Essential & Practical Circuit Analysis: Part 1- DC Circuits

hey how's it going everyone this is the solid state workshop with essential and practical circuit analysis this is going to be a two part series on circuit analysis so put on your thinking cap and let's go so what exactly is circuit analysis anyway well certain alysus is a toolkit for understanding and designing more complex circuits so if you want to get into the world of electronics or you want to understand electronic circuits well you're going to need to understand the concepts brought to you in circuit analysis first and that's because circuit analysis is a layer of abstraction it is based off of other abstractions while other abstractions are built off of it so you can look at this process chart here of how that works so you start off in physics and you learn about the electrons and protons and all the small low-level things and then you move on to circuit analysis and learn how that works and then you move on to electronics which is based off a circuit analysis and then you can move on to electronics systems which is based off electronics so in order to be successful in each area you first need to understand the area that precedes it the layer of abstraction before it all right cool what will be covered in this video well a whole range of topics we're going to start from the very beginning with some simple circuits and then we're going to move up into the world of more complex circuits but along the way we're going to discover methods which simplify the complex circuits and make performing an analysis on them easier so here are the topics that we're going to talk about today we're going to start by talking about linear circuit elements and then we're going to move on to Ohm's law and series and parallel circuits which you may have already seen before in a physics class or something then we're talk about voltage and current dividers and move on to the big two Kirchhoff's current law and its associated nodal and Kirchhoff's voltage law and to sociated loop analysis and we're talk about some methods for analysis that are a little more abstract but are definitely still really cool and those include source transformation Thevenin or and equivalence and finally superposition today we're we dealing with linear circuit elements in all of our examples and what are the linear circuit elements that we might encounter well we have the humble resistor with units ohm and symbol Omega the capacitor with units farad and symbol F inductor with units Henry and symbol H the voltage source with units volt and symbol V and the current source with units ampere or amps and symbol a and now what makes a linear circuit element linear well say we would write out a relationship between current and voltage for any of these elements more specifically resistors capacitors and inductors but say we write out current as a function of voltage we would find that the current through these devices these elements is going to be a function of voltage times some constant and so that makes it a linear element because it there is a linear operator in that relationship in a nonlinear circuit element the operator is something like an exponential operator e to the power of something but we won't be encountering nonlinear circuit elements today so don't worry about those and more specifically today we're going to be dealing with exclusively resistors voltage sources and current sources before we go any further let's discuss the different parts of a circuit and the names we give to these parts here's the circuit that we'll use as an example and now the first part is called a node and the node is a junction of connecting wires it's where circuit elements are joined together so in this case we have four nodes because there are four points of connection in this circuit we have a node connecting the voltage source to a resistor here we have two resistors and a voltage source connected together and here we have a voltage source and a resistor connected together we also include this big node at the bottom and every point on this big node is at the same electric potential and well how could that be well there's nothing in between any of the points on the node here that could alter its potential such as a resistor or a source and if say we did add one of those things well then this whole thing is no longer node maybe only this part would be a node next part we have is called a branch and so a branch is really just a fancy name for any of the circuit elements between two nodes so here are our law all our branches in the circuit it's really as easy as it sounds it's the element between nodes I guess the difference between a branch and just a circuit element is that a branch is an element that is part of a circuit by definition while a circuit element on its own can just be a standalone thing that we can talk about by itself all right so finally let's talk about loops and a loop is a closed path that begins and ends at the same node and a loop can't pass through any node more than once so here we have three different loops in this circuit we have a small here we have another small loop here and then we have one big loop that encompasses the entire circuit and these are all loops so try to become familiar with these terms because I'll be using them to describe points on a circuit and plus you get to sound smart when talking to your friends so that's a plus alright so Ohm's law the most fundamentally basic and essential relationship in all circuit analysis now before we talk about the actual equation let's talk about the variables that it is consisted of so we have V for voltage I for current and R for resistance and now remember voltage is an across variable meaning you take the voltage between two points or across some element in the circuit and I or current is a through variable meaning it's the current through an element and resistance is simply a constant assigned to a particular resistor now Ohm's law says V equals I times R I equals V over R or R equals V over I so it makes it possible for us to solve for an unknown current voltage or resistance granted that we already have two of those elements of that equation so if we have a circuit that looks like this with a voltage source of 30 volts and a resistance of 75 ohms we can determine the current that must flow through that resistor and that's easy because in this case I equals 30 over 75 and that equals 0.4 amps circuit elements can be connected in two basic fashions either in series or in parallel so let's talk about series first in a series circuit the elements in a circuit are connected one after another meaning the front of one circuit element is attached to the back of another element and the front of that element is attached to the back of another element that's what makes it a series circuit in a series circuit there's only one path for current to flow and that's because there's only one loop here where else because the current possibly flow and how could it possibly be different at different points in the circuit so we could test for current here and it'll be the same as current the same value for current will be found here and the same value for current we found here now if we have multiple resistors in our circuit then we can simplify multiple resistors down to one equivalent resistance and we can do that using a very simple equation that says our equivalent equals r1 plus r2 plus all the way up to RN which is the enth resistor so in our particular instance we have an hundred ohm resistor and a 50 ohm resistor in series so the equivalent resistance that would be formed between this point and this point would be 150 ohms now that we know the equivalent resistance basically we can rewrite these two resistors as one resistor then we can solve for the current just as we did previously so applying Ohm's law we see that the current is equal to 30 volts the source voltage over the equivalent resistance of 150 ohms so dividing 30 by 150 gives us 0.2 m/s in a parallel circuit like this one all resistors share a common voltage and we can see that because the 30 volts source here is attached across both 150 ohm and 100 ohm resistor simultaneously and in the same place but the difference between parallel and series is that the currents through each of these resistors can be different and in this case they are different because the values of the resistances are different so there's going to be less current flowing through the 150 ohm resistor than in the hundred ohm resistor and say we wanted to find the current up here the green current how can we do that well we can turn these two resistors arranged in the parallel configuration into one equivalent resistor like we did in series circuits and we're going to use this equation 1 over R equivalent equals 1 over R 1 plus 1 over R 2 all the way up to 1 over R N and so in this case we find that the equivalent resistance that will replace this circuit block right here is equal to 1 over 150 plus 1 over 100 and then whatever answer you get there you just flip it so this arrangement of resistors can be replaced by 160 ohm resistor and so therefore we can find what that green current would be and we're going to choose Ohm's law and you divide 30 volts by our equivalent resistance of 60 ohms and we find that the green current would be 0.5 amps now say we wanted to find the currents through each of the resistors how would we do that well we would simply divide the source voltage since both share the same source voltage by the resistance so for the pin current through the 150 ohm resistor that's equal to 30 volts over 150 which equals 0.2 and the blue current through the 100 ohm resistor is equal to 30 volts over 100 ohms and that equals 0.3 amps so some takeaways from here basically we can say that the green current flows into this node here this junction of components splits some of the goes down one chute and some of it goes out another chute here and then back at the bottom and down at the bottom these currents recombine back into the green current and rinse and repeat a voltage divider is a circuit that creates a voltage which is some fraction of its voltage source so here is an example of a voltage divider circuit it is a series circuit where the output is observed or taken across the second resistor so in this case that's that 10 ohm resistor and it's a series circuit because the 10 volt source and the 40 ohm resistor and the 10 ohm resistor are all in series with each other we can kind of ignore these wires or lines over here because when we're taking the output we're just observing the output we're not really intruding upon the circuit so it doesn't really contribute to the circuit at least in the ideal case we'll see more about that later but anyway in a series circuit like this with multiple resistors and a voltage source or any source the voltage across each resistor will be different as dictated by Ohm's law V equals IR so where I the current is the same for all of them but the resistance kindig-it will dictate the voltage across each resistor so if we want to find V out in this case we're going to first start with finding the equivalent resistance so that's that's easy that's 40 plus 10 so that's 50 ohms and now we can find the current through these resistors and that is point 2 amps easy enough and then to find V out we can simply multiply the current which is the same for both resistors because it's series circuit and multiply it by the resistance of r2 and then this case that's 10 ohms so you just do point 2 times 10 ohms and that will give you the voltage across the 10 ohm resistor ah now technically we can also take the voltage across the 40 ohm resistor and that would be 0.2 amps times 40 ohms which would give us 8 volts so as you can see 2 volts and 8 volts would add up to 10 volts so that makes sense I hope all right cool what we just did is all fine and dandy except it takes a couple of steps about three to reach our answer so we want to write a general equation that describes the output of any two resistor voltage divider so here is a general schematic of a voltage divider so we have a vs and we have an r1 and we have an r2 and we also have a current that flows through the circuit it's the same current everywhere in the circuit because this is a series circuit and we have a voltage output AV out which is observed across r2 so let's put all these things into mathematical equations so our equivalent equals r1 plus r2 because it's a series circuit and the current i is equal to vs / r equivalent and V out is equal to the current through r2 times the resistance of r2 and we know that the current is the same for everywhere in the circuit so that current through r2 is simply just I and so now we can substitute in to our V out equation we can substitute in vs / R equivalent for I and then we can rearrange and come to the final equation of vs times r2 over r1 plus r2 and that will give you the voltage output the V out for any two resistor voltage divider with any voltage source and any two resistors now most generally we can write an equation that says V out equals vs times our x over R equivalent and what is our X Rx is the resistor you would like to find the voltage across and our equivalent is again the total resistance so we can find the voltage across any of the resistors and we can have any number of resistors we can have three four five six resistors and all you have to do is put the value of that resistor in the numerator and over top of the equivalent resistance and multiply it times the source voltage and you will get the voltage across that resistor rx in practice the output of the voltage divider is going to be hooked up to something we want V out to do something for us so in circuits that means we're going to attach a load to it but when we attach a load to it which is in this case a resistor we fundamentally change the nature of the voltage divider hmm so let's look at this circuit this is the new circuit the new voltage divider with a load resistor RL attached across its output and RL takes the place of what was before an open circuit and we're going to learn that attaching a load RL reduces the dividers output voltage and well how could that be well let's look back to the original equation that says V out equals vs times r2 over r1 plus r2 and then if we look back at the circuit we would notice that our two an RL are in what configuration well they're in parallel so what used to be just r2 is now really r2 parallel RL the equivalent resistance so we can rewrite the V out equation to take this into account so the old r2 now becomes r2 parallel RL and we can substitute that in any time we see r2 and we know that based on our knowledge of parallel resistors in parallel it's in general that are two parallel RL is going to be less than the original value of r2 and why is that well think here if you have r2 which is some value and then you attach another resistor in parallel with it you're effectively creating another path for current to flow through so the equivalent resistance is going to be less because now there is another path for current to travel through alright so if the equivalent resistance up here is less than the original r2 that means that this voltage output is going to be less than the original voltage output now if the load resistance RL is much greater than a resistance of r2 then the output voltage only drops by a little bit and why is that well thinking back to the equation for parallel resistors again adding a big resistor in parallel with a comparatively small resistor will only reduce the equivalent resistance by a small amount so this resistance will only be decreased by a small amount so if this is only decreased by a small amount then V out is only a little bit smaller than the original V out another way to think about this is that the current through RL the current through the load the load current should we must should be much less than the current through the divider the current through r1 and r2 so if we knew that our load was going to draw 10 milliamps here then we should ensure as a rule of thumb that the current through r1 and r2 is at least 10 or maybe even a hundred times greater than the 10 milliamps flowing through RL and that's going to ensure that the voltage across RL is close to what we intended it to be originally I wasn't going to include a section on current dividers because most real-world circuits employ voltage dividers and not current dividers and that's because most sources in the real world more closely resemble voltage sources however understanding current dividers is still definitely a good thing to know so a current divider is a circuit that creates a current which is some fraction of its current source so it's kind of the opposite of a voltage divider which creates a fraction of its voltage source and this is what one looks like this is the general form the general schematic of a current divider and the current divider is a parallel circuit where the output current is observed in one of its branches so in this case it's the current through the branch consisting of Rx the current I out going through this branch and so the first step to finding out is to first find this value R T and R T is the equivalent resistance of all the other branches excluding Rx so it's the equivalent resistance of these two branches and not including the rx branch and if there's only one other branch other than rx so say r2 wasn't here then it would simply be the value of r1 that would be the equivalent resistance but since we have two branches here we would follow the formula for a parallel circuit for parallel resistances which is RT equals 1 over r1 plus 1 over r2 and then flipped now to find I out you're going to follow this equation which says I out equals is times RT over rx plus RT let's compare this to the voltage divider equation in the voltage divider equation what was in the numerator can you remember right it was the resistor which we were concerned about so r2 generally but in the current divider it's the opposite it's the equivalent resistance of everything except for the resistor that we are interested in and