 Hopefully everything's working. So my thought is I'm going to talk a lot about terminal voltage. I'm going to do some examples from my older reviews that you guys haven't seen that will allow you to do the newer review as practice. I will take questions from the newer review as well, but where possible I'll try and find a similar question that we can do and still allow you to do the newer review. So I'm just going to scan through here and look at some of the terminal voltage type questions. And I can tell a terminal voltage type question when I see the battery with the little dotted line around it. So I'm scrolling, glancing at the circuitry diagrams right now. Haven't seen a terminal voltage question yet. Ah, number 17 must be a terminal voltage question. I see the little dotted line around it right there. You're going to glance at them really quickly. And in fact, number 17 asks, what is the terminal voltage in this particular cell? That's a great question. Terminal voltage, in terms of my ski hill definition, terminal voltage was what the voltage was after you skied down the bump. There was an equation. I'll be honest, I don't fall back on the equation as much as I know that the terminal voltage mat is my EMF, 4.5, minus whatever I lose going through there. That's the terminal voltage. Oh, unless we're recharging, and it'll be very obvious that you're recharging because it'll be a big, huge battery somewhere else, then it's not the EMF minus that. It's the EMF plus that. That's your terminal voltage. What is the terminal voltage physically, Greg? It's what you would actually measure on either end of the battery, because we said once you actually start running a battery, you don't get the EMF, which is the voltage printed on the side that never changes. You're going to get slightly less than that. So here, I would say, well, do you remember when I'm analyzing a circuit diagram, what's the first thing that I always want to try and find, do you remember? Total current. Oh, they told me it, right there. So here's what that means. These are each 10, which means I can make the assumption that the current splits up exactly in half. It's a convenient little shortcut we did. But more importantly, I know the current here is 0.8 amps going through this resistor. How many volts do I lose going through this resistor then? Whatever 0.8 times 0.63 is, 0.504. So going through this resistor, I lose 0.504 volts. I started at 4.5 volts. I lose 0.504 volts. You know what your terminal voltage is? You know what you'd actually measure? 4.5 minus 0.504. You would measure 4.0 volts. That's terminal voltage. And you notice I didn't use the, but technically I did do that. I said it's your maximum EMF minus your voltage, IR's voltage. But it's the voltage in the internal resistance traditionally symbolized with a lower case R to show you that it's not that internal resistance and total current. Is that okay? If not, we'll get there. Yes? Okay. Let's try looking at a few more complicated ones. Blah, blah, blah, blah, blah, blah, blah, blah, no terminal. I am going to ask you a question about the position or the placement of instruments, specifically an ammeter and a voltmeter. We said an ammeter to measure current, remember it's measuring skeers. We need to put it in such a way that it's in series. So it says if we want to measure the current through the light bulb, where we would need to put the ammeter, we would need to put the ammeter so that the number of skeers that go through the bulb also go through the ammeter. Not here because here there's two different currents joining together. You know what? Right there is correct, right there is correct. Right there is not because the skeers could go down the ski hill without even going through the ammeter. So right now the answer is B or D. And they also want us to measure the terminal voltage of the battery. The terminal voltage of the battery is the difference in height between here and here. It's chairlift minus whatever bump you get when you get off the chairlift. The answer to this would be D. Right, for instruments. Let's keep looking for a terminal voltage question, blah, blah, blah, here we go. So the circuit shown in the diagram below consists of a 9 volt battery. That's the maximum theoretical voltage, what we call the EMF symbol, little funky E. And a 3.5 watt bulb, A. If a current of 0.4 amps, ooh, they told me total current. So what's the current going through this light bulb, Gary? What's the current going through this light bulb? That's volts. No, what's the current going through this? So current is symbol I measured in amps, okay? And current can split up, but can it split up anywhere from here to here? So all the current must go through here. This is Kirchhoff's loss. A message for Karlu to place Mr. Flaw at 204 or to call Mr. Flaw at 294. Do I know two? Then I know four. We also know that power equals VI. So how many volts must I lose going through this resistor? Oops, let's get the V by itself, Mr. Duk. Voltage is power divided by current. It's 3.50 watts divided by 0.4. Greg, what'd you get? 8.75. So the voltage loss here is 8.75 volts. By the way, what's the terminal voltage of this? Isn't it also 8.75 volts? Because chairlift minus bump gives me only one hill. If this hill is 8.75 volts, that must mean these two combined had to have a height of 8.75 volts, otherwise you couldn't start and end up at the same height again. Does that make sense? So I gave you the formula for terminal voltage, and great. I use the concept more than I actually use the equation. Now that's not what this question wanted. So what does A want me to find? Okay, here it is. What do I know? Oh, what's the current going through here? 4. Oh, how many volts must I lose going through here if the chairlift is 9 and the only ski hill is 8.75? What must I have lost getting off the chairlift? Must have lost 0.25 volts. Do I know 2? Then I know 3 or 4. So for part A, little r, well, let's see, v equals i times r. So r equals v over i. We lose 0.25 volts. We have a current of 0.4. What is 0.25 divided by 0.4? 0.625 ohms. That's how big the internal resistance is. And there was probably a part B to this. Let's go see 0.63, there you go. The light bulb is now replaced by a lower resistance light bulb. Okay, copy. Let's go looking back up here. What is the current, what is the resistance of this light bulb right now? Because if I know 2, I know 3. Oh, if resistance is voltage divided by current, can someone go 8.75 divided by 0.4? Oh, they didn't tell me. I can figure it out. What is the resistance of this? Sorry? 28.9? 21.9. 21.9? Okay. It gives us a number. We could do this numerically if we needed to. We could make up a resistor of 10, let's say, I wonder whatever. So one way Miguel to do this would simply to make that 10 ohms, 5 ohms smaller, I probably wouldn't do 20. I'd want to make it very noticeable, and then crunch the numbers and answer the question. The question is asking will the terminal voltage be greater, less than, or the same? Or I can maybe do this now algebraically. Let's see. If we make this smaller, by the way, right now, what's my total resistance right now? I think right now my total resistance is 21.9 plus 0.625 because Matt, this and this are in series. Right now my total resistance is 22 plus 6.25, sorry, 22.625. So if I make this smaller, what's going to happen to our total? Our total will decrease. Now that our total, this time it includes the internal resistance. If our total decreases, what's going to happen to your current? Yeah, I think current will increase because you're stuck with an EMF of 9.00. Lower that, this gets bigger. So you have a bigger number than 0.4 coming through this little internal resistance of 0.25. I think the total current times R is going to increase and since terminal voltage is your EMF minus that, if this number here gets bigger, what's going to happen to your terminal voltage? It's going to get smaller, right? You can crunch the numbers as well. There's an algebraic one. Let's do a few more. Let's do a bunch more. Terminal voltage. Oh, here's another one. What's the terminal voltage there? Let's see. They gave me the EMF. This is my maximum theoretical voltage. Did they give me the total current? Nope. Did they give me anything with a pair? No. I'm going to have to do this the old-fashioned way. I need to rewrite this as an equivalent circuit. What's the total resistance of this circuit? Well, Miguel, these two are in series. The total resistance is that mathematically this is the same as and the 4.5 ohm resistor where this is 6 volts. Is that okay, Jacob? Mathematically. I know there's an internal, I don't care. I'm saying, you know what? For what it's worth mathematically, it's like that. What I want to know is what's the total current? Well, that's going to be the total voltage divided by the total resistance. It's going to be 6 divided by 4.5. How many amps are running through this circuit? Greg? 1.3. 1.3 repeating? How many amps right here then, Greg? Because the total current goes through there. Oh, how many amps right here? How many volts do I lose going through here? 1.33 times 0.5, right, good old, I should put units, I times R, 0.66. So here's what you're saying. Share lift, bump at the end. So what's your actual terminal voltage, 5.33? Now there's another way you could have got that. You could have said, since this is my only ski hill, if I go 1.33 times 4, that has to equal share lift minus bump because the height of this has to cancel out that, otherwise you couldn't get back to where you started from. Mathematically, yes. Because the circuit doesn't work though. And mathematically I said, well if I want to find total current, total current is total voltage divided by total resistance. What's the total complete resistance? It includes what's in the battery. Oh, as long as you use the EMF, which is the voltage before any resistance was taken to account anyways. Is that right? I'll be doing a few more, it'll come. I still see a few blazed looks. I'm going to nuke that one because it actually got tossed. Here we go. Terminal voltage. And a part A and a part B. So let's try this one. Okay? What's the terminal voltage? And by the way, Miguel, you can tell when they're asking about terