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Circuit Current Calculation · Acedra
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Circuit Current Calculation
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0:00
So some guy comes up to you in a dark alley and he says "Hey kid, you want a dollar?" And you say "Sure!" and he says "Well get me the current going through every one of these resistors." And so he draws the following circuit diagram for you. There's a resistor there and then there's the funky point right here, it doesn't look so funky yet, but he's just dropping resistors all over the place and you know it's going to be a long evening. But you only got fifteen minutes to solve the AP problem
0:29
so you better get working, I mean the problem posed by the strange guy on the street, who's gonna give you a dollar if you get it right. We got resistors all over the place. So this is our first picture of this circuit diagram and he starts giving you some numbers. Turns out he's a pretty nice guy you talk about his kids everything's fine. There's a 12 volt battery powering the circuit and he wants to know, want current, and I'm gonna
0:57
choose myself a color here, he wants current through each thing and voltage across each resistor. So that's the language that we use the current goes through a resistor and the voltage is across a resistor. Voltmeters would come on here like,like this. I'll just draw you a voltmeter here's the voltmeter that comes on here and you would connect like that 'cause see a voltmeter, an ideal voltmeter has, infinite resistance and no current given the choice the current come out here the
1:32
current's like "oh, I see infinite resistance or some, finite, resistance" it chooses to go through the resistor. An ammeter is the opposite sort of thing. An ammeter right here would, excuse me, an ammeter would be in line with, here we go it's an anarchy meter. No, just kidding, it's an ammeter and the current's going through the ammeter. Now ideally the ammeter has no resistance at all so I should draw this ammeter green to try to stay with my
1:58
with my program there. The ammeter will log the resistance through this nasty nasty nastily drawn resistor. Let's see if I can clean up this resistor a little bit and it would give us the current going through that resistor there. So here's the procedure, let's make this thing a real problem by giving you all kinds of resistor values. This one he says is sixty 8 ohms and this one he says is 5 ohms he wants this one to be 12 ohms. I don't know where he's
2:33
getting these 'cause these are not standard resistances, but somehow he managed to get all these, 30 ohms, set up this problem yourself so so that you can also do this and pause me whenever I get ahead of you. But I'm going to use equivalent resistance is to find the overall resistance. This is my battle plan: I'm gonna find the overall resistance of the circuit and then I'm gonna say hey if I've got a voltage
2:57
and I've got some resistance over here how much current will be going through the circuit as a whole, and then on notice that that current is present here and it goes through this resistor so the overall cuurrent will go through this one, and we'll come back and go through through that one when I get back into the battery but you see the current forks right here it's got three different paths that it can take so this is going to be a little complicated but my intention is to solve
3:17
the entire problem with V=IR are and equivalent resistances in parallel and in series so this is series equivalent resistance is R one plus R two plus R three et cetera and for parallel I know that one over in the equivalent resistance is the sum of the inverses of the individual resistances and we will play with those so that you get a feel for how this pattern works. Okay, my plan is to use these 3 simple relationships to solve the entire problem.
