[Music] [Music] hi everyone in today's lecture we wanted to show you some uh simple ways of solving these you know circuits that we have been seeing over the last few days specifically we are we have been using you know kvl KCl as part of mes analysis nodal analysis to solve some very simple electrical networks I thought in today's class I'll start off with some maybe a couple of very simple techniques that will help you get practice and then we'll go through one practice problem to see how to solve it using superposition because that is the latest technique that we have learned so we will first look at the two techniques that will keep appearing again and again when you solve these types of electrical networks we will solve them long hand in long form and then learn to use it in you know some applied problem so the first one is what happens when let us say you have a voltage and two resistors so I'll draw it in this form so let us say I'm now going to start with numbers and we'll abstract it out next so let us say I had a 1vt voltage source and um let us say I had a 2 kilm resistor and maybe another 2 kilm resistor let's make it the easiest possible now I want to find out for this network what the voltage across the bottom 2 km resistor is maybe I'll just call it some unknown voltage volage VX well we know how to solve this it's actually fairly simple you have a single mesh I'll call the current through that I and we can say that this current I is nothing but the voltage going through the loop right which is 1 volt the voltage increase over the sum of these two resistors so in other words if you are let me maybe take one step back and even write out the mesh starting from here and ending back at the same point going clockwise so I have a voltage increase of 1 volt so + 1 volt I have a voltage drop of 2 kilms * I minus 2 kiloohms * I and again a second voltage drop - 2 kilohms * I and I come back to the same point so KV L around the loop tells me that the potential difference is is zero around the loop so this tells me that the current I is nothing but 1 V over 4 kiloohm which is25 milliamp what is the value of VX well VX is simply 2 kilms * I in this polarity so is simply 2 kiloohms * I so this is 2 kilohms time25 milliamp so this is nothing but half a volt in fact the voltage drop across the other 2 km resistor is also half a volt and of course the sum of those those two voltages has to be equal to the input battery voltage but now imagine I decided to change this to some other resistance value how would you find out so we need a more systematic way of doing this so what I'm going to do I'm going to abstract it out so I'm going to say that I have a voltage source V and I have two resistors R1 and R2 and I want to find the voltage across one of these so let us say I'll call this as the voltage across R2 and I want to find out what VX is we'll follow the same procedure we will start from here go around the loop and come back clockwise and let us say that this current is I I think you can see where we are going with this so we will say that V is the input voltage maybe I'll give it a name so that we don't get confused so I'll call this V1 so the input voltage V1 over R1 + R2 what did we have previously it was current was 1 volt over 2 KMS + 2 kilms what is VX VX is simply R2 * I so VX is simply R2 by R1 + R2 * V1 it is simply R2 * I what is the value of v y if I were to call this as v y what is the value of Vy VY is nothing but R1 * I so v y is nothing but R1 by R1 + R2 * V1 what is V1 V1 has to be equal to v y + VX clearly the sum of these two voltages is nothing but R2 by R1 + R2 + R1 by R1 + R2 which is R1 + R2 over R1 + R2 which is just one so the sum of VX + v y is simply equal to V1 so what this tells you is that there seems to be some relationship between the input voltage applied and the voltage across some resistor in this pair of resistors the voltage across any particular resistor seems to be the value of that resistor over the sum of the resistors in fact you can now extend this so this is a very useful thing to know because you can can quickly write down voltages and uh in in networks so for example I now give you this network I tell you that there are three resistors and I tell you that input voltage is V1 and I tell you that the resistance values are R1 R2 and R3 and I want to find out the voltage across R2 well I can tell you that this voltage has to be equal to R2 * the current and the current through this is nothing but this potential difference V1 over the sum of all these resistors because remember resistors in series add up we have seen this before so the total resistance across V1 is R1 + R2 + R3 the current drawn from V1 is V1 by R1 + R2 + R3 so the current current through the loop is R1 V1 by R1 + R2 + R3 the voltage across R2 is nothing but R2 by R1 + R2 + R3 * V1 so I'm guessing that you can see the pattern here if you have many resistors in series placed across a voltage source the voltage across any one of them is the value of that resistor to in this case over the sum of all the other sum of all the resistors including that particular resistor times the voltage across all of them now you should imagine that I can extend this out to a current source as well because this doesn't have to be only for a voltage source I should imagine in fact you will see this happening again and again any particular property that you see with a voltage source you should automatically imagine that there exists some similar or analogous property with current sources for example I take a current source and I have a resistor now of course the voltage if I take a current source i1 and a resistance R1 the voltage across the resistance is simply i1 * R1 because I know this current has to flow through R1 now the situation with the current source is slightly different so a similar property exists if you put another resistance in parallel because remember just like the sum of the voltage drop across R1 R2 R3 is equal to the input voltage V1 you