their variables, and their direction. I chose I1 to be the current flowing in this branch here and to be flowing into the node, and then I3 and I2 to be flowing in these branches here out of the node. Now, that might not be what you did because you might have chosen, you know, this current to flow in the direction of the source. And that is fine. Based on what you do, that would reflect in your nodal equation due to Kirchhoff's current law. And then when you substitute with Ohm's law and you do it properly, you'll get the same thing I get. You could define. the current directions and variables how you want them to you could call them whatever you want to call them you could say they're going in whatever direction they're going in yeah it will affect your nodal equation it will be different based on what you choose but you will get the same thing once you do the analysis correct this part yeah so according to what i choose because i can you know i i choose these I1 is equal to I2 plus I3. And that's my nodal analysis. The Kirchhoff's current law applied to this node here. Now I want to substitute using Ohm's law. I1 is the current flowing through this branch here. I know the voltage at this point here. So again, assuming this to be a zero potential, this will have to be 32 volts, this point, because 32 minus zero would give you the potential difference 32. So 32 minus this voltage would be the voltage drop across the 4-ohm. So 32 minus Vth divided by 4 would be I1, the current flowing through the 4-ohm. So somebody had asked, you know, where's the ground point? I didn't have to put the ground point in the circuit. You could have assumed it. But there's only really one assumption to make to get the actual answer. The reason being is these two sources have the same negative point. So naturally, this would be the zero potential. Yeah. So this is zero volts. This is 32 volts. 32 minus Vth, the unknown voltage resolvent, for divided by 4 ohms is I1. By the starting voltage, where the current coming from, minus where the current going over the resistance, it gives us the current flowing through that resistance. So that's why we have this expression from I2 is VTH minus the earth potential 0 divided by 12. That's what we have here. I3 is 2 amps. But I define the current to be going out of the node and into this positive current source terminal here. This current source is defined. It is sending current out of its positive terminal in the direction of this symbol here. So I3 is minus 2, not 2. If I had defined I3 to be going into the node in the same direction of the current source, then it would have been positive 2. yeah but based on my designated branch current i3 it is minus two if i had put positive two here i would have gotten something different and it wouldn't have been the answer i guess i would have gotten it wrong yeah so using this formula here which only has one variable i could simplify and find the variable to be certain Cool? Question. I don't want to do any intermediate math there because I don't have time for that. Yeah, you know how to do mathematics, you know how to get out of the city boat. What is very, very important here... is for you to know what happens when you take out the load resistor which voltage you're looking for and to do the nodal analysis properly based on the directions you choose the branch current direction remember for nodal analysis you d you define those currents you obtain the nodal equation you substitute using ohm's law the voltage drops across the resistances in those branches and then you you simplify in this case you know you didn't need to find a second node of equation because you do have another node the other node in this circuit is the zero potential that's all you only have one variable and one equation and i did say in in the lecture that you will have you you need to analyze you know the number of nodes based on how much variables you have now we're saying this here is the only other node and that is not really a node because it's just the ground potential you can't analyze that though yeah if you analyze this node here it will be the same thing not so i3 going in sorry yeah i3 going in i2 going in and i1 going out so it's the actually the same node just on the return path which is not a node to be considered because it's actually the same node here so 30 volts is the answer we got it using nodal analysis Let's do it using mesh analysis. Yeah? Remember, this comes off. This is the circuit we have here. And we have two loops. We could see it here. I chose, for mesh analysis, you have to choose your mesh currents. I chose them to be I1 and I2 going clockwise in the two side meshes, I guess you could call them, the left mesh and the right mesh. You can very well do the analysis using the left mesh. the outer mesh or the right mesh and the outer mesh. There are three loops in this circuit and you need to choose two. Why? Because you have two currents to define. Yeah? So using this first mesh you need to apply Kirchhoff's voltage law. we chose it to go clockwise we know that the voltage drop did i no so did i i didn't put the voltage directions but you know the voltage direction across this 32 volt not so it's upward and if the current is flowing clockwise the voltage drops across the resistors in that mesh is going to be opposite to that clockwise current so it's 32 minus I1 by 4 minus 12 I1 minus I2 so it it it resembles the uh the example we had in the lecture yeah it it's actually you know very similar yeah it's just have a it just have a a current source instead of an exported so so yeah that's how I got this yeah i put the minus terms on the right hand side of the equation don't let that confuse you uh the drops are on the side and the source is on this side yeah and now let me do it for i2 now we can't find a mesh equation for i2 why because we don't know what the voltage is across the current source we can't know what the voltage is across our current source because we don't have the resistive value it is not a resistor we don't have that resistive value to use ohm's law to find a drop we can't say i2 by 2 amp that that wrong it has to be a resistance to use this and is not a voltage source to just put in the equation of the into the voltage equation but the aim of this is to find i2 not so mesh current solves for current so we know i2 here is If I chose it to go clockwise here in the opposite direction to the current source, then I2 is just minus 2 amps. I2 is flowing through this branch here and is contributing to this branch here with I1. Now, so it's flowing here. If I chose that, then this source and this source is in this branch, this current source. I already know the current. I2 is equal to minus 2 amps. So substituting minus 2 into this first equation we got here for mesh 1, I could calculate I1 which is minus 5 amps. Remember mesh analysis solves for currents and we want this voltage here across the 12 ohm. Is that a problem? No, we have the currents. We can multiply by this resistance and get the voltage. Remember the voltage across here is 12 multiplied by I1 minus I2. 12 by minus 2 minus 0.2 is 30 volts. Sorry, that's nonsense. 