(II) Determine the magnitudes and directions of the currents i...

Question

(II) Determine the magnitudes and directions of the currents in each resistor shown in Fig. 26–57. The batteries have emfs of ε₁ = 9.0V and ε₂ = 12.0V and the resistors have values of R₁ = 25 Ω, R₂ = 48 Ω, and R₃ = 35 Ω.
(a) Ignore internal resistance of the batteries.
(b) Assume each battery has internal resistance r = 1.0 Ω.

(a) Ignore internal resistance of the batteries.

(b) Assume each battery has internal resistance r = 1.0 Ω.

Accepted Answer

Welcome back. Everyone in this problem. An electrical heating system in a building is powered by three resistive heating elements. Each connected to separate voltage sources. The voltage sources labeled V one and V two supply 120 volts and 240 volts respectively with resistances of R one as 10 ohms, R two as 15 ohms and R three as 20 ohms. As shown in our diagram, the heating elements form a mixed series parallel circuit with the voltage sources determine the magnitudes and directions of the currents I one I two and I three flowing through each heating element, ignoring any internal resistance of the voltage sources. For our answer choices. A I one is 0.92. Pi I two is 7.4 pi and I three is 6.5. Pi B says they are 0.927 0.4 and 5.5 amperes respectively. C 1.27 0.4 and 6.5 amperes and D 17 and 6.5 amperes respectively. Now, before we start, let's just label our resistances and voltage sources here. So we know that V one is 120 volts. V two is 240 volts. Our one is 10 homes R two is 15 ohms. I write it above it and R three is 20 ohms. Ok. So, so that's all the information we have. And from it, we want to determine the magnitudes and the directions of the currents. I one I two and I three flowing through each heating element. Now, let's label our circuit with currents and we know that if it's positive, that means it is flowing in that direction. But if it's negative, it's just flowing in the opposite direction. OK. So let's say here that we have I two, let me put it in red. Actually, I two flowing through R two, I one going in this direction through R one and I three going through R three. OK. Now we can apply Kerk Goff's current and voltage laws to help us figure out the values of I one I two and I three. Let's start with Kirk's current law. OK. Now, by Ko's current law, the KCL, OK. Then at the junction, let's say at this junction, OK. Then we can say that I too is going to be equal to I one plus I three. OK. So let's call that equation one. Now we can also apply Kuros voltage law to help us figure out our potential differences throughout each loop in our circuit. OK. Now, by Kuro and I'm going to write this one out. OK? Off's voltage law, but I could have easily written KCL but I just have some KVL. Sorry, but I just have some more space here. We're going to look at two separate loops to figure out the voltage to, to get some potential differences rather for our loop. OK. Now let me share this in red here. Let's take our loop with V one and let's look at the loop for V one starting at the negative terminal of V one and moving counterclockwise. OK. So let's say we were moving counterclockwise, let me put that in blue as well to differentiate it. Now, when we're moving counter clockwise, then we should find that we get V one minus I one R one minus I two R two to be equal to zero. We can substitute the values that we know. And for now we're going to ignore our units of measurement. So we can rewrite this as 120 minus 10 I 110 ohms is the value of R one minus 15 I two to be equal to zero. And now if you notice here, if we divide through our entire equation by five, since all of them have five as a common factor, then we could get this as 24 minus two I one minus three I two to be equal to zero. Thus, we can say that two I one plus three I two equals 24. And now we will say that as our second equation. Now let's apply Kho's voltage law to the bottom loop with V two. OK. In other words, let's go down here and again, we'll start with the negative terminal of V two. So when we do that, that means we'll be going in the clockwise direction. OK. So for our second loop, you know, I should actually have put that in blue, OK? For our second loop going clockwise, starting from the negative terminal of V two, then we should get V two. OK? V two minus I three R three minus I two R two to be equal to zero and V two minus I three R three minus I two R two. So now in this case, then if we substitute our values, so that's 240 minus 20 I three minus 1582 equal to zero. OK? And again, we can go through here and factor out or divide through our equation by five to get uh sorry to get uh 48 minus four I three minus three I two to be equal to zero, which means that we can say four I three plus three I two equals 48. And we can label this as our third equation. Now we have three equations, three unknowns and we would like to solve for I one I two and I three. So for starters, let's substitute equation one into both equations. Two and three. Let's start with equation two. So if we substitute equation one into equation two. OK? Let me bring this down a bit. Then we should find that we get two, I one plus three multiplied by I one plus I three. Because equation one tells us I two equals I one plus I three being equal to 24. I know that tells us then that if we expand and combine five, I one plus three, I three will be equal to 24. OK? So we can call this equation four. Next, if we substitute equation one into equation three as well. OK, then now we're going to get, well, it's going to eventually work out to be three I one plus seven I three. For, for the sake of time here, I can't do it all but three I one plus seven I three being equal to 48. And now we have two equations, equations four and five in terms of I one and I three. So we can eliminate and then solve for I three. OK? So let's do that. Let's eliminate by multiplying equation three or sorry, multiplying equation four by three and multiplying equation five by five. OK. So now when we do that, OK, our equation is gonna become 15 I one plus three I three, sorry plus nine I three rather equal to 72 and 15 I one plus 35 I three equal to 240. And now if we subtract the first equation from the second OK, then that should give us 26 I three. OK. Being equal to 168 volts, which means that I three will be equal to 168. Oh, remember we're ignoring units. So it's equal to 168 divided by 26 which would be equal to 6.46 pires or approximately 6.5 pi So this is the value for I three. Next, let's substitute I three into one of our previous equations. OK. So then if we solve I three equal 6.46 pires into let's say equation equation equation four here. OK, then we should get five, I one plus three multiplied by 6.46. OK? Being equal to 24 volts. OK? And now when we go ahead and work this non and go ahead and solve A I one, we should eventually get I one to be equal to 0.924 amperes or approximately 0.92 amperes. Finally, we can find I two because now remember I two equals I one plus I three. So that's going to be 6.46 amperes. Well, let me write it the other way around 0.924 S plus 6.46 pi which is gonna be equal to 7.384 pi or approximately 7.4 pi. So now what does this tell us? Well, from our solution I one is 0.92 si two is 7.4 Pires and I three is 6.5 Pires. Therefore, a is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Ohm's Law

Kirchhoff's Laws

Series and Parallel Circuits

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