Professor Anderson

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Hi everybody. How you doing? Good. Let's take a look at resistors a little bit more, we started talking about resistors in series and resistors in parallel yesterday, let's try one more example of that and see if we can make it pretty complicated, and how we might simplify it Again, if you are watching this live at home feel free to chime in, we've got Hilda here manning the moderator station, and so if you have a question just fire it off to her and she'll relay it to me. Okay let's try this example right here. So we have a voltage V, and now we have a very complicated set of resistors. And let's label these R1 R2 R3 R4 and R5. And remember the goal here is to get that entire circuit into something that looks like this. One voltage and one resistance, R equivalent. And we need to find that R equivalent, what is the equivalent resistance of this circuit? And then once we do that, of course, we can determine the current in the circuit. Okay. How do we do that? What we said last time, let's start from the innermost point, the most complicated innermost set of resistors, and it looks like that's going to be these two in parallel right here. So let's redraw the circuit as the following: R1, and then it splits, we have R2, and then those two in parallel, we are going to replace with Rp. And we'll call it R p 3 4, which means 3 & 4 are added in parallel. The other side of that wire went down to R5, and then the wires came back together and off they went back to the battery. Alright, that's the first step. How do I identify that R p 3 4? Well we know that they add in parallel. So 1 over R p is gonna be 1 over R 3 plus 1 over R 4. So yesterday, we said at this point you could just plug in some numbers and then take the inverse of that to calculate R p 3 4, that's perfectly acceptable. Alright so now we have a simpler circuit, and now we have to, again, go into the innards and it looks like this 2 in series right here is going to be important. So let's redraw the circuit, battery, R1 and then it splits, and we're gonna have some R s in series on that side. We still have R 5 over on the other side, wires come back together and goes back to the battery. Alright, so you're going to get a value for R p 3 4 from this one, how do you get the value of R s over here? Well those are just two added in series that's not too bad. So it's R 2 plus whatever we found in the first step, RP 3 4. Alright. Now we're dealing with this circuit, and let's wrap around the board here and we'll draw it over on this side. Just like in the old game asteroids right? You go around one side of the screen you immediately come back in on the other, Anybody remember that game? That was like, the popular arcade game when I was a kid, we used to go down to the 7-Eleven, put in our quarters into the asteroids game,. And it was always one guy that had like way more points than anybody else, he would sit there for like three hours playing the game and everybody else stood there, drinking their Slurpees going wow he's so cool. Okay, what are we talking about? Alright, yeah, circuits. Right, so let's take that circuit and bring it back over here, and let's simplify it again. So R1 is going to stay where it is, but now we have two that are, again, in parallel. So we need to draw a parallel resistor and it's going to be parallel combining s and 5. And what is this R p s 5? Well again we're adding them in parallel, so R p s 5 is going to be 1 over R s plus one over R 5. And we found RS from this step and so you can plug it in there, R 5 would be given to you, plug it in there, you take 1 over those, add them up and then you take the inverse to calculate R p s 5, and now we're at the last step. The last step is our basic circuit, one resistor, and now we can call that R equivalent. That's what they mean when they say find the equivalent resistance, simplify it down to one circuit element. and that's your R equivalent, and now R equivalent is those two things added in series. So it's R 1 plus R p s 5. So, there is a lot buried in that R p s 5 of course, it has all the other resistors buried into it in some complicated fashion, but if you go through step by step and just calculate numbers, at this point you would just have a number for R p s 5 and you can plug it right into there. And now if you're trying to find something like the equivalent current, well we just go back to Ohm's law. I is V divided by this equivalent resistance. This would be the resistance of the circuit, this would be the current in that circuit. And like we said before, you can do this for extremely complicated arrangements of resistors. You can always reduce it to a simple circuit.

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Hi everybody. How you doing? Good. Let's take a look at resistors a little bit more, we started talking about resistors in series and resistors in parallel yesterday, let's try one more example of that and see if we can make it pretty complicated, and how we might simplify it Again, if you are watching this live at home feel free to chime in, we've got Hilda here manning the moderator station, and so if you have a question just fire it off to her and she'll relay it to me. Okay let's try this example right here. So we have a voltage V, and now we have a very complicated set of resistors. And let's label these R1 R2 R3 R4 and R5. And remember the goal here is to get that entire circuit into something that looks like this. One voltage and one resistance, R equivalent. And we need to find that R equivalent, what is the equivalent resistance of this circuit? And then once we do that, of course, we can determine the current in the circuit. Okay. How do we do that? What we said last time, let's start from the innermost point, the most complicated innermost set of resistors, and it looks like that's going to be these two in parallel right here. So let's redraw the circuit as the following: R1, and then it splits, we have R2, and then those two in parallel, we are going to replace with Rp. And we'll call it R p 3 4, which means 3 & 4 are added in parallel. The other side of that wire went down to R5, and then the wires came back together and off they went back to the battery. Alright, that's the first step. How do I identify that R p 3 4? Well we know that they add in parallel. So 1 over R p is gonna be 1 over R 3 plus 1 over R 4. So yesterday, we said at this point you could just plug in some numbers and then take the inverse of that to calculate R p 3 4, that's perfectly acceptable. Alright so now we have a simpler circuit, and now we have to, again, go into the innards and it looks like this 2 in series right here is going to be important. So let's redraw the circuit, battery, R1 and then it splits, and we're gonna have some R s in series on that side. We still have R 5 over on the other side, wires come back together and goes back to the battery. Alright, so you're going to get a value for R p 3 4 from this one, how do you get the value of R s over here? Well those are just two added in series that's not too bad. So it's R 2 plus whatever we found in the first step, RP 3 4. Alright. Now we're dealing with this circuit, and let's wrap around the board here and we'll draw it over on this side. Just like in the old game asteroids right? You go around one side of the screen you immediately come back in on the other, Anybody remember that game? That was like, the popular arcade game when I was a kid, we used to go down to the 7-Eleven, put in our quarters into the asteroids game,. And it was always one guy that had like way more points than anybody else, he would sit there for like three hours playing the game and everybody else stood there, drinking their Slurpees going wow he's so cool. Okay, what are we talking about? Alright, yeah, circuits. Right, so let's take that circuit and bring it back over here, and let's simplify it again. So R1 is going to stay where it is, but now we have two that are, again, in parallel. So we need to draw a parallel resistor and it's going to be parallel combining s and 5. And what is this R p s 5? Well again we're adding them in parallel, so R p s 5 is going to be 1 over R s plus one over R 5. And we found RS from this step and so you can plug it in there, R 5 would be given to you, plug it in there, you take 1 over those, add them up and then you take the inverse to calculate R p s 5, and now we're at the last step. The last step is our basic circuit, one resistor, and now we can call that R equivalent. That's what they mean when they say find the equivalent resistance, simplify it down to one circuit element. and that's your R equivalent, and now R equivalent is those two things added in series. So it's R 1 plus R p s 5. So, there is a lot buried in that R p s 5 of course, it has all the other resistors buried into it in some complicated fashion, but if you go through step by step and just calculate numbers, at this point you would just have a number for R p s 5 and you can plug it right into there. And now if you're trying to find something like the equivalent current, well we just go back to Ohm's law. I is V divided by this equivalent resistance. This would be the resistance of the circuit, this would be the current in that circuit. And like we said before, you can do this for extremely complicated arrangements of resistors. You can always reduce it to a simple circuit.