it the resistor that we're interested in this case is rx and you'll notice that R T is the equivalent resistance of r1 and r2 but not rx interesting so here we are the real meat and potatoes of circuit analysis it's getting good so Kirchhoff's current law or KCl for short says that all currents entering a node must equal all currents exiting a node basically what goes in must come out and in mathematical terms the sum of all currents in node equals 0 can also express this as a summation of I and where J is a particular instance of current just tells you what current we're talking about and so here are some illustrations to help you understand KCl the yellow and the green currents represent currents entering a node while the red blue and pink currents represent currents exiting a node and in the chart below we can see that the sum of all the currents entering the node equals the sum of the currents exiting the node regardless of what what the individual currents are they all add up to the same current in practice we write a KCl equation for a node in the form I 1 plus I 2 plus I 3 just like we have in this equation here and we set that all equal to 0 as we learned before because the sum of all currents in a node equals 0 and I 1 and I 2 and I 3 are just all the currents coming in out of the node if you didn't catch on so we use the convention that a current entering a node is a positive I because you can think of it as adding to a node and a current exiting a node is a negative I because you can think of it as being subtracted from the node so let's do a problem with this so we're given a circuit that looks like this and we're also given some currents initially so we're given that there there's 4 milliamps going to the right 12 milliamps go line up here and two milliamps going to the left and we're also given the currents i1 and i2 as arbitrarily going to the right but we're not actually sure if they go to the left or right quite yet and so we're also set up nodes so we're going to call this node right here node a and this node right here in the middle node B and now we're asked find I 1 and I 2 using KCl and the first thing we should ask ourselves is do I need to find one current before I can find the other current so if I want to find I 1 we first need to find the current i2 and it doesn't make sense hopefully that make sense in a minute or two so let's write the node equation for node B first because that's correlated with I - so the KCl at node B what is that going to be well here's what it is and we'll explain why well we have a positive I two entering the node so we write positive i2 and we have a positive 12 milliamps also entering the node so we write positive 12 and we have 4 milliamps exiting the node so we write minus 4 milliamps and that's it and then we set that equal to 0 and then we can easily solve and find that I 2 equals negative 8 milli amps and we'll talk about what a negative current means in a second so now let's move on to node a over here so what's the KCl equation at node a well here it is and we'll talk about that so we have a negative I 1 coming out of the node because it's exiting it's a negative i1 and you have I - also exiting the node and then we have a positive two milliamps coming up from the bottom here into the node a so we have a negative i1 - and i2 + 2 milliamps and set that equal to zero and we know that I 2 is negative 8 milliamps so we can simply plug that right into I 2 and we can then solve for I 1 and we find that I 1 equals 10 milliamps and so what did that negative 8 milliamps from before mean the I 2 equals negative 8 milli amps well in the direction that we've specified so we just said that yeah I 2 is going to go to the right it means that it's actually it's actually a positive 8 milliamps going to the left so all we have to do is if we get a negative current is we just say that the real current is going in the opposite direction and is a positive current so a negative current just tells you to flip the direction of the current and make the value of the current positive building off of KCl is something called nodal analysis and nodal analysis is a process that uses KCl determine node voltages and so here's the basic setup for nodal analysis to use nodal analysis we first need to create a reference node and so a reference node in our example circuit diagram over here is the strip of wire on the bottom that's our reference node now how do we choose which node we want to make a reference node in a circuit well rule of thumb is to pick the node in your circuit with the most connections to it that'll make your life a little easier now remember in a circuit of voltage is defined as a difference in potential between two points so if I was to say that the voltage at some point is 5 volts well then I also have to specify that it's 5 volts with respect to what other point so I would have to give that other point so when we set up a reference node we're providing that other point in the circuit we're saying that all of our voltage measurements are referred or referenced to that reference node so to this node down here so if we look at our example what is the voltage at this point right here well it's VX with reference to ground or with respect to ground any way you want to say and the voltage here is V Y with respect to ground all right moving on and so we know that a current through resistor is described and governed by Ohm's law which says the difference in potential across a resistor over the resistance gives us the current so in this case that's going to be VX minus V Y over R and why is it VX minus V Y well because the current is going in the direction to the right meaning that there would be a lower voltage at this point then there would be at this point VX because a resistor creates a voltage drop and now VX and V Y or what we call node voltages and we might be interested in solving for those and so to solve for these we can write the KCl equation for each node in the circuit except for the reference node here we exclude the reference node in our equations and so we can write out the standard KCl equations like we did before in our previous examples except now instead of writing I 1 I 2 I 3 equals 0 we write in the form VX minus V Y over r+ v z- v be whatever / r whatever you want whatever points there are in the circuit so we use this form the X minus V y over R instead of just I alright the moment we've all been waiting for the nodal analysis sample problem all right now no one was waiting for it alright anyway here is the sample nodal analysis problem that we're going to attempt here hopefully I don't get it wrong we're gonna try to find the voltage vo here the V out whatever you wanna call it and before we get started let's note a couple things about the circuit the reference node is already set up for us and it's this node down here how do we know that's the reference node because there's a ground symbol attached to it that basically signifies that it is the reference node and then we have two nodes up top here that we are interested in node 1 here and node 2 here and we also have two currents that we're interested in and that's I 1 which goes to the right and I 2 which also goes to the right those are set arbitrarily I can have made them go to the left if I had chosen to alright so let's get started let's start at node 1 let's write out the KCl equation at node 1 what's that going to be well we have three currents associated with node 1 and those are I 1 I 2 and is so let's write out the equation for them since I ones cowan into node it's a positive i1 since i2 is coming out of the node it's a negative I 2 and since is is pointing into the node it is a positive is and now this equation here is just a standard KCl equation we've done this before but now to do nodal analysis we want to put it into the form of a difference in voltage across a resistor over the resistance and so which ones can we do that for what we can do it for i1 because I one associated with the current through r1 and i2 is associated with the current through r2 so we can write this out we can write those two out in that form however we can't write it out like that for is because is does not have a resistor associated with it so here's what that new equation looks like and we have V s which is the source voltage over here minus v1 which is the voltage at node one over the resistance r1 and in a very similar fashion its v1 minus v2 over r2 and now why didn't just for for instance why why is I two defined as the voltage v1 minus v2 and not the other way around well because the current is going from left to right and then is we just tack on at the end here and set it all equal to zero now we can just do some kind of algebraic clever rearranging and we can get it into a form like this where we have the one times some constant plus v2 times some constant usually these constants are the are constants for resistance and then we're going to set that equal to the other constants that we know here all right so just let that be for now we have an equation with two variables so in order