00:01
In this question, we are going to analyze a super fun and challenging looking resistor network that has a variety of combinations of parallel and series resistors in order to find the current in each resistor, the power dissipated in each resistor, and the voltage across each resistor.
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So to do this, we're going to start from the end.
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So i like some color coding, excuse me, to help me determine how things are connected.
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So we're going to start with red goes to our positive value, our positive voltage, whatever that was.
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And then our purple, we're going to use purple to get to our negative battery terminal.
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Terminal and then everything connected to that wire is at the same potential.
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We'll then have a color again, it's just a qualitative tool, a drop from red to orange as we pass from r1 and our circuit splits into r2 and r3 and everything else.
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So we can see pretty clearly that r2 is is going to be in parallel with everything else.
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Because if we take the path through r2, we go orange to purple.
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And if we go over to r3 and through anything else, we're going to end up starting at orange and ending at purple.
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So we now know r2 is in parallel with everything else in the circuit.
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And we're going to have a change from orange to yellow as we pass through r3 and have our most complicated branch.
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Branch.
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By the way, i renamed these for myself, point c and d as the junction after r1 and after everything else has come together.
01:56
And i called e the junction that splits to r4, r6, r8, and r9.
02:03
I called f the one in between r5 and r7.
02:09
All right, so after our orange, orange, we're going to have, or sorry, after our yellow, well, we're going to have another drop across e, or after e, across eight, and then we'll have, you can tell we're going to need two more colors.
02:32
We're going to need a color here for these, and a color here for these.
02:36
So i am going to use a lighter blue, we'll go super light blue for the connection between 10 and 9 and 7, and then we'll do a dark blue between 6, 7, and 5.
03:00
So this helps us now kind of figure out what's connected we can see very clearly that all right so we can see very clearly that our 8 is in series with our 10 so i'm gonna call that our 810 and our 8 is 10 ohms and our 10 is 2 ohms so that is 12 ohms and then we can see very clearly that 810 is in parallel with r9.
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So i'm now going to call it r8910.
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We'll find with 1 over r9 plus 1 over r810.
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And we'll take that to the negative first power.
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And so r8910 will be found with 1 over 6 plus 1 over 12 and invert it, and we get r8910 is 4 ohms.
04:29
So what we are now saying is this.
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Between point e and point f, this circuit has the direct connection through r6, and then we have r7 over here along the bottom, but 8, 9, and 10 all just collapsed into this r8, 9, 10.
05:03
Well, swell.
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Now, we can say, i'm going to call it r7 through 10 will be found with r7 plus r8, 9, 10.
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So r7 through 10 is going to be equal to 8 ohms plus 4 ohms, which gives us 12 ohms.
05:32
So now we know that we have 12 ohms in parallel with r6.
05:39
So i'm going to call it r6 through 10 equals 1 over r6 plus 1 over r7 to 10 raised to the negative first power.
06:03
So i have 1 over r6, which is 6 ohms, plus 1 over 12.
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And we're going to invert it.
06:12
Hey, hey, we've seen that before.
06:13
So my r6 through 10 is going to be another 4 ohms and it's helpful now to sketch again what does this part of our circuit look like so we have point c that's going to have our connection down to d going through r2 we're still going to have r3 over here to the right of point point c, but we've now, actually give me just a second, and let me connect from d up to e along the diagonal that will be r4, but we've just collapsed everything between e and f to this r6 through 10, and then of course we still have between points d and f, we have r5.
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Notice i'm not redrawing the whole circuit.
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I'm just drawing the section that i need to look at.
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Now this helps us because we could automatically see that my r6 through 10 is in series with r5.
07:34
So r5 through 10 is going to be r5 plus r6 through 10.
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R5 of course was four ohms and we just had six through 10 was was 4 ohms.
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So that gives me 8 ohms in this area right here.
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And now we can see that r5 through 10 is in parallel with r4 because they are two different paths to get between e and d.
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So i'm going to call it, not surprisingly, r4 through 10 will be equal to 1 over r4 plus one over, oops, not r10, but r5 to 10 raised to the negative first power.
08:36
And so my r4 through 10 will be equal to one over eight ohms plus one over eight ohms, which will be four ohms.
08:52
All right, so we've now collapsed 5, 6 through 10 and 4 into being something that will be in series with r3 because they are two different paths to get from c to d.
09:21
So i just said with r4 through 10, we're going to have to scroll down a bit here.
09:29
And so, not surprisingly, we're going to be able to call it r3 to 10, which will be r3 plus r4 through 10.
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R3 is 4 ohms, and then we have 4 ohms.
09:47
Oops, not plus.
09:48
So we then have equals 8 ohms for my r3 to 10.
09:58
And we've now basically collapsed our circuit down to, you know, we still have point a and r1 over here.
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And we've got point b down here.
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But we've now basically collapsed our circuit.
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Here's point c.
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Here's point d.
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We've now collapsed our circuit to r2 that connects from c to d, and then r3 through 10, you know, whether you want to draw as a triangle, it doesn't matter, or a curved line, but now r3 through 10, we can see very clearly, is in parallel with r2.
11:00
I'm going to call this, it's, excuse me, r2 through 10.
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We're going to find with 1 over r2 plus 1 over r3 through 10 raised to the negative first power.
11:23
So 1 over 8 ohms plus 1 over 8 ohms raised to the negative first power gives us 4 ohms.
11:31
And then finally, finally, this is in series.
11:38
Let us say series, by the way.
11:40
Let me fix that this is in series with r1 so finally our equivalent resistance of the whole circuit is going to be r1 plus r2 through 10 and we're going to get 6 ohms plus 4 ohms and we finally get 10 ohms as the equivalent resistance of this entire circuit all right and now we're we're only halfway through the fun of this.
12:09
Most of the difficult work is done because now we're going to talk about our currents.
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And then after we have our currents and our resistances, we can calculate power and voltage really quickly and easily.
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We are going to have to work our way backward through this.
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So here's the thing.
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What i notice is that without a given voltage or starting current, we can't do the rest of this question.
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So what i'm going to do is i'm going to make up.
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We're going to pretend that we have a voltage between a and b equal to 20 volts.
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And now, now we are able to say that for part b, our voltage between a and b will be equal to our total current times our equivalent resistance.
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So our current will be equal to the voltage between a and b divided by our equivalent resistance.
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So we have 20 volts divided by not two, but 10 ohms, which gives us two amperes.
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All right.
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So now it's time.
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You know, i'm actually going to pause for a moment and upload a new image of the diagram so that we can kind of sketch where our currents are splitting.
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So give me just a moment.
14:09
So picking back up with a way that we'll be able to draw out our currents...