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And welcome to my walkthrough of chapter 21, problem 9 for college physics and ap courses.
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So this problem is all about circuitry, and really the first thing i want to note when you do any kind of circuitry problem is try not to get overwhelmed by the sheer amount of components and all the equations for each component.
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Just keep in mind that these are all individual things you can break down, and they all make up the total.
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So we're given a couple things in the problem here.
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We're given our source voltage of 120.
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We're given the value of our first resistor here.
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We're given as well the nominal voltage of this light bulb by our second resistor here.
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So nominal voltage is really just saying that it's ideally what it's supposed to be.
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When you go to home depot or wherever you go to get light bulbs, it's going to be what's on the back of the package, 75 watts.
00:53
And we're also given the total current in the entire circuit provided by our motor here.
00:58
For my motor diagram i'm not exactly an artist so uh before we really jump into it i also have remember that power can be written in three different ways current voltage and resistance and we're going to be using all of those in this problem here so just keep that in mind um so part a is really asking about the real the actual usage uh the power usage of this light bulb here it's gonna it's gonna it's gonna be less than 75 because that's we're speaking ideally here so in order to really get that, we need to start with knowing what our r2 value is.
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And to calculate our r2 value, we are allowed to use our ideal numbers.
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So we're going to use p equals v squared on r, that equation here.
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We're going to isolate r because we're solving for it.
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And we have the nominal power usage already.
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So we have our r2 up top is our source voltage squared over that power usage, that nominal power usage.
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So plug in our numbers in, 120 squared over 75.
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We're going to get about 192 oms of resistance from that r2.
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All right.
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And so moving forward, we're going to actually shift our focus a little bit.
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We're going to go back to our r1.
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Now that we have the resistance up here, we're going to find, we need to go this, sorry, we need to go piece by piece.
02:23
When you start from our voltage source, calculate the voltage drop from a, r1.
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And then remember, voltages are shared between parallel components, right? so once we have the voltage drop from our r1, we can apply it to both of our parallel components here because we lose some voltage because this resistor is in a series with these other two.
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So we have to lose something there.
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All right.
02:50
And in order to figure out that voltage drop, we'd use a very standard oms law, v equals ir.
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We're calling this v1.
02:56
We have our current...