12 by 0.5 minus minus 2 is 3. sorry five and uh therefore sorry 2.5 and therefore 12 by 2.5 is 30 volts question so you can choose to be i2 to be going in the same direction correct yeah so intuitively you could do something better that is more intuitive to the circuit you could choose your mesh currents to go in the direction of the positive terminals yeah and the positive terminal i did that for this one but the positive terminal here is this way so i could say you know i2 is is counterclockwise but i wanted to do this intentionally to show you if you go against you know current sources or voltage sources what would happen how it would affect the circuit you just have a negative number yeah and this very important thing the end correct answer so the variations here could be different loops you could choose the outer loop and one left loop you could choose different current directions you could change the variable name if you want it's up to you to name this what you want i suggest you name it i something but you know it don't matter you can use x y whatever so this using mesh analysis we got the same answer 30 volts so you're your journey in trying to understand this is is taking a circuit and doing it in different ways and seeing how the equations change yet the answer remains the same yeah last thing you know the superposition theorem uh how do we use the superposition theorem the superposition theorem states that you know consider one source at a time find the parameter you want and then when you have all the param that parameter for the different contributions of sources then you could just add them and that would be the total contribution so superposition says that the sum of the parts is equal to the whole circuit yeah the contribution of the whole circuit so the sum of the contributions individually is equal to the contribution of the whole circuit or all the sources in the circuit so let's consider the voltage source only if you're considering the voltage source only you have to remove the current source removing the current source means pulling out this current source here open circuit in it and therefore we have this remember we are no one home again right and we all we now pulling out the current source so this open circuited here no current flowing through here you have no uh element inside here this is just nothing yeah so when we remove that we eliminate this loop basically so all we have is this simple circuit here a source of all its source in series with a 4 1 or 12 ohm you could use the voltage divider to find the voltage we need which is the voltage across the 12 32 by 12 over the sum of the two in the loop 4 plus 12 24 volts now let's consider the current source only if we consider any current source only we have to remove the voltage source how do we remove our voltage source we short it and that is what we have here so that zero volts is provided so now we do that we actually have a parallel circuit here now with a current source supplying the total current that splits in these two parallel branches using the current divider theorem i could find the current flowing through the 12 ohm it's 2 the total current multiplied by the resistance in the other branch divided by the sum of the two branches 4 plus 12.5 amps now that i have the current flowing through the 12 ohm i could multiply it by the resistance value 12 and i'll get the voltage across it six volts now i take in the two contributions the two are calculated voltage for vth for the different circuits considering uh one source each sum them and i'll get the total voltage across the 12 ohm considering both sources 30 volts so the i guess the only difficulty using superposition is when removing these sources uh interpreting the circuit properly because errors could be made when you know removing these things if you do some madness here if you You had a short circuit in this instead of an open circuit in this, you get something totally different. if you had open circuit this instead of short circuit in it you'd get something totally different you're not getting your wrong answer and then uh here's zero is everybody safe with these three methods uh and and what these seven and equivalent uh voltages cool so i go forward the resistance which is something interesting finding rth means you have to remove the sources from the original circuit and see what you are left with if you remove the current source that was here you'd have a open you have to open circuit it removing the voltage source that was here means short circuiting it so you are left with this yeah So before you had that 1 ohm there and you ignored it because no current flowed through it. But now it is. You could see that it is included here because it is at it is connected to this terminal here. And I am considering all resistances in the circuit between these terminals. Yeah. So this comes back in. It was always here. It wasn't just it was just not analyzed in terms of. the to find the open circuit voltage because it has no current flowing through it and therefore no voltage across it so it doesn't contribute when looking for voltages and current but if you're looking for equivalent resistance between these two terminals here then it definitely still here remember is our practical resistance circuited there whether we analyze it or not depends on if it's energized but what we're looking for here now is the equivalent resistance between the two terminals. So it's definitely there. So after open circuiting this current source, this branch is not, you know, it doesn't have any resistor or anything and it's not connected to the circuit anymore and this was short, the voltage source was short circuited. So in short circuiting this voltage source, what we did is we ended up with 4 ohms in parallel with 12 ohms. do you all see that remember uh components in parallel share the same node on either side and that's what happening here with the four and the twelve ohm so we have a four ohm in parallel with the twelve ohm to do that you consider uh you know one over req is is is one over four plus one over twelve or you could use uh product over some you can do that or you could use you could simplify and you'll realize this product over some 4 by 12 divided by 4 plus 12. So y12 is 48, 4 plus 12 is 16, that would give us 3. Yeah? So the equivalent resistance of these two parallel resistors is a 3 ohm resistor. And the equivalent resistance of these two when resolved would be in series with this one. It's not in parallel because it's not sharing the two nodes on either side, but it is sharing this node on one side. So it's in series with the equivalent. So 3 plus 1 is 4 ohms seen from the output load. tournament so i know y'all would have some questions here anybody uh confused yet all right let me move forward i only have 10 more minutes yeah Good. So how about the Norton equivalent circuit for the circuit below? To find the Norton equivalent circuit, you have to remove the load. Step one is the same for both. Remove the load. When we remove the load for Norton equivalent, remember the Norton current source is the short circuit current through the terminal. So after removing the