to solve for both of those variables we're going to need another equation so that we can get two equations two variables and then solve using some method so let's move on to KCl at node 2 and the KCl no.2 is kind of weird but it will hopefully make sense so KCl no.2 is i2 minus i2 equals 0 and you might be saying well obviously that's that's weird I don't know why you're writing that out but bear with me so let's because the I because there's two currents associated with no two we have the current flowing into i2 from here and the current coming out of i2 down to the reference node and those currents are in magnitude the same but we can define them differently so what do I mean by that well the current going through r2 here is technically v1 minus v2 over the resistance r2 and because no current can flow anywhere else that same current also flows through our three so that's why they're the same but the current through R 3 can be defined as v2 minus zero for ground over the resistance value of r3 so that's where this comes from so yes they are the same current but we can define the current through each one of these resistors which are participating in this KCl equation differently all right so now again we can do some rearranging like we did before and land on an equation like this so now we have two equations and two variables and that means we can solve for it with some method of our choosing for me I'm going to put it in matrix form and so here's what the matrix form is for these two equations so we have a 2 by 2 matrix here which includes all of our our variables and here is our matrix for V 1 and V 2 which we're trying to solve for and here are some of our constants in this matrix and then however you want to solve for this is up to you you can use system of equations you can plug things in whatever is most comfortable for you is what you should use for me I can pump this into my calculator pretty easily so that's probably what I'll do I would do the inverse of of a this matrix times this matrix and that should give me V 1 and V 2 pretty easily and so let's just before I get ahead of myself here we can rewrite this matrix with real numbers so I've converted all the R 1 s and R 2 s and R 3 is into real numbers here and I've converted this expression into a real number over here and so I can solve for it and so I find that V 1 equals 14 volts and V 2 equals 6 volts and because V 2 and V out here are the same point electrically because they're on the same node it means that V 2 equals V o or V out and so V out equals 6 volts across here well that's kind of a lot of work so I'd be lying to tell you that this was the easiest way you could have done this problem so bear with me and we'll now look at a an easier way to do this so I just blew through that last example at like Mach 3 and I don't know if that made it seem easy or if I just confused you and hopefully I didn't confuse you but the way we did that problem was not very smart we kind of went into it blindly in a brutish brute force method we just kind of collected as much data as we could and then prayed that we could find an answer based off that data so let's try to be a little smarter about it and we'll do that in this method again let's start with a KCl at node 1 so the KCl node 1 is going to be the same I 1 minus I 2 plus is and you should know why we've have negatives and positives there and as we start expanding it out into the nodal analysis form the beginning it starts off the same the S minus V 1 over R 1 which written which corresponds to the current I 1 here but now I 2 we define as V 1 over R 2 plus R 3 why did we do that well it's because we lumped together R 2 and R 3 as one resistor because they are in series as we learn before and we can disregard this whole wire kind of flying off here because there's no current in that wire it's not going anywhere it's not doing anything so you can completely disregard that wire and just call this a series connection of r2 and r3 and now what is the voltage across r2 plus r3 well it's the voltage at node 1 up here v1 so it's v1 with reference to ground is the voltage across r2 plus r3 so that's why we have that term there and is is simply just tacked on the end like we did before because we have no other way of defining it all right so now what we can do is we can split up all our fractions and move things over to other sides of the equation and we get all our V ones on one side of the equation and our constants on the other side of the equation and now we can plug in for r1 r2 and r3 vs and is and we can find v1 just like that so we found the voltage at this node is 14 volts of course with respect to ground and so now how can we find the voltage at v2 if we have the voltage up here and we have these two resistors in series how could we do of course we're going to use the voltage divider equation because this is a voltage divider we have two resistors in series we can pluck off the voltage across that second resistor by using the voltage divider equation so that is super simple and all we have to do is say that v2 equals the the source voltage or so-called source voltage in this case it's v1 here times the resistor of interest r3 over the combination of r2 r3 so v1 times r3 over r2 plus r3 and we simply solve for that and we find that v2 equals 6 volts and we know that v2 is the same thing as V out and now without those are the same answers that we got before and it took a lot less work to get to the to get to those two answers so use your head when you're doing these problems try to pick out these circuit elements like circuit elements these circuit blocks like voltage dividers voltage dividers come up everywhere so look for those first look for things like that and it'll make your life a whole lot easier Kirchhoff's voltage law is the complement of his current law and states that all voltage drops must equal all voltage Rises in a closed loop and well what does that mean so imagine that you're going to go skiing you can imagine a voltage rise as a ski lift that carries you to the top of a mountain and as you get higher on your journey to the top of the mountain your potential energy like you know from physics also increases your potential increases so that's a voltage rise and then once you get to the top of the mountain you're up at the top yeah you're going to start skiing down and skiing down the mountain is like passing through resistors in a circuit that's one resistor right there and then you get to this flat point you can think of that as a node even though that really wouldn't be flat but just for the sake of this argument and then the next hill is like another resistor and then you hit some flat point and then the next Hill is like another resistor hmm well let's think back about skiing again so as you as you skilled as you ski down your first hill right here and you get to this flat point you're at a lower potential than you were before up here so you've experienced a potential drop and same thing goes for here you're at a lower potential at this second flat section then you were at the first flat right here so you experienced another potential drop and eventually you get to the bottom where you experience one more potential drop and so you notice that once you've reached the end here you've reached the same potential that you started at at the very beginning before you even started your ascent up the hill over here you're at the same potential potential of zero and so if you're an electron in a circuit and these are all resistors and this was a voltage source which kind of energizes you then if you're if you're an electron you would after you finish your descent you would get immediately back on the lift and start all over again because it's an electric circuit and it keeps going right all right so maybe that's kind of a weird example here's the circuit I was trying to model using the skiing analogy so we have the voltage source here which is the voltage rise and then we have three resistors in series and those would have been the three little hills and those are all voltage drops because if we have a current going in this direction then they're all going to be voltage drops and then this will be a voltage rise through the voltage source so we said that the drops must equal the Rises right that is the sum of all voltages in a closed loop equals zero that's another way of saying that and once again we can express this in a in the summation notation that rhymed as the summation of all the V's equal to zero and again J just tells you J just tells you what voltage we're talking about so voltage v1 v2 v3 whatever one we're talking about and now before we wrote out an equation for each node and ciseaux ciated currents that was in KCl but now we're dealing with a VL and so we're going to write out an equation for each loop and all the voltages associated with that loop so we write that out in the form v1 plus v2 plus v3 and on and on up to V N and set that equal to zero and that's how we write an equation for a loop and that is a KVL equation alright just to clarify a few things