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Refer to Figure 21.7 and the discussion of lights dimming when a heavy appliance comes on. (a) Given the voltage source is $120 \mathrm{V},$ the wire resistance is 0.400$\Omega$ , and the bulb is nominally 75.0 W, what power will the bulb dissipate if a total of 15.0 A passes through the wires when the motor comes on? Assume negligible change in bulb resistance. (b) What power is consumed by the motor?

$P _ { 2 } = 67.7 \mathrm { W }$$P _ { m } = 1642 \mathrm { W }$

Physics 102 Electricity and Magnetism

Chapter 21

Circuits, Bioelectricity, and DC Instruments

Direct-Current Circuits

University of Michigan - Ann Arbor

University of Washington

Simon Fraser University

University of Winnipeg

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and welcome to my walk through of Chapter 21 Problem nine for college physics and AP courses. So this problem is all about circuitry. And really the first thing I want to note when you do any kind of circuit you problem is try not to get overwhelmed by the sheer amount of components and all the equations for each component. Just keep in mind that these are all individual things you could break down. Ah, and they all make up the total. So we're giving a couple things in the problem here were given our source voltage Ah, 1 20 were given the value. Our first resistor here were given as well the nominal voltage of this light bulb by our second resistor here. So nominal voltage is really just saying that it's ideally, it's what it's supposed to be. When you go, it's a Home Depot. Wherever you go to get lightbulbs, it's gonna be with on the back of the package 75 watts and were also given the ah total current in the entire circuit provided by our motor here. Sorry for my motor diagram. I'm not exactly an artist, so Ah, before we really jump into it. I also have remember that power can be written in three different ways current voltage resistance. And we're gonna be using all of those in this problem here, so just keep that in mind. Um, so part A is really asking about the rial, the actual usage, the power usage of this life here. It's gonna It's gonna be less than 75 because that's what speaking ideally here. So in order to really get that, we need to start with knowing what are our two value is and Teoh calculator are to Valley. We are allowed to use our ideal numbers. So we're gonna use p equals V squared on our that equation here, we're gonna set isolate are because we're solving for it. Um, and we have the nominal power usage already. So we have our two up top is our source voltage squared over that power usage that nominal power usage. So plug in our numbers in 1 20 squared over 75 we're gonna get about 192 OEMs of resistance from that are too All right. And so moving for we're gonna actually shift our focus a little bit. We're gonna go back to our one. Now that we have the resistance up here, we're gonna find we need to find, um we need to go this. Sorry, we need to go piece by piece. When you start from our voltage source, calculate the voltage drop from our one. And then remember, voltages are shared between parallel components. Right? So once we have the voltage drop from our are one, uh, we we can apply it to both of our parallel components here because we lose some voltage because this resistor is in Siris with these other two. So we have to lose something there. All right? And in order to figure out that voltage drop, we use a very standard OEMs law vehicles ir recalling his view one, we have our current total. And then we were given the resistor. So again, these were really all just given values 15 times. Point Forum is going to give us six volts. And so they were gonna combine these steps here a little bit. We have a voltage drop is going to be our total bolted source, minus that six volts. So we're gonna have about 114 volts shared between these two parallel components. All right, we're gonna call that value. There are parallel voltage, V p. And now, in terms of a we're almost done. We just take Rvp. We need to get the actual power usage from our bulb. Right. We're gonna go back up to our power equations and use another. Um, sorry. We're gonna go back up to our power equations and used b squared on our again, but we're not gonna isolate Are because we're solving for P this time. So we have. Sorry, we have p two is equal to that parallel voltage squared over the resistance of our to which we calculated up top. So actually meant something. After all, 1 92 owns. Um Then there we go. So we're have a parallel vultures squared over that second resistance is gonna give us about 67.7 watts. So, as you can see, it's a little bit less than 75. Um, that's just because those numbers are really idealistic and they're pretty much never actually that exactly. So it's good PR value for our part of 67 up next were asked to find the power usage of our motor here, which isn't bad, because we Ah, now that we know our power usage from this component Well, and we can find out the power usage from this component, it's gonna the only thing that's left. Um, so it's just in terms of ah total minus a bunch of stuff. So we have power. Total is power of 12 and three combines or specifically, in this problem, our power total is going to be the power from the first resistor, plus the power from our, uh, lightbulb plus the powerful motor, which is what we're solving for. So we're gonna isolate that that value Peace. Obama. Excuse me. Piece of em over here. So piece of em is equal to total power. Minus that. Resist the power from the first, her sister minus the power from the bulb. And as I said, we're going to use all three of these voltage equations. Are these power equation? Sorry. Ah, in this problem and that we could see that right here. So our total power is gonna be our total current times are on our total Voltan source. Um and we have the power from resistor. One is gonna be the same current. We don't have any other thing any other currents to work with. And current is consistent across, um, para Components and Siri's. So this time it's OK to keep the same current. So we have the same current squared times are one value minus the power from the bulb. But fortunately, we calculated the power from involved in part A 67.7. And so, uh, once we do this, it's just a matter of plugging your numbers in. Whoops. Sorry about the scrolling. 15 times, 120 15 amps from our current times 120 voltage source, minus 15 amp squared Times the 150.4 home resistance minus that is 67.7. And so that's going to give us a total of 1642 watts or personally, I to use scientific notation almost always so 1.642 times 10 to the third kilowatts. And that's gonna be the power usage from our motor. And that'll that'll wrap up our problem here. We have figured out everything here and, uh, thank you

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