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Referring to Figure $21.6 :$ (a) Calculate $P_{3}$ and note how it compares with $P_{3}$ found in the first two example problems in this module. (b) Find the total power supplied by the source and compare it with the sum of the powers dissipated by the resistors.

$P _ { 3 } = 7.1188 \mathrm { W }$$P _ { \text {tot } } = 28.22 \mathrm { W }$

Physics 102 Electricity and Magnetism

Chapter 21

Circuits, Bioelectricity, and DC Instruments

Direct-Current Circuits

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So in this problem, we have a set up. Ah, like this. So we have a battery here, and one resist is in serious connected over here. And we have to the other one, they're in peril. Okay, so this is our one are two in our three. This is B and we have very, very equal. 12 volts are wise. One home aren't too is, uh, six homes and our three is the 13 homes. So in part A, we want to find out the P three, which is the power over the third resistant are free. So firstly, we neither determining the resistance off our two and our three So are 23 equal are to our three Divide by our two planets are three. Okay, because these two, they're in peril. You can say that this equal 4.1 homes and so the total current I equal v divide by our one plans are 23 right, Because our wise in Siri's was this too, was that with the art, two are free. So this is a total current, which is 2.35 years into. Then we can determine the eye three, which is a current through our three. So this equal are two are two plus on three times I were eyes the total current and aren't too Uh ah. And so we can find on the video this equal points 74 piers Okay, so we can determine p three p three equal. I three squared times are three. So this gives you a 7.2 warts. And if you look up your example 21.1 and 21.2 so you can find out a P three under these thio circuits. So this is in the 21.1 are one are two and our three and you can see that the are three. The P three for this circuit is, uh, 4.50 68 wants, and there's another one, which is a parable. So 123 Okay. And this is 21.2. And in that example, they found that the power p three is ah, 11.1 walls. So clearly we can see that our our p three, now that we have this relation p Siri's So this is a piece serious. That's a it quot s and this sequel. Peopie So this p label the parallel So P s smaller than p three p three is our p three. Okay, it's smaller than pee pee and, uh, in part B. We want to testify that Ah, the power ah dissipated for the battery is the same as the power for the anticipated for the three resistance. So the power consumed in the battery he according to the definition, it's a V times I right v is the voting of the battery and eyes the current. So we have V is the 12 volts and I's a 2.35 according to pot, Eh? So this gives us 28.2 walls and p one. He wanted the power dissipated inning or one. So this equal I square times are one right eyes, the total current. So this actually gives you 5.50 to waltz and to figure out p two waiting to determine the current through our two first. So now that we already have the current in the main branch, so I to equal I'm minus I three, right? So this gives you 1.61 peers so that we can determine P two equal. I too square times, aren't you? So this gives you a 15.55 watts and P three, as we obtain, impart a So this is 7.2 warts. So, uh, we have p one plus p two plus p three equal 28 point to waltz so this equal to the power consumed on the battery, okay?

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