say we have like a single loop circuit like this one the first thing we want to do is set up some arbitrary loop current in this case it's called I right here and usually we set up I to go in the clockwise direction and if we had multiple loops in our circuit they'd all go in the clockwise direction and that just helps to simplify the problem so set them all up in the clockwise direction and now if we were to go through a voltage source from negative to positive as we would since we're going in the clockwise direction then we are experiencing a voltage rise and interestingly enough a voltage rise in the KVL equation by convention is actually a minus V hmm got weird so in the KVL equation this would be minus v1 and now if we had a voltage source say in the same place here except we had the positive and the negative switched and we went through that voltage source from positive to negative then we would have a voltage drop and in the KVL equation of voltage drop is a positive V and then as we keep going along and we go through the we go through all these resistors right here all these resistors will have a voltage drop because they are resistors and resistors always have voltage drops and that's a plus all right let's do a problem okay here's our example circuit and we're asked to find VAB across this 2 ohm resistor using KVL so the first thing that we should do is to find the current around the loop so the current that goes around this loop and the current will be the same everywhere because everything is in series with each other so every point in the circuit will have the same current so to do KVL we need to set up our arbitrary loop current in the clockwise direction so we call I that Loop Current and then we want to start at some point in the circuit and perform KVL by going through each one of the elements in the loop and doing that until we return back to that point where we started so this for this circuit we're going to start at point B down here we're going to go through all the elements in the clockwise direction and add up all of their voltages until we get back to B and here is what the KVL equation would look like and we'll explain why so we start at B we go clockwise we're going to go through a voltage source from positive to negative that means it's a voltage drop so you write a positive 15 in the equation and then we get to this 3 ohm resistor right here and we don't really know the voltage across it but we do know that there is a current I going through that resistor so V equals IR as per Ohm's law so you can just simply write plus 3 times I since it's a voltage drop it's a positive 3i and then you go through that and then you come to this 1 ohm resistor and it's also going to be a voltage drop because it's a resistor and it's 1 ohm times the current I so 1 I and then you get up to this 6 volt source and you're going to go from negative to positive on your six volt source so that's in fact a voltage rise so you write a negative six in your KVL equation and then you come to the three ohm resistor and do like you did before and do 3i plus 3i and then you come to your two ohm resistor and do plus 2i and then you return back to your starting point B and once you've reached your starting point you know you've completed the KVL equation so now what we can do is combine all our like terms and we can solve for our current and so we see here that the current I equals negative one amp and what is negative one happening it means it's actually one amp in the opposite direction that we specified so yeah it is in fact in the other direction so now we want to find VAB how can we find VAB well of course very easily we know that V equals IR and so we just multiply the current times the resistance in which a B is across and a B is across this 2 ohm resistor here so we do V ay B equals negative 1 amps times 2 ohms and we get that VA B would be negative 2 volts from A to B which is the same thing as a positive 2 volts from B to a so loop analysis with KVL it's kind of the parallel to nodal analysis with KCl loop analysis is a process that uses KVL to determine loop currents so here's the example circuit that we're going to look at so if we have a circuit like this the first thing we want to do is to assign loop currents to all the independent loops and in this case we have two independent loops and they are these smaller sub loops with inside the circuit those are independent loops and when you see a problem that should be solved with loop analysis it'll be pretty obvious what are the loops there should be like two or three of them maybe four if you're lucky so we want to do is assign the loop currents to both of these so we'll call them I 1 and I 2 and they'll both be in the clockwise direction and that's because that makes life easier all about making life easier here you know all right so to solve for these loop currents we want to write a KVL equation for each of them and then solve using system of equations or matrices or whatever floats our boat all right so now a couple words of wisdom even though I'm not that wise but a couple words of wisdom about solving these if we're in loop one with current i1 then say we're to go through this middle resistor here what's the value of the voltage across that resistor well you would initially say well it's it's R times i1 right close but look I 2 is also passing through that resistor we're not in I 2 by I 2 has to be considered so what is the voltage across this resistor well it's the resistance value R times i1 which is in the positive direction because we're going in the positive direction of I minus i2 because i2 is going any op direction of i1 so i1 is in the positive direction so we have a positive I 1 minus I 2 because it's going in the opposite direction of i1 so R times I 1 minus I 2 is the voltage across this resistor and now say we were in the I to loop it would be basically the opposite I 2 is now in the positive direction because we're going around the loop like this and now I 1 is going in the opposite direction because I 1 is going down through the resistor and I 2 is going up through that resistor and we're going to say that I 2 is in the positive direction because that's the loop that we're in so R times i2 minus i1 for that particular loop hope that makes sense all right so let's work on this circuit using loop analysis we're asked to find the current i1 through this 1k resistor and we're going to set up three different loop currents because we have three independent loops in this circuit so we're going to set up an i1 and i2 and an i3 loop and so let's start up in I 1 and write the KVL equation for that so the KVL for I 1 here it is and how do we get there well say we started in the lower right-hand corner of the eye one loop right here and we went in the clockwise direction as per eye one is in what we get well we go through this voltage source from negative to positive so therefore we have a negative 15 as per the convention and then after we get the negative 15 we come across this 2k resistor right here and what is the voltage across this 2k resistor well it's going to be the current through the 2k resistor times 2k the resistance of the resistor so what is the current through that 2k resistor well it's obviously going to I won because we are in the i1 loop but what other current is associated with it well also I - right because i2 is going through that 2k resistor also now what direction are we end what we're in the direction of i1 so i1 is going to be positive and i2 is going in the opposite direction of i1 so it's going to be negative so how would we write this well just look up here 2 K times the quantity I 1 minus I 2 and so i1 is going from right to left and i2 is going from left to right and so that's kind of justification for why we can write I 1 minus I 2 they're going in opposite directions alright so after we're done with the 2k resistor we keep going up in our loop and nothing there nothing there and we come across the 1k resistor what is the voltage across the 1k resistor well it's going to be the current through the 1k resistor multiplied by eights resistance so what's the current through this 1k resistor well it's simply just I 1 it is the only associated current with this 1k resistor so the voltage is simply 1 K times i1 and then we keep going we keep going we keep going and we get back to our original starting point so we know we've finished and so we finish off our equation by setting it equal to 0 and so now a good next step for us is to look at these resistance coefficients here of 2 K and 1 K and find the greatest common factor between the two of them and then divide by that so in this case we're going to divide by a 1k so let's divide everything in this equation by 1 K and and then simplify further so here's what we get now how did we go from a 15 over 1k negative 15 over 1k to negative 15 milliamps well remember if you divide something in volts which fifteen is in negative 15 is in by something in kilohms you will get something in milliamps maybe something good to remember makes your life easier now 2 K and 