3:56
And to do it I'm gonna keep track of which sheets I'm working on and I'm gonna ultimately come back to this sheet and fill in every bit of information V, I, and R for each resistor given the resistances, alright. So I wanna simplify this this messy thing right here and I'm planning to draw it just a little bit differently I want to draw myself a battery and a resistor right here and then you see I've got these three paths, see my three paths that I have right here,
4:24
so I'm gonna take one down right here and then bring it back, well shoot, there's a resistor right here. Sorry, this is a little bit sloppy. I'm gonna try to do a little better next time. Now this intersection here has got three of them so I'm gonna draw one coming out here and then down that way and the third one is going to be over here and that way and then that way and then back do you see how that is the same map as what we've already done. So now I label these guys
4:57
this is picture 2 by the way, but we've got 68 ohms and 5 ohms and 30 ohms and 5 and 5 ohms in this sucker's 12 ohms in that sucker there his 12 Volts. Now I first need to recognize before actually simplify, that was just cleaning up, I first need to recognize what is the combination of resistors that I can analyze first. It looks like these three resistors are in series but they're not because there's something in parallel with it,
5:28
it looks like these two resistors are in parallel but they're not because there's something else in parallel with them. In addition, it looks like these three are in parallel with each other, but it's not that simple because these two are in series with them so I'm looking for the absolute simplest combination of resistors that I can find and I think it's these two fives together. So if I use the equivalent resistance formula for series
5:51
resistors I'm just going to find that the equivalent resistance is 5 plus 5 so that's 10 ohms and I love finding series resistors 'cause I can simply add them. So as I go to the next slide I want to point out that this will be 3 and my system now looks like this: I've got myself but 10 ohm equivalent resistance here and here I've got thirty ohms still and here I've got five ohms still, and then over here I've got 68, and the battery is there, and over here
6:31
I've got what I say 12 ohms, here and 68 ohms there, and 12 volts. Okay, so there's slide three. In slide three I now see very clearly that these three are, in fact, in parallel and I try to find the equivalent resistance for those guys right there let me point to those well, let's see. I'll put it down here R equivalent is one over the sum of all those 1 over 5 plus 1 over 30 plus 1 over 10. Wow adding more down here actually makes my answer woah,
7:07
makes my answer what, bigger or smaller? Well you'll have to figure that out. Play with that and make some sense of it. But right now I'm going to do that calculation so I can get myself the resistance. *humming to self* *he's still humming* Boy I hope i don't make a mistake that'd be dreadful, I got three. I got three ohms so check this out this resistance of the three of these guys if they are acting as a single resistor is less than
7:38
all of the individual resistances I think that's cool and I think we should move to slide slide 4 at this point. Slide 4 has still as I simplify this further it's still got a 12 volt battery and we come over here to a 68 ohm resistor and then there's this other resistor here we just found this to be only 3 ohms and then it comes in here our circuit continues over here with the 12 ohm resistor now this equivalent, woah sorry, this equivalent resistance is 3 ohms
8:13
and so now we don't have to write the 5 and 30 and the 10 that sort of like if you find, um, you know if you if you've got a vector this direction and a vector this direction instead of using these two which might be the components of an overall vector you're going to use only the overall vector itself the R vector. So instead of using individual resistances as we're using just the overall resistance now we see that these three are simply in series so then I say
8:42
the equivalent resistance here is, well it's just 68 plus 3 plus 12 I get my calculator, just kidding, I think I get 83, hi. Then the final diagram looks like this and then we'll work our way back up, filling in on all the information that we need. The final diagram looks like this: there's a 12 volt battery, and the 12-volt battery is sending current through an 83 ohm resistor, and now I'm going to use these two colors to fill in
9:15
the voltage across the resistor and the current through the resistor. So it's a very simple circuit and I say V is is IR and so, well I like to go around the circuit need to find some voltages That's 0 and it's equipotential for this so that's also 0 and this is 12 and its equipotential for this so that's also 12 volts and, what did I say I'd use for voltage? I think I wanted to use blue for voltage if I'm trying to find the voltage well. The voltage will be 12
9:49
volts, cool! And what about the current? The current through this resistor I total, that's the total current through the circuit the current through that resistor, well I guess it will be uh, V divided by R so like I got 12 I need to divide that sucker by 83 and I find that the total current is going to be a small number in fact will be working with very very small numbers. 0.14458 amps, alright that's a reasonable amount of current for circuit. We don't want
