 In this video, we're going to look at an unbalanced bridge circuit. Now, let's just take a look at what we're dealing with right here. We've got three resistors in series, this one, this one, and this one, and those three resistors in series are in parallel with these three resistors in series, this one, this one, and this one. Now, things can get a little bit funky, especially when we add a bridge to our circuit. By a bridge, I mean basically putting a jumper across. Let's do that. What I mean by that is, if I took a jumper and I went from B to C and I put that jumper there, that's going to throw everything into whack, because typically what normally would happen before I did that, let's just get rid of that there, is current would flow through here and after that node, it would also split and go across there. Now, by adding that jumper in, it messes everything up in the sense that current can now go through here, through here, and can either go up this jumper and out that way, or it can go through this jumper and go down this way. It just messes everything up. What we need to do first off is we need to redraw the circuit so we can understand a little better how this all works. I've kind of neatened the circuit up there for you. I've got the same resistors are all in place, except now I've just made it so it looks nice and straight and easy, kind of tidied it up a bit. Now as with all circuits, the best thing we can do to it is throw some actual numbers at it. So let's pick some numbers here and see what happens. So let's just pick a nice easy voltage. We like that this voltage of 120 volts. Then let's go down here and pick some easy ohmic values that say 5 ohms here, 10 ohms here, let's say, I don't know, let's say 40 ohms here, just to be a little crazy. Down here we're going to go absolutely nuts. We're going to say, let's say this is 150 ohms, I know it's crazy, 700 ohms and 500 ohms. So that's our first step as we got these ohmic values. Now the next step when we're dealing with these is you notice that this is basically going to be a series parallel circuit in the sense that current can come down to this node, split across here, and then it's got to also when it leaves or enters this node we've got to go through these two resistors and when we get down here we have to go through those two resistors. So let's go ahead and add these together. So when I do that I'm going to get 10 plus 5, and I'll do the math for you, 15, and then 700 plus 500 down here is going to be 1200. So we're going to get rid of these and get rid of these. I'm dealing with these circuits and I say this to my students all the time is it's so very important to redraw these because you can get yourself really behind or really confused if you just keep trying to work off the same drawing. So what we're going to do now is I'm going to take us to a new drawing and see what it looks like. So here we go. I've taken those two resistors that were in series, I've crushed them into one, and those two resistors that were in series and crushed them into another. So let's get our values written in again. So we have 120 volts was the voltage we like to deal with. Then we're dealing with down here 15 ohms. This was 40 ohms. This is 150 ohms and this was 1200 ohms. So we've got the series out of the way. Our next step now is to crush the parallel. Now with this branch in here that means that these two resistors are in parallel and these two resistors are in parallel because current can go through here and out or current can go through here and through straight through. So these two will be crushed together and these two will be crushed together and then therefore these ones will be in series with these ones. So our first thing is to add these together. So we're going to use the reciprocal 1 over RT equals 1 over R1 plus 1 over R2. So we're going to go 1 over 40 plus 1 over 1200 gets us our answer. And we will get our answer of 38.7 ohms crushed together. So 1 over 40 plus 1 over 1200 equals 1 over 38.7. I will do the same thing over on this side. We're going to go 1 over 15 plus 1 over 150 equals 13.6. So let's just get that drawn in there. 13.6 ohms. So again 1 over 15 plus 1 over 150 gives you 13.6 or 1 over 13.6. 1 over 40 plus 1 over 1200 gives you 1 over 38.7. And again like I tell my students we want to crush this down and redraw it. So let's see what it looks like redrawn. So I've crushed it even down to a very simple circuit. And again this was 120 volts that never changed. Down here this one became 13.6 ohms and this was 38.7 ohms. And stick with me because this all kind of builds back out. So we're going to go 13.6 plus 38.7 and that gets me an overall RT. So my total resistance in this circuit right now is 52.3 ohms. From there we can get our current because I'm just going to go 120 divided by 52.3 and I get a current of 2.29 amps. Let me just change my pen color here and it just so it looks a little easier. So this is going to be 2.29 amps. We can use that current now to figure out what the volt drop is across each resistor. So we're going to get the volt drop across this resistor and the volt drop across this resistor. So if I do that 2.29 times 38.7 equals this voltage here is going to be 88.6 volts. When I do the same thing for the 13.6 ohm resistor I'm going to go 2.29 times 13.6 equals that will be 31.1 volts. It's actually 31.144 but who can we just want to round and make it fairly easy for ourselves. So the next step is with any kind of combination circuit is we're going to kind of pull our circuit back so we're going to start stretching it out again. So this 88.6 that's going to go across each of the voltage from before from the side that we had before where we had the two in parallel in series with the other two in parallel. So let's see what that looks like now. So this is what we had before all I've done is kind of uncrushed it. So these two that were in parallel are now back in parallel so they were crushed before to give us our voltage before and our ohmic value before. So now we're just going to transfer those voltages over from the previous side that we were dealing with. So this one here these voltages were 88.6 volts. This is going to be 88.6 volts because they're in parallel and over here we have 31.1 volts and over here we have 31.1 volts. So we're getting the same voltages on each one because they are in parallel from the previous slide. We're getting close to being done so still hang on with me here. What we've got now is we can definitely have enough information to figure out what the jumper current is going to be at this point. We don't have to stretch it out any further from here. What we're going to do is work out what the current is across each one of these resistors and then we're going to see what happens. So I'm going to go off and just calculate out what the currents are and put them in here but basically I'm going to be going 31.1 divided by 15 to get this current 88.6 divided by 40 to get this current 88.6 divided by 1200 to get this current and 31.1 divided by 150 to get this current. So there you go I've got the math done for you. So 31 volts divided by 15 gave us roughly around 2.07 amps 88.6 divided by 40 gave us 2.2 amps 88.6 divided by 1200 gave us 74 milliamps and 31.1 divided by 150 gives us 208 milliamps. So what we need to do now is look at these nodes because they are going to come into effect and be very important when we're trying to determine what the current is through this jumper. So let's get our nodes drawn in as we always should do. So there's a node there and a node there. If we look at this I have 2.2 amps coming along here and then suddenly I have 2.07 amps so something had to drop off I lost some current through this branch and if I did the calculation I basically lost 130 milliamps going down through this. So there's 130 milliamps that is traveling through this jumper to get down here. Now let's see if that plays out because whatever happens down here better add up also on this branch. So on this branch I have 74 milliamps and then I have 130 milliamps and so therefore it adds up to be roughly around 208 milliamps so it all works out. I have current dropping off and then I have current picking up across that bridge so we can easily say that the current in the bridge right here is 130 milliamps and it is traveling from this point down to this point. So what we can do now is we can go back to our original drawing and I can put that current in there and we can talk about which way the current is flowing. So now if we go back to our original circuit from B to C we've determined that we have a current through there of 130 milliamps and we know that current now is traveling in this direction and current travels from negative to positive in electron current flow therefore it's traveling from negative which would be B to the more positive point of C and that is how you calculate the current in a jumper in an unbalanced bridge circuit.