load. resistor, we have to short circuit them, the terminals. So we have to draw a line, make a connection, a short circuit between A and B after removing the load resistor. When we do that, we will realize that the 5 ohm branch will be bypassed when the load terminals are short circuited. So how do I know that? Well, I could take this node here. I could consider this branch as 5 ohms and this one has 0 ohms. because zero ohms is our ideal short circuit and the current coming into this node will split based on the resistance and since zero is much less than five ohms all the current that coming into this node will go through this branch that is what a short circuit does so what i want you all to take from this is once you short circuit an inactive component you divert all the current from that component through the short circuit, according to Kirchhoff's, or sorry, the current divider rule. Because this short circuit ideally has a zero ohm resistance, meaning all the current going to this node will flow through the short circuit, leaving no current to flow through the shorted resistor or component. Meaning if no current flows through this 5 ohm resistor here, it will be bypassed and hence can be removed from the circuit. If no current flowing through the component, no voltage drop, no contribution to the circuit. Now it's there, but we don't need to analyze it or include it in the circuit now when we're doing the analysis to find the short circuit current. Remember the current we find in is the current through here. That current is the node of, is the current I n, the current of the Norton's equivalent current source. Yeah. So what we have remained is this circuit here. The 5 ohm was bypassed because of the short circuit. So we want to find I short circuit, which is equal to the 7 n, you know, sorry, the Norton's. current source value so we're using mesh analysis i uh i need to go a little quicker uh i define the currents in each loop the left and the right loop to be clockwise i1 and i2 for mesh analysis of loop one we can't find a voltage across this because it's a current source but we know the current yeah so i1 is just two amps For mesh 2, we can apply that Kirchhoff's voltage law. It would be 12, this voltage drop would be in the positive direction, and the drops would be in the opposite direction of I2. So downward, right to left, left to right. So all these voltages would be negative with respect to this source voltage. So 12 minus 4, I2 minus I1. minus 8 I2 minus 8 I2 or you know you could put those drops on the right hand side of the equation and you'll get this. So what you'll notice here is this 8 ohm and this 8 ohm becomes in it comes in series with each other because of this short circuit. So that's why we have 8 I2 plus 8 I2. You could actually add these because they're in series and get 16 because they have the same current through them. They're in the same branch because of the short circuit. And so we have 16 I2. But this is fine how we do it here. It will simplify to that. Anyhow, substituting for I1 is 2. Enter here. We will get I2 to be 1. I2 is the current we're looking for. Yeah, the short circuited current. between the terminals a and b and so the the northern current source is one amp yeah using nodal analysis on the the circuit we got you know bypass five ohm we could analyze for this node here this is the only node we have I designated the current in this branch to be... in the direction of the current source I1 and the other two to be flowing out of the node, I2 and I3. So I1 is equal to I2 plus I3. Immediately we could see I1 is the 2-amp from the current source. I3, sorry I2 is V at the positive potential here is 12 because this the negative potentials or terminals of the sources would be 0 so this voltage is 12 V n is what we solving for so where the current coming from minus where it going on the opposite side of the passive element is the voltage drop divided by that value of the resistor is the current I2. So Vn minus 12 over 4 is I2. Yeah, the current direction says Vn is Vn minus 12. I3 is Vn minus 0 divided by 8 plus 8. Remember, these two are in series because of the short circuit. So the current I3 through these two resistors here, we could sum these two. It would be through our equivalent 16 ohm resistor is Vn minus the ground potential here, 0, divided by that 16. And we could see immediately I3 is the current we're looking for. So the current in this branch is the short circuited current. which is the value of the Norton current source. Yep, simplifying to this because we only have one unknown variable. We get Bn, the voltage at this point, this node, to be 16 volts. And if we want the current through this branch, we know the voltage. We know the zero potential is here. So 16. minus 0 over the 16 ohms would give us the current through this branch here, I3. So 16 divided by 8 plus 8 is 1 ohm. So that is from the theorem, 7 ohms, sorry, Norton's theorem. Norton's theorem states that we could simplify the circuit to an equivalent current source in parallel to the resistor. But to find that value of that equivalent current source, we have to short circuit the terminal and find the current to that short circuit. So it is from the theorem. It is from the basic law. yeah so you do need to short circuit that and whatever that current is through the short circuit is that equivalent uh current source value using superposition i'll quickly quickly go through it uh using the uh considering the voltage source only open circuit the current source end up with this loop alone of voltage sorry the current through could be easily found using ohm's law because this is just a series loop and you want to find the current through it so it's just v divided by all the resistances four plus eight plus eight and you'll get 0.6 amps not even my divider theorem have to be used yet it was simplified as simple series circuit yeah where we have to find the current considering the current source only we remove the voltage source short circuited We end up with this. And we can use the current divider theorem to find the current in this branch here. It's 2 multiplied by the current in the other branch, 4, divided by the sum of the resistors, 4 plus 8 plus 8, 0.4. 0.6 plus 0.4 is 1. The equivalent of Norton's resistance is done just like the Thevenin's resistance. We remove the sources. short circuit the voltage source open circuit the current source and look at the equivalence through these two terminals here what is the equivalent this is the equivalent circuit here after shorting uh after opening the current source just remove these connections here and you could see that the eight the four and the other eight ohm resistor in series with each other and that in series that equivalent value is in parallel with the 5 ohm and what you'd get is 4 ohms right and that is that is the norton equivalent there the one amp we get using all the methods of analysis and the in parallel with the four yeah so that's it that's it now uh We'll continue next time. Make sure the recording will be available. Yes. Make sure you go through the other pre recordings and time for the next feedback session on Wednesday. Yeah.