can say that there needs to exist a similar property for currents except remember that currents distribute across nodes so remember you apply KCl the sum of currents entering the node is equal to some of the currents leaving the node so you should imagine that a similar property exists if you were to take resistors in parallel because now if I thought of this current as i1 this current as I2 clearly clearly the input current I I'll call the input current as I the current I has to be equal to i1 + I2 if you were to apply KCl here and the second thing you can say is that because R1 and R2 are in parallel they have the same voltage VX across them what is that voltage VX VX simp simply is either i1 R1 or I2 R2 they are both equal by definition why do I say they're equal by definition very simply if you start from this you get a potential increase i1 R1 you get a potential drop I2 R2 you come back down to the same node so clearly the potential drop across R1 has to be equal to the potential drop across R2 so in other words i1 R1 has to be equal to I2 R2 I now have actually two equations so I can solve for i1 and I2 let us say that I'm interested in the current through R1 let us say I'm interested in i1 so the I want to solve for i1 I have two equations i1 R1 is I2 R2 and of course the second equation is that i1 + I2 is I this equation tells me that I2 is nothing but R1 by R2 * i1 all I have to do is take this and put it back here so this tells me that i1 plus R1 by R2 * i1 is I or so R2 + R1 * i1 / R2 is I in other words i1 the current through R1 is equal to R2 by R1 + R2 * I now this is interesting this is telling you that the current through resistor R1 in some way depends on both the resistors R1 and R2 through the denominator but you can see that in the numerator it depends on R2 and this is a common observation basically if you have two resistors one of the being let's say much larger than the other for example let's say R2 was much larger than R1 this seems to show that if so I'll write it down in fact if R2 were much larger than R1 this seems to tell you that I2 will be much much smaller smaller than i1 and this should be obvious because the potential drop across the two resistors has to be the same the potential drop is nothing but the product of the current and the resistance if you have a larger resistance connected in parallel with a smaller one you should expect that the current through the larger resistance is smaller so that the potential drop is the same across both and that is what this equation is also telling you this is telling you that the current through the resistance R1 is actually R2 by R1 + R2 * I this is slightly different from this expression for the voltage across R2 the voltage across R2 was R2 by R1 + R2 * V1 whereas the current through R1 for a parallel combination of resistors is R2 by R1 + R2 * I so if you have two resistors in parallel the numerator on this side will actually be the other resistor so there are SE many properties such as these there are also some properties that will help you actually simplify Networks and I just want to tell you two of them the first one is to realize that if you have a connection like this and then you are connecting to something else some Network n let us say I'm going to abstract out that Network n some voltage V and some resistor r as far as the network n is concerned this is exactly the same as having a voltage source without the resistance R so if I call these nodes some N1 and N2 in other words a resistance across a voltage source really does not affect the rest of the network why do I say this very simple the voltage across the resistance is of course V the voltage across the network is also V therefore in both cases the current through the network does not change I'll call that I in now the current drawn from the voltage source changes because now I have some current IR but as far as the network is concerned nothing changes this is also a very useful property the current drawn from the voltage source in the two cases is of course different but as far as the network is concerned it's exactly the same now you can now imagine as I told you earlier you can draw similar properties for current sources so now you can imagine I have a network uh n Prime I am going to connect a current Source I across two nodes N3 and N4 and let us say this is some current I and some resistance R what is the current flowing through N3 it is I what is the current coming out of N4 it is I and therefore you can say that this is exactly equivalent to having this current flowing through this into this network n Prime at N3 and N4 without the resistance you have the same current I flowing into N3 and coming out of N4 as far as the network is concerned there is no change at all what changes between these two networks again in the earlier case the current drawn from the voltage source was different you can now imagine that if there was a voltage drop V here or I'll call it VN VN Prime you will of course have the same voltage drop VN Prime here but in the first case you had a voltage drop VR across the resistor and that voltage drop sum of those two voltage drops will now appear across the current Source but remember that we are talking about ideal voltage source ources and ideal current sources here what was the characteristic of an ideal voltage source the characteristic if you look at the vi characteristic of an ideal voltage source it was parallel to the x-axis let me call this voltage some V1 the voltage source ideal voltage source can actually Supply any current that is required without changing its voltage as long as the voltage applied to the network n stays the same the state of network n will look the same and similarly if you take the current Source the current source i1 will have a characteristic like this as long as that same current i1 is being fed to the terminals of NP Prime here the state of NP Prime will remain the same so it turns out that this is a very useful property to solve Networks