10:24
hundreds of amps or anything like that. Now I'm planning to take this current here and see where it goes on this circuit. See this circuit right here, where's that current gonna go? What can I do here? Throw out some of this nonsense. Um, that current that goes through this equivalent resistance will be going through this resistor and that resistor and that resistor so I can write that green down all over the place I can call it here I've got my I, remember there's no,
10:55
there's uh there's no backing up of current so we upon the principle of diarrhea this will be exactly the same as all these others .1458 .14458 amps I is 0.1445 8 in fact there's something pretty interesting going on here. We again say here I is 0.14458, as soon as I get back to a resistor that actually exists in my diagram one I can actually fill in all the information from that resistor so this 68 ohm resistor is the 68 ohm resistor from diagram 1 and the 12 ohm resistor down here
11:36
is the 12 ohm resistor down there this 3 ohm resistor a combination of all four of those resistors we can't work with those yet but I do know the current here and if I know the current then I can know the voltage drop because V is I R, so we put that down right here and point out how I'm going to get V for this so this is 12 volts here. And then there's a voltage drop across that resistor and that voltage drop has to be the current going through the resistor which we said is this amount
12:03
going to the whole circuit times that resistance so I've got that number right there in my calculator and I'm gonna multiply it by 68 and I find that the voltage drop across this resistor is 9.831 3 volts. What does that mean this voltage is right here? Well it doesn't mean it's 9.8 blah blah blah blah blah it means it's 12 minus all that business right there so I'm gonna take 12 and subtract my answer and I find that this voltage is, ooh, this voltage is 2.1 6 9
12:37
volts we might have some rounding things as we continue on but well what do you think the voltage is over here? We're gonna have a voltage drop and it's going to be equal, the voltage drop across this resistor is equal to the voltage at this resistor times the current that's going through it. You see how everything's going to work out here let's try that again I'm gonna go back up and get my .14 458 and I'm going to multiply it by 3 and I find that the voltage drop here the voltage drop is 0.43
13:08
3 74 volts which means if i take 2. 1 69 and I drop the voltage going through here that's what the current does the voltage decreases 'cause the potential of the charges that are moving through here oh, pretend it's positive charges right. Well the potential the charge is going to be decreasing, so I take 2.16 8674 whatever blah blah blah and I subtract point well I subtract that answer which was .4 3374 and I find the voltage at this location right here this voltage gonna be 1.7 3
13:44
49 volts. Now I want to know the voltage drop from here to here and buy gum I hope its 1.7 349 volts because this has to be 0 if that's 12 because batteries cause a voltage rise not a voltage drop so let's see what we get. I'm gonna take this resistance and multiply it by this current and see what it is I got 12 times 0.1445 8 enter equals *gasp* it is it is that is the voltage across this resistor we find that to be 1.73
14:25
5 volts and so this if I take 1.73 5 volts and I subtract 1.7 35 volts, I find that that is 0. And that's a happy thing for battery starts at zero pumps up to some voltage and that's the reason that the charges move the battery is giving the charges energy and the charges me to the circuit as a result, I'm not going to be in so much detail when I go to Step 3 because see I can immediately fill in as we were saying I can immediately fill in the voltage
14:52
and the current through this resistor because it's the same resistor, and I can do the same thing for this 12 ohm right here I've got all that information so I know at the voltage here is 9.8313 that's the voltage drop across that resistor, and I know the voltage drop across this resistor down here this voltage is 1.7 35 volts, and I also know the current through each of these resistors this current is, I'm on my number one so this is my final answer I'm already filling some
15:27
information in on that and I'm about to get my dollar, .14458 amps. And this guy right here is also experiencing the same current 'cause all the currents going through there hasn't been any forking yet at this resistor over here all the currents have come back together and I'm gonna get the same current thats going through the battery and it's the total current of the circuit as well so that 0.144 5 8 amps. Now we'll come back to situation one in a little bit
15:56
but we now have to figure out what's going on because there's a voltage drop across this resistor but it's an equivalent resistor and has a current going to this resistor but it's an equivalent resistor in fact it was equivalent to these things a 5 a 30 and a 10 so what do we know then for each these 5, 30, and 10? Do we know that they each have this current, or do we know that they each had this voltage drop? And answer that question is contained in these wires.