However, the simplest of complex circuits. Yeah. For your example.

So what we need to do is find a VTH and RTH because we want to reduce this complex circuit to its 7N equivalent. Its 7N equivalent is an equivalent voltage source of a value VTH and a series resistance, equivalent resistance of a value RTH. Yeah. so according to the rules the first step of solving this problem of the reducing this circuit to its equivalent uh counterpart is removing the load resistor we have the load resistor here i labeled it rl it's definitely the load resistor because it's between the terminals of this particular circuit the output terminals yeah So we have to remove it before we go forward with the analysis.

So when we remove that, we have the circuit on which we have to perform circuit analysis using one of the complex methods of analysis. Yeah. So what is VTH?

Thevenin's theorem states that that voltage source as equivalent voltage source is the open circuit voltage. across the terminals. So after we remove this load, the voltage VAB is the voltage we need to find. And that voltage is the value of the Thevenin's voltage source.

And then after we have to find RTH, the Thevenin's resistance. And the Thevenin's resistance is simply the equivalent resistance looking in from these terminals here. When I say looking, I mean, At these two points, what is the equivalent resistance, or between these two points, sorry, what is the equivalent resistance of the circuit? And to do that, we have to turn off the sources. Yeah?

How do we turn off sources? We need to short-circuit voltage sources so that the terminals of the voltage source no longer give us a potential difference of 32 volts. It gives us a potential difference of zero. because a short circuit is the same point you are bringing these points together when you short circuit thereby applying a zero volt at this point instead of 32 volts yeah and to take off the current source you know usually this current source will send it two amps to this branch and to the rest of the circuit taking it off means sending zero amps to the rest of the circuit so open circuit you have to open circuit the current source so that it would send zero arms through this branch yeah so let's go forward and i want to do so is there no stepping through here okay i take off the animation all right so let's find the evidence equivalent circuit for the circuit below so we have the original circuit we want to find bth the first thing we need to identify is the 7 and equivalent voltage okay so the first thing we're doing is the uh we're removing the load resistor yeah and we need to find out what voltage the open circuit voltage is so the 7 and equivalent voltage is the voltage v a b the voltage across this two points these the potential difference across a and b Yep, and what we can see here when we remove this load is that we open-circuited this loop.