1 K are going to cancel to just at 2 and 1 K and 1 K are going to cancel to a-1 and of course 0 divided by 1 K is going to be 0 alright so now we can simplify this a little more and come across our final equation for the KVL for the loop I 1 so we have two variables and one equation so we're going to need at least one more equation to solve for both of those variables so let's move on to the KVL for i2 and we're going to follow a very similar process to find our KVL for i2 so let's start in the same place that we kind of started in I 1 will start in this lower right-hand corner and will move clockwise so what's the equation going to be so here it is and how do we get there so if we're in our loop and we move clockwise the first thing we come upon is the 2k resistor what's the voltage across this 2k resistor well it's simply going to be again the current through that resistor times its resistance so what's the current going through that resistor well it's only going to be i2 there's no other Associated currents nothing else is flowing through the 2k resistor so the voltage is simply 2k times i2 as we have here and then we keep going clockwise around the loop and we come across the second 2k resistor and now the voltage across this 2k resistor is going to be basically the opposite of what we found in the I 1 loop when we were in I 1 loop so up here remember we found that the voltage across that same resistor was 2k times i1 minus i2 the quantity of but now we're going to find as the voltage in this loop is going to be 2k times i2 minus i1 which is basically the opposite and so why is that well now remember that we're in the direction of i2 i2 is our positive direction so it's going to be a positive i2 and i1 opposes I - it's in the opposite direction so we're going to have a negative I 1 because i2 is going from left to right and i1 is going from right to left so that's why we have positive and a negative all right so then after the 2k resistor we hooker right here and we meet up with our 4k resistor so what's the voltage across the four carriers this they're going to be well again it's going to be the same thing it's resistance times the current through it so we're in the loop I - so that takes the positive direction I too is in the positive direction so I 2 is positive again and now we have I 3 also going through that fork a resistor and I 3 is in the opposite direction of I 2 so we write a negative I 3 and after we were done with that resistor we come back to where we started so we've know we know we've finished so our equation equal to zero and let's do a similar process to what we did in I 1 so what we're going to do well find the greatest common factor of all the resistances which in this case is going to be 2k and divide everything by 2 K and so a little more simplification and we land upon something like this so how do we get this five milliamps how did that sneak in there well look at our look at our I three loop over here what is in our I 3 loop we have a current source a 5 milli amp current source so whenever you have a current source in one of your independent loops the value of the loop current i3 right here will automatically take on the value of the current source so actually in this case I three is going to be a negative five milliamps and why is that well you guessed it it's because the five milliamps orse is going in the counterclockwise direction and i3 is in the clockwise direction so to account for that we make I three a negative five milliamps because it's going in the opposite direction of I three of the five milliamps ORS so that's why when we have an i3 right here it's a negative i3 you plug in and negative five milliamps and a negative and a negative make a positive and that's why we get positive five milliamps right here set that equals zero and let's continue our simplification process and we come across our final equation for the KVL for I 2 and now we have two equations two variables we're good to go let's solve for I 1 and I 2 and well actually right now we're just gonna solve for I 1 because that's the only thing we're interested in okay so we can use whatever method we like the best I'm going to use some simple system of equations and land upon I 1 hopefully all right so we take the first equation from I 1 right here and we're gonna multiply it by 2 so that the I 2's right here go away I'm sorry I mean multiply this by 2 so that the ITU's go away negative 4 plus a positive for I 2 it's going to be of course 0 I 2 and then just continue adding everything here as you know how to do I'm sure so we get 5 i1 equals 20 milliamps so of course I 1 equals 4 milliamps and that is our answer so the current through this one care sister is going to be 4 milliamps and that's how you do it using loop analysis all right so we're on the homestretch now and I promise that the rest of video won't be too painful so now let's talk about something called source transformation and before we can do that let's get some definitions out of the way so first thing is called a feminine circuit what's a Thevenin circuit well Devon circuit is just a circuit that takes on this form so that's a voltage source in series with a resistor and the second thing is a nor in circuit and the Norton circuit takes on this form it's a current source in parallel with a resistor and so the slide is titled source transformation so what's that well source transformation is a method that allows us to easily convert between these two types of circuits so that means if I have a Thevenin circuit I can create a Norton circuit which will behave identically in a circuit so what's the conversion process well the first part is super easy the resistor for both of them is the same so our Thevenin and our norton are the same value so from now on we can just refer to the resistance as R because they're the same so if I have a Norton circuit you can find the equivalent Thevenin voltage by using Ohm's law and multiplying the norton current i know by the resistance and that would give you the equivalent Thevenin voltage and if you have a Thevenin circuit you can find the equivalent norton current by dividing V Thevenin by R and it's literally that easy and that's how you can do source transformation turning a Thevenin circuit like this into a Norton circuit like this and vice versa they can really help you simplify down circuits and make them easier for you to analyze using the methods that we've already learned so mister Thevenin and mr. Norton both created theorems that say very similar things basically they say you can take any black box circuit consisting of voltage sources current sources and sisters and can convert them into a Thevenin circuit or a Norton circuit and a a black box is basically just some system or some circuit where you don't really care about the actual components inside of it you may know the components and you may know the schematic for the black box circuit but you don't really care about them because all you care about is how the black box interacts with the outside world what are its inputs and outputs like so you can reduce a super complex circuit to just one source and one resistor and that's all you need to describe the behavior of one of these circuits and as we just learned about source transformation you can convert this black box circuit and to say a Thevenin equivalent circuit and then once you have it in the form of a Thevenin equivalent circuit you can easily convert it to a Norton equivalent circuit and of course you can do it the other way around you can you can convert first to a Norton equivalent circuit and then you can convert to a Thevenin equivalent circuit so it's really up to you to find the Thevenin or an equivalent circuit one of the things we need to find is the value of resistor R and so for what we're doing we're going to be presented with some multi element circuit and we're going to be asked to reduce it down to its equivalent so how do we find our the first thing we want to do is to detach the load resistor if there's one present and how do you identify the load resistor well it's the resistor the output voltage is being taken across or the output current is going through next thing to do is to set all sources equal to zero that means you're going to short-circuit all the voltage sources and you're going to open circuit the current sources and why is I set them equal to zero remember that a voltage is a difference in potential between two points in a circuit and that's what a voltage source creates a difference in potential and so if you were to connect those two points together as you do in a short circuit there's no longer any potential difference therefore the voltage is zero and in a current source by open circuiting in it you prevent the possibility of any current flowing so therefore there is ZERO current and so that's how you set source is equal to zero and that's why and lastly you want to find the equivalent resistance of this new circuit looking in quote-unquote looking in from the two output terminals and you can use your knowledge of series and parallel