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Hello and welcome to chapter 18, problem number 22 of holt physics.
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So in this problem, we're given a sort of graphic of three equivalent light bulbs hooked up to a battery.
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And i think a difficult part of this question is really just converting that graphic to a recognizable circuit diagram that we're all familiar with.
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But if you look at it, you'll see that b and c are connected in series, and then the b -c equivalent resistance is connected to a.
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Via parallel connection.
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So this is what's going to look like when you draw your circuit diagram with our points of interest, de and f.
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And we also need to keep in mind that the brightness of each light bulb is going to depend on the power usage of each light bulb.
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And we can manipulate the power usage with three different equations by changing our current voltage and resistance here.
00:55
So to start things off, it asks us what happens if we take out a in our circuit.
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So right off the bat, we need to establish that a and then the b, c, equivalent resistance, they're going to share the same voltage.
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So taking out a won't affect our voltage.
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It will, however, affect our current because the current is not shared between components in parallel.
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So if we look at p equals iv, we can see that our current goes up.
01:23
Therefore, b and c also go up in their brightness, and a is completely removed from the circuit, so it's not part of our question here.
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Up next, it asks us what happens if we take out light bulbs c.
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So the next important thing we need to look at is the fact that b, c, is an equivalent resistance.
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So it's going to, taking out c, we'll change that r equivalent resistance here.
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So we're going to use p equals v squared on r.
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And we'll see that that that equivalent r is now just the resistance of b...