Do you all see that? If we remove RL, we open-circuited the loop, meaning no current is flowing through the 1-ohm resistor because we pull out that load resistor. So the 1-ohm resistor will be inactive in the circuit when the load is removed. because current could only flow in a full loop a closed loop where branches make a full a closed path for current to flow so the seven ends equivalent voltage if no current is flowing through the one ohm is also equal to the ab is also equal to the voltage across the current source do you all agree because if no current is flowing through one the one ohm resistor there is no voltage drop across it and therefore the voltage across AB is equal to the voltage across the 2-amp current source.

And since the current source is in parallel with the 12-ohm resistor, the Thevenin's voltage is also equal to the voltage across the 12-ohm resistor. So what we are looking for is the voltage at this point here and this is one node which is the voltage across the 12 ohm or the voltage across the current source this point here you're following me so far this one ohm resistor left the circuit why because no current is flowing through it and you could listen to this and and always apply it whenever current isn't flowing through our branch or a number of particular components a particular number of components you could ignore those components because if no current is flowing through them they're inactive in the circuit and hence doesn't need uh you to include them for the circuit analysis no current flowing to them no voltage drops across them they're not contributing to the circuit that's why this part came out and this is what we have here as the equivalent circuit when we remove the just the source in series with four ohm in parallel with the 12 ohm and then that 12 ohm is in parallel with the current source yeah everybody gets over so now that we know so one end the biggest things is to find what we need to find yeah and what we need to find is this voltage here to find the seven ends equivalent voltage which is the open circuit voltage across the terminal taking all the load did something to the circuit so all this is included in your skills of circuit analysis it's not just taking a circuit and doing it you know performing methods of analysis you have to make changes and you have to roll with those changes you have to in in uh infer what those changes mean yeah so we need this voltage here using nodal analysis we could find this voltage it's still a complex circuit because the two sources there we have to use complex methods of analysis we have three options i'll show you the three on this suit yeah To find this voltage here using nodal analysis, we need to analyze this node. We need to analyze the node, and to analyze that node, we need to choose the current branches, their variables, and their direction.

I chose I1 to be the current flowing in this branch here and to be flowing into the node, and then I3 and I2 to be flowing in these branches here out of the node. Now, that might not be what you did because you might have chosen, you know, this current to flow in the direction of the source. And that is fine.

Based on what you do, that would reflect in your nodal equation due to Kirchhoff's current law. And then when you substitute with Ohm's law and you do it properly, you'll get the same thing I get. You could define. the current directions and variables how you want them to you could call them whatever you want to call them you could say they're going in whatever direction they're going in yeah it will affect your nodal equation it will be different based on what you choose but you will get the same thing once you do the analysis correct this part yeah so according to what i choose because i can you know i i choose these I1 is equal to I2 plus I3.

And that's my nodal analysis. The Kirchhoff's current law applied to this node here. Now I want to substitute using Ohm's law.

I1 is the current flowing through this branch here. I know the voltage at this point here. So again, assuming this to be a zero potential, this will have to be 32 volts, this point, because 32 minus zero would give you the potential difference 32. So 32 minus this voltage would be the voltage drop across the 4-ohm. So 32 minus Vth divided by 4 would be I1, the current flowing through the 4-ohm.

So somebody had asked, you know, where's the ground point? I didn't have to put the ground point in the circuit. You could have assumed it. But there's only really one assumption to make to get the actual answer.

The reason being is these two sources have the same negative point. So naturally, this would be the zero potential. Yeah. So this is zero volts. This is 32 volts.

32 minus Vth, the unknown voltage resolvent, for divided by 4 ohms is I1. By the starting voltage, where the current coming from, minus where the current going over the resistance, it gives us the current flowing through that resistance. So that's why we have this expression from I2 is VTH minus the earth potential 0 divided by 12. That's what we have here.

I3 is 2 amps. But I define the current to be going out of the node and into this positive current source terminal here. This current source is defined. It is sending current out of its positive terminal in the direction of this symbol here.

So I3 is minus 2, not 2. If I had defined I3 to be going into the node in the same direction of the current source, then it would have been positive 2. yeah but based on my designated branch current i3 it is minus two if i had put positive two here i would have gotten something different and it wouldn't have been the answer i guess i would have gotten it wrong yeah so using this formula here which only has one variable i could simplify and find the variable to be certain Cool? Question. I don't want to do any intermediate math there because I don't have time for that. Yeah, you know how to do mathematics, you know how to get out of the city boat. What is very, very important here...

is for you to know what happens when you take out the load resistor which voltage you're looking for and to do the nodal analysis properly based on the directions you choose the branch current direction remember for nodal analysis you d you define those currents you obtain the nodal equation you substitute using ohm's law the voltage drops across the resistances in those branches and then you you simplify in this case you know you didn't need to find a second node of equation because you do have another node the other node in this circuit is the zero potential that's all you only have one variable and one equation and i did say in in the lecture that you will have you you need to analyze you know the number of nodes based on how much variables you have now we're saying this here is the only other node and that is not really a node because it's just the ground potential you can't analyze that though yeah if you analyze this node here it will be the same thing not so i3 going in sorry yeah i3 going in i2 going in and i1 going out so it's the actually the same node just on the return path which is not a node to be considered because it's actually the same node here so 30 volts is the answer we got it using nodal analysis Let's do it using mesh analysis. Yeah? Remember, this comes off. This is the circuit we have here. And we have two loops.

We could see it here. I chose, for mesh analysis, you have to choose your mesh currents. I chose them to be I1 and I2 going clockwise in the two side meshes, I guess you could call them, the left mesh and the right mesh. You can very well do the analysis using the left mesh. the outer mesh or the right mesh and the outer mesh.

There are three loops in this circuit and you need to choose two. Why? Because you have two currents to define.

Yeah? So using this first mesh you need to apply Kirchhoff's voltage law. we chose it to go clockwise we know that the voltage drop did i no so did i i didn't put the voltage directions but you know the voltage direction across this 32 volt not so it's upward and if the current is flowing clockwise the voltage drops across the resistors in that mesh is going to be opposite to that clockwise current so it's 32 minus I1 by 4 minus 12 I1 minus I2 so it it it resembles the uh the example we had in the lecture yeah it it's actually you know very similar yeah it's just have a it just have a a current source instead of an exported so so yeah that's how I got this yeah i put the minus terms on the right hand side of the equation don't let that confuse you uh the drops are on the side and the source is on this side yeah and now let me do it for i2 now we can't find a mesh equation for i2 why because we don't know what the voltage is across the current source we can't know what the voltage is across our current source because we don't have the resistive value it is not a resistor we don't have that resistive value to use ohm's law to find a drop we can't say i2 by 2 amp that that wrong it has to be a resistance to use this and is not a voltage source to just put in the equation of the into the voltage equation but the aim of this is to find i2 not so mesh current solves for current so we know i2 here is If I chose it to go clockwise here in the opposite direction to the current source, then I2 is just minus 2 amps.

I2 is flowing through this branch here and is contributing to this branch here with I1. Now, so it's flowing here. If I chose that, then this source and this source is in this branch, this current source. I already know the current. I2 is equal to minus 2 amps.

So substituting minus 2 into this first equation we got here for mesh 1, I could calculate I1 which is minus 5 amps. Remember mesh analysis solves for currents and we want this voltage here across the 12 ohm. Is that a problem?

No, we have the currents. We can multiply by this resistance and get the voltage. Remember the voltage across here is 12 multiplied by I1 minus I2.

12 by minus 2 minus 0.2 is 30 volts. Sorry, that's nonsense. 12 by 0.5 minus minus 2 is 3. sorry five and uh therefore sorry 2.5 and therefore 12 by 2.5 is 30 volts question so you can choose to be i2 to be going in the same direction correct yeah so intuitively you could do something better that is more intuitive to the circuit you could choose your mesh currents to go in the direction of the positive terminals yeah and the positive terminal i did that for this one but the positive terminal here is this way so i could say you know i2 is is counterclockwise but i wanted to do this intentionally to show you if you go against you know current sources or voltage sources what would happen how it would affect the circuit you just have a negative number yeah and this very important thing the end correct answer so the variations here could be different loops you could choose the outer loop and one left loop you could choose different current directions you could change the variable name if you want it's up to you to name this what you want i suggest you name it i something but you know it don't matter you can use x y whatever so this using mesh analysis we got the same answer 30 volts so you're your journey in trying to understand this is is taking a circuit and doing it in different ways and seeing how the equations change yet the answer remains the same yeah last thing you know the superposition theorem uh how do we use the superposition theorem the superposition theorem states that you know consider one source at a time find the parameter you want and then when you have all the param that parameter for the different contributions of sources then you could just add them and that would be the total contribution so superposition says that the sum of the parts is equal to the whole circuit yeah the contribution of the whole circuit so the sum of the contributions individually is equal to the contribution of the whole circuit or all the sources in the circuit so let's consider the voltage source only if you're considering the voltage source only you have to remove the current source removing the current source means pulling out this current source here open circuit in it and therefore we have this remember we are no one home again right and we all we now pulling out the current source so this open circuited here no current flowing through here you have no uh element inside here this is just nothing yeah so when we remove that we eliminate this loop basically so all we have is this simple circuit here a source of all its source in series with a 4 1 or 12 ohm you could use the voltage divider to find the voltage we need which is the voltage across the 12 32 by 12 over the sum of the two in the loop 4 plus 12 24 volts now let's consider the current source only if we consider any current source only we have to remove the voltage source how do we remove our voltage source we short it and that is what we have here so that zero volts is provided so now we do that we actually have a parallel circuit here now with a current source supplying the total current that splits in these two parallel branches using the current divider theorem i could find the current flowing through the 12 ohm it's 2 the total current multiplied by the resistance in the other branch divided by the sum of the two branches 4 plus 12.5 amps now that i have the current flowing through the 12 ohm i could multiply it by the resistance value 12 and i'll get the voltage across it six volts now i take in the two contributions the two are calculated voltage for vth for the different circuits considering uh one source each sum them and i'll get the total voltage across the 12 ohm considering both sources 30 volts so the i guess the only difficulty using superposition is when removing these sources uh interpreting the circuit properly because errors could be made when you know removing these things if you do some madness here if you You had a short circuit in this instead of an open circuit in this, you get something totally different. if you had open circuit this instead of short circuit in it you'd get something totally different you're not getting your wrong answer and then uh here's zero is everybody safe with these three methods uh and and what these seven and equivalent uh voltages cool so i go forward the resistance which is something interesting finding rth means you have to remove the sources from the original circuit and see what you are left with if you remove the current source that was here you'd have a open you have to open circuit it removing the voltage source that was here means short circuiting it so you are left with this yeah So before you had that 1 ohm there and you ignored it because no current flowed through it. But now it is.

You could see that it is included here because it is at it is connected to this terminal here. And I am considering all resistances in the circuit between these terminals. Yeah.

So this comes back in. It was always here. It wasn't just it was just not analyzed in terms of.

the to find the open circuit voltage because it has no current flowing through it and therefore no voltage across it so it doesn't contribute when looking for voltages and current but if you're looking for equivalent resistance between these two terminals here then it definitely still here remember is our practical resistance circuited there whether we analyze it or not depends on if it's energized but what we're looking for here now is the equivalent resistance between the two terminals. So it's definitely there. So after open circuiting this current source, this branch is not, you know, it doesn't have any resistor or anything and it's not connected to the circuit anymore and this was short, the voltage source was short circuited.

So in short circuiting this voltage source, what we did is we ended up with 4 ohms in parallel with 12 ohms. do you all see that remember uh components in parallel share the same node on either side and that's what happening here with the four and the twelve ohm so we have a four ohm in parallel with the twelve ohm to do that you consider uh you know one over req is is is one over four plus one over twelve or you could use uh product over some you can do that or you could use you could simplify and you'll realize this product over some 4 by 12 divided by 4 plus 12. So y12 is 48, 4 plus 12 is 16, that would give us 3. Yeah? So the equivalent resistance of these two parallel resistors is a 3 ohm resistor.

And the equivalent resistance of these two when resolved would be in series with this one. It's not in parallel because it's not sharing the two nodes on either side, but it is sharing this node on one side. So it's in series with the equivalent. So 3 plus 1 is 4 ohms seen from the output load.

tournament so i know y'all would have some questions here anybody uh confused yet all right let me move forward i only have 10 more minutes yeah Good. So how about the Norton equivalent circuit for the circuit below? To find the Norton equivalent circuit, you have to remove the load.

Step one is the same for both. Remove the load. When we remove the load for Norton equivalent, remember the Norton current source is the short circuit current through the terminal. So after removing the load. resistor, we have to short circuit them, the terminals.

So we have to draw a line, make a connection, a short circuit between A and B after removing the load resistor. When we do that, we will realize that the 5 ohm branch will be bypassed when the load terminals are short circuited. So how do I know that?

Well, I could take this node here. I could consider this branch as 5 ohms and this one has 0 ohms. because zero ohms is our ideal short circuit and the current coming into this node will split based on the resistance and since zero is much less than five ohms all the current that coming into this node will go through this branch that is what a short circuit does so what i want you all to take from this is once you short circuit an inactive component you divert all the current from that component through the short circuit, according to Kirchhoff's, or sorry, the current divider rule. Because this short circuit ideally has a zero ohm resistance, meaning all the current going to this node will flow through the short circuit, leaving no current to flow through the shorted resistor or component.

Meaning if no current flows through this 5 ohm resistor here, it will be bypassed and hence can be removed from the circuit. If no current flowing through the component, no voltage drop, no contribution to the circuit. Now it's there, but we don't need to analyze it or include it in the circuit now when we're doing the analysis to find the short circuit current.

Remember the current we find in is the current through here. That current is the node of, is the current I n, the current of the Norton's equivalent current source. Yeah. So what we have remained is this circuit here. The 5 ohm was bypassed because of the short circuit.

So we want to find I short circuit, which is equal to the 7 n, you know, sorry, the Norton's. current source value so we're using mesh analysis i uh i need to go a little quicker uh i define the currents in each loop the left and the right loop to be clockwise i1 and i2 for mesh analysis of loop one we can't find a voltage across this because it's a current source but we know the current yeah so i1 is just two amps For mesh 2, we can apply that Kirchhoff's voltage law. It would be 12, this voltage drop would be in the positive direction, and the drops would be in the opposite direction of I2. So downward, right to left, left to right. So all these voltages would be negative with respect to this source voltage.

So 12 minus 4, I2 minus I1. minus 8 I2 minus 8 I2 or you know you could put those drops on the right hand side of the equation and you'll get this. So what you'll notice here is this 8 ohm and this 8 ohm becomes in it comes in series with each other because of this short circuit. So that's why we have 8 I2 plus 8 I2.

You could actually add these because they're in series and get 16 because they have the same current through them. They're in the same branch because of the short circuit. And so we have 16 I2.

But this is fine how we do it here. It will simplify to that. Anyhow, substituting for I1 is 2. Enter here. We will get I2 to be 1. I2 is the current we're looking for. Yeah, the short circuited current.

between the terminals a and b and so the the northern current source is one amp yeah using nodal analysis on the the circuit we got you know bypass five ohm we could analyze for this node here this is the only node we have I designated the current in this branch to be... in the direction of the current source I1 and the other two to be flowing out of the node, I2 and I3. So I1 is equal to I2 plus I3.

Immediately we could see I1 is the 2-amp from the current source. I3, sorry I2 is V at the positive potential here is 12 because this the negative potentials or terminals of the sources would be 0 so this voltage is 12 V n is what we solving for so where the current coming from minus where it going on the opposite side of the passive element is the voltage drop divided by that value of the resistor is the current I2. So Vn minus 12 over 4 is I2.

Yeah, the current direction says Vn is Vn minus 12. I3 is Vn minus 0 divided by 8 plus 8. Remember, these two are in series because of the short circuit. So the current I3 through these two resistors here, we could sum these two. It would be through our equivalent 16 ohm resistor is Vn minus the ground potential here, 0, divided by that 16. And we could see immediately I3 is the current we're looking for.

So the current in this branch is the short circuited current. which is the value of the Norton current source. Yep, simplifying to this because we only have one unknown variable. We get Bn, the voltage at this point, this node, to be 16 volts. And if we want the current through this branch, we know the voltage.

We know the zero potential is here. So 16. minus 0 over the 16 ohms would give us the current through this branch here, I3. So 16 divided by 8 plus 8 is 1 ohm. So that is from the theorem, 7 ohms, sorry, Norton's theorem. Norton's theorem states that we could simplify the circuit to an equivalent current source in parallel to the resistor.

But to find that value of that equivalent current source, we have to short circuit the terminal and find the current to that short circuit. So it is from the theorem. It is from the basic law.

yeah so you do need to short circuit that and whatever that current is through the short circuit is that equivalent uh current source value using superposition i'll quickly quickly go through it uh using the uh considering the voltage source only open circuit the current source end up with this loop alone of voltage sorry the current through could be easily found using ohm's law because this is just a series loop and you want to find the current through it so it's just v divided by all the resistances four plus eight plus eight and you'll get 0.6 amps not even my divider theorem have to be used yet it was simplified as simple series circuit yeah where we have to find the current considering the current source only we remove the voltage source short circuited We end up with this. And we can use the current divider theorem to find the current in this branch here. It's 2 multiplied by the current in the other branch, 4, divided by the sum of the resistors, 4 plus 8 plus 8, 0.4.

0.6 plus 0.4 is 1. The equivalent of Norton's resistance is done just like the Thevenin's resistance. We remove the sources. short circuit the voltage source open circuit the current source and look at the equivalence through these two terminals here what is the equivalent this is the equivalent circuit here after shorting uh after opening the current source just remove these connections here and you could see that the eight the four and the other eight ohm resistor in series with each other and that in series that equivalent value is in parallel with the 5 ohm and what you'd get is 4 ohms right and that is that is the norton equivalent there the one amp we get using all the methods of analysis and the in parallel with the four yeah so that's it that's it now uh We'll continue next time.

Make sure the recording will be available. Yes. Make sure you go through the other pre recordings and time for the next feedback session on Wednesday.

Yeah.

Transcript for:Understanding Thevenin and Norton Circuits

Transcript for:
Understanding Thevenin and Norton Circuits