resistors and combinations of series and parallel to simplify down to just one resistor let's just hone in on seven and equivalent circuits for now we'll get to Norton equivalent circuits a little later but for now we need to find the second part of our feminine equivalent circuit we've already found R now we need to find the Thevenin so what are we going to do well here's our known circuit just for example and the resistor in purple over here rosette represents the load resistor so what's the first step the first thing I want to do is to detach the load resistor again as we've done before in the previous step finding R and the second step is to find the open circuit voltage vo C across the circuit two output terminals so it's really as simple as that now that process can be pretty complicated because you might have to use nodal loop voltage divider or source transformation any of those processes to find vo C but that's really all you have to do is have to detach the load resistor and then find what the voltage across the output terminals would be and this includes all the sources of course the sources are not set equals zero when you're finding V Thevenin as usual let's do an example to really understand what we just learned so we have a circuit and we're asked find VAB the voltage across these two output terminals using Thevenin theorem so what's the first thing that we're going to well let's find our if evident or just our whatever one you want to call so here's how we find our Thevenin we've in this new circuit over here well what have we done well we've removed the load resistor of six K and that's the load resistor because a B is across it and then we've short-circuited our three volt voltage source and we've open circuited our two milliamp current source and now all we have to do is find the equivalent resistance of this new circuit looking in from these terminals so we get an equivalent resistance of three kilohms easy now the next part is me a little more difficult but it's going to be finding v7 so what are we going to do well here is a circuit we're going to work on we've moved this circuit down here and we've removed the six kilo ohm load resistor and I've labeled this as node 1 and the voltage at node 1 being v1 so there's a bunch of different ways you could you could do this to solve for V OC but here's what I'm going to do it let's do the KCl at node 1 so we have 2 milliamps going into the node and we can't have a current here because this end is disconnected so we can only have a current through the 2 K and the 1k like so so the KCl is simply going to be 2 K into the node and then out of the node comes a current of v1 since that's the voltage here over the equivalent resistance of 2 k plus 1 K which is 3 K and then and set that equal to 0 because those are the only two currents and then we can simplify and we get that the voltage at v1 is 6 volts so basically you can think of it as a six volt currents or a voltage source between here and here so now how can we find V OC let's try a KVL and we can do a KVL on this loop right here even so vo C is totally an open circuit you can still do a KVL on it you can you can create a loop that includes that voltage so here's how the loop turns out and how do we get a negative six volts we'll remember if this is technically a voltage source between here and here you're going to go through the negative to the positive here of that voltage source set so that would be negative six volts and then you get a negative three volts going through the voltage source here and then you get a positive vo C going through the voltage source and then you return back to where you started so that's how we get that KVL and then you can simplify and add and you get that V Thevenin which is the same thing as vo C equals nine volts but we're not done because we just found vo C and or we found V Thevenin and we found RF Devin but now we need to find V a B which is the voltage across this 6k resistor right here so let's let's move on to the right over here and this is our circuit now including the load resistor because we can't just disregard that load or sister it's still part of the circuit here's our equivalent R Thevenin equivalent circuit includes the nine volt source right here and the 3k resistor in series and then we add in the 6k load resistor so we're asked to find the voltage across that six k-loader sister and well how can we do that well it should be obvious by now if we're going to use a voltage divider and that's real easy because all you have to do is multiply that source voltage of nine volts butter's by the r2 over r1 plus r2 combination and that's going to be a V a B of six volts so as you can see kind of a lot less work than doing it in a more traditional manner but we still get to the same result and we can probably do it quicker Noren equivalent circuits so to find I nor in which is the parallel to V Thevenin we follow a very similar process with some small changes so we have a known circuit and again the first step we're going to do is to detach the load resistor just as we did before and now the second step is to short-circuit the output terminal so we didn't do this before before we had an open circuit but now we're going to short-circuit the output terminals meaning we take a piece of wire and directly connect those two terminals together and the third step is to find the short-circuit current is see through the shorted output so the current through that piece of wire that we just put in there whereas before we found the open circuit voltage between those two terminals now we're finding the short circuit current and now you can use again any form analysis to find that short circuit current nodal loop voltage divider whatever is the quickest and the easiest or whatever you understand the best let's do the same problem we did before using Thevenin theorem and now let's use Norton's theorem so here's the circuit from before and we're asked to find V a B and now we're going to use Norton's theorem to find a VAB how is the process going to be similar and how is it going to be different well the first part finding our Norton or RF Evan the same thing is going to be identical so we just short circuit our voltage sources open circuit our current sources and remove the load resistor find the equivalent resistance and that's what we did here three kilohms same as before now we have to find I nor in so how do we do that well I redrew the circuit down here except now I removed the six kilohm load resistor I have short circuit the output next I'm going to do is just for me visualizing I've moved the voltage source over there just so I could see a little better you don't have to do that personal taste and now what do we have here well we have a current source in parallel with a resistor what can we do with a current source in parallel with a resistor that we just learned of course we can do source transformation on it and make it a voltage source in series with a resistor and that's what we did here so we had a two milliamp current source in parallel with a 3k resistor to find the V Thevenin of that you just have to multiply two milliamps times 3k and you get six volts and now you put that 3k resistor and you put it in series with that new six volt voltage source and now we have a one loop circuit with two voltage sources and one resistor what could be easier and what we want to find is the current through that loop because the current in that loop is going to be ISC and that's what we're interested in so let's do the KVL on that circuit so let's start up here on the top on the upper right and we go through the three volt source so that's going to negative three come around here go through the six volt source and that's going to be a negative six and we come across this resistor and what's the voltage across this resistor you guessed it it's 3k times the current ISC and we set that equal to zero because we know we're finished and so we can reduce and simplify and we get that is C equals three milliamps so that is our short-circuit current three milliamps and that's also equal to I Norton so now let's rewrite our circuit in its Norton form and now we can solve for V a B so here it is we have our three milliamp source which we just determined was three milliamps in parallel with the 3k resistor and now again as we did before we're going to tack on that six K load resistor right here because that has to be included in our analysis to find VAB and now how can we find the voltage across a be real easy find the equivalent resistance of the 3k and the 6k in parallel which turns out to be 2k and then do V equals IR and that is going to be three milliamps times the 2 kilo ohm resistor the equivalent resistance between 3k and 6 K which is 2 K so 2k times 3 milliamps and we get 6 volts across a B just as we got before using Thevenin theorem and now we've just used Norton's theorem to find the same answer so it's really up to you which one do you like better and use whatever one makes more sense for for you alright so there's light at the end of the tunnel now thank you for sticking with me we're almost done the last topic for today is superposition theorem and it's really a clever theorem so in a circuit you can think about voltage sources and current sources as driving forces each which contribute to the circuit in some way so superposition basically says you can consider the circuit one source at a time find each sources contribution to the circuit and then add up all those contributions at the end and that will be the actual contribution so here's the general process for doing that superposition says that a circuit with multiple sources can be solved by this process first thing we want to do set all sources equal to zero except for the one that we're consider and we know how to set all sources equal to zero then you want to solve for the necessary currents and voltages depends on the problem of course using only that one source which is actually going to be relatively easy and then you want to repeat step two with the next source and then the next source and however many sources you have and then at the end you want to quote superimpose the solutions on to each other basically the sum of all the solutions now you're going to have a bunch of different equations with seemingly the same variable it's going to be like you know I 1 I 2 or whatever but you need to differentiate between the variables from the different equations because they're not the same so what you want to use is the prime symbol to differentiate variables with the same name so here's the general process we have a circuit looks like this one of these double loop circuits classic right and here's what we're going to do we're going to take out one source so in this in this case right here we've taken out this voltage source and we're going to solve for I 1 single Prime and I 2 single Prime and then weave in in the right circuit over here we've taken out this voltage source and we solve for I 1 double Prime and I 2 double Prime and then at the end the big finale to actually find the real value of I 1 all you do is add up I 1 single prime plus I 1 double Prime and you get the actual value for I 1 same thing for I 2 so that's the process now of course let's do one last example last hoorah alright last problem for the day let's try to find vo the voltage across this 6k resistor using superposition so what's the first thing that we're going to do we're going to draw the circuit with one of the sources removed and the first source we're going to remove is the 12 volt source so that's going to get short-circuited so that's what happens here and now you're probably wondering well where did the where did this where these two 6k resistors go remember now if i short circuit this 12 volt source what am i also short-circuiting I'm also short-circuiting these two 6k resistors so they are also short-circuited so they go away and you just replace this entire circuit block right here with one short circuit which is what you see right here and now we're interested in solving for vo single prime right there what have I done to help us out a little bit I put in a current i1 and so what type of circuit are we looking at right here well we're looking at a classic current divider and how are we looking at a current divider because we have a current source and these two branches right here they're in parallel so that's a classic example of a current divider so if you can think back to the equation of the current divider we're going to find that I 1 is defined like this and it's equal to everything that we're not interested over the total resistance times the current source and that's what we have here so we're interested in the current i1 which is going through that one 6k resistor so what we put up top is the 2 6k resistors in series that's the equivalent resistance of that branch and you put it over the equivalent total and that's 6 k + 6 k + 6 K and then multiply that by the value of the current source and you get the current i1 lovely so simplify that down you find that I 1 equals 4 milliamps now how do we find vo single Prime well easy V equals IR but now you just have to take into a fat into account that I 1 is flowing from negative to positive of V o single Prime so that we basically just flip the voltage so it's a negative 24 volts negative 4 times 6 equals negative 24 volts so that's vo single Prime now we have to find vo double Prime and so what's how you can do that well we're going to take the original circuit and instead of deleting the voltage source now going to delete the current source so let's draw that out so it's going to look like that so we have an open circuit where the current source used to be and now we call this voltage vo double prime and upgrade out these two 6k resistors and these these two six characters really have some bad look wide I gray those out well if I have a single loop here yet - if you really notice right here you have a single loop right here what happens in this loop has no bearing on what happens in this loop so we're just going to disregard whatever happens in that loop treat this as a single loop and if I want to find the voltage across this 6k resistor what can I do well I can do a number of things I can do case I can do KVL or probably something else but if right now I'm going to use voltage divider I'm really into using my divider right now so how do I use voltage divider well I just do my source voltage of 12 volts times the resistor that I'm interested in which is this 6k resistor but it would be the same actually for any of them over my equivalent resistance of all the resistors so that's 6 K + 6 k + 6 K and that will give me the voltage across that 6 K resistor the voltage vo double Prime and that equals 4 volts so I've just calculated that vo single prime equals negative 24 volts and vo double prime equals 4 volts how do I find out the real value of vo real easy all you have to do is add the two together vo single prime + vo double prime equals vo and so of course that is going to equal negative 20 volts real easy and it was pretty quick all right great work here are my ending remarks for this video first thing the key to solving circuit problems quickly and correctly is practice by far the most important thing you can do is keep practicing second thing each technique that we've learned today is not terribly difficult on its own the challenging part is really identifying which technique or combination of techniques is appropriate for a given circuit so someone can ask you solve for this current using this technique and it'll probably be pretty easy because you're going to know exactly what you have to do but if someone just says solve for this current and they don't tell you what technique to use that's where the challenge lies that's really important to be able to identify which technique makes the most sense third thing experiment with circuits on your own can't stress this enough play around with circuits and see how changing things in a circuit will affect it and you can purchase the components that we used in this video even though we didn't use real components we just put them up on a screen but we can you can purchase the real components from distributors like Mouser and digi-key and they're pretty inexpensive probably only a couple cents each for those resistors and you can create a voltage source using batteries or if you have a power supply that's even better and you can see how changing things around will affect the circuit in a similar way you can use simulation software like ltspice which is free and you can download from linear technologies website to verify your findings and to learn more about circuit behavior so if you have homework to do or if you're just analyzing some circuit and you want to make sure you did it right you can put it in ltspice and check if your answers match of course you can always play around and see how changing values will change your circuit all right so I really encourage you to do those those things and keep practicing and I'm sure you'll be great at circuit analysis as always thanks for watching I know my videos are long and if you sat through this whole thing thanks so much but I hope it was helpful for you I'd love to hear your feedback if you have any what would you like to see more of what did you not enjoy so much I'd like to hear that too and if you liked it give it a thumbs up and if you want to see more videos definitely subscribe to my channel alright guys have a great day

Channel: Solid State Workshop

Views: 3,386,939

Rating: 4.7227306 out of 5

Keywords: Circuits, Circuit Analysis, Electronics, Network Analysis, Computer, Physics (Field Of Study), Circuit Theory, Electrical Engineering, Arduino, Linear Circuit Analysis, Linear Circuits, Electrical Network (Invention), Analog, Technology, Math, KCL, KVL, Kirchhoff's Circuit Laws (Field Of Study), Superposition, Voltage Source, Current Source, Voltage Divider, Current Divider, Nodal Analysis, Loop Analysis, Mesh Analysis, Source Transformation, Thevenin, Norton, Ohm's Law

Id: sqxzQkAdJm0

Length: 96min 51sec (5811 seconds)

Published: Sun Mar 22 2015