16:23
These wires are equipotentials which means that if there is a potential here then it's the same here and here and if is a potential here that it's the same here and here because they're all the same potential so I can do this in a number of ways I can start by my battery, I'd like to do that. Let's start from my battery I'll say that this is zero and this is 12 volts, this location right here is 0 the location there is 12 volts and that means that this location we still got the same
16:50
voltage drop here so this location up here is 2.1 69 volts and the location down here on the other side that equivalent resistance well its all of this area here but that's 1.7349, volts so the voltage drop across here is the same for all of them and I simply have to subtract those two guys before I can look over here because I wrote it down at 0.43 374 and I'm gonna write that down sideways V equals 0.43 3374 volts .43374 okay V equals 0.43 37
17:33
4 volts that's the voltage drop across each one of these resistors. And now my plan is to figure out how the current is you want to guess which one of these resistors has more current how 'bout I ask you to guess that right there which one of these resistors will have more current if they all have the same push on them? I suppose the one that's the easiest to go through will have the most current hmmm the one that's easiest to go through is this one so we predict this will have the most current
18:06
in order to find current I'm gonna say current is V over R *gasps* R is in the denominator of course the one with smaller resistance will have a greater current and let's figure it out. I'm planning to take this voltage and divide it by that resistance so here we go I got 0.433 74 and I'm dividing it by 5 and it's an even smaller current the current through this guy right here 0.0 8 6 7 amps that's the current here,
18:42
and the current through this guy is, well, it's that same voltage but divided by 10 so I guess it's going to be half as much current as right there and that's true it's 0.4 sorry, it's 0.04 3 3 7 amps and, oh shoot! no that's my 10 sorry that's my 10 over here it should be through this resistor 0.043 374 amps. But problem is this was an equivalent resistor, so we'll have to get back to that on the next slide and the final thing that we need to do is divide the whole thing
19:20
by 30 and we'll find that there's even still less current going through this really really resistive thing. That's just 0.01 44 5 8 amps right there. So all these currents together though they will add up to the current that we had in the previous slide. The current hat we had and slide 4 was 0.14 458, and if you're feeling up spunky I would suggest that you add this and that and that together I hope that you will get that answer
19:53
right there. That would be awesome right, because what we're seeing is the current that's coming in here is forking in three different paths, and it's not forking evenly it prefers most of the current well the plurality of the current anyway, most current yeah in fact even most the current is gonna be going through here and a tiny bit going through there 'cause there's a lot of resistance but still a little bit going to here. So as I present more and more paths
20:19
even if they're very high resistance a tiny bit of current will go through them and it will enable the total current of the circuit to get larger. That's why additional parallel paths decreases resistance, excuse me, because it increases current. I mean that's one way of looking at it anyway. Do we have this all figured out? We were going to um copy some more numbers on here, no I think we've got everything that we need to go to slide 2, in slide 2
20:44
and really slide 2 is their final slide because slide one was just a slight tweak on this sucker. I'm gonna take that information and write it down here but I don't need to. I mean we've already figured out this 5 completely, check. We've already figured out the 30 completely, check. We figured out the 68 completely, check. And we've figured out that 12 completely, check. Now our final conclusion has to be what's going on with these two 5s, remember we
21:09
created a 10 out of the two 5s because the principle that adding up two resistances in series will give you the equivalent resistance. I know the voltage at this location is 2, oh shoot! It's supposed to be blue, I'm so sorry the voltage at this location is 2.16 9 volts, and the voltage at this location is 1.7 349 volts. So that means, well, I mean I guess I could ask you what the voltage is right there, what do you think?
21:38
These are equal resistances here so you're kinda guessing it might be the average of those two. You'd be right if you do that, but how can we do this more carefully? I'll tell you. We know the current that's going through this resistor and the current going through that resistor. In fact it's the same as the current that's going through the equivalent resistance *weird shushing noise*. Notice, and you have to be very careful you have to know
22:00
whether I saying the current is the same or the voltage is the same. I guess if it's parallel the voltage is the same and if it's series the current is the same. So you'll really start to get a feel for that as you do a few of these problems. You have to do them to know what the heck's going on though. The current through this guy 0.043374 amps the current through this guy's 0.04 3374 amps, sure. Now what is the voltage drop across this resistor?
22:29
Well it's going to be the current times the resistance 'cause V is I R. So I'm gonna take 0.04 3374 and multiply it by 5 and I find that voltage drop the voltage drop across this resistor is 0. 2168 7 volts. That means, and I'm going to do the same math to get this other one and I find the voltage drop from there to there to also be 0.2 16 87 volts. Sure, which means that that is the average of the two of them and I guess if I multiply that by 2, if I multiply this number by 2
23:14
I get the voltage drop from here to here and so if I take that number and add it to 1. 73, I'm adding it now to this number down here 1.73 49 volts, I find it to be,oh, this number up here, 2.169 volts. Okay cool, so that's the voltage across each of these resistors and the current through each of these resistors is the same as the current going through there, and if we wanted voltage and current across everyone one of the resistors in our original problem
23:44
we have, in fact, done it. And presenting it in a neat way is not really possible because there's an enormous amount of information and several decimal points because I chose the resistor values at random but that's how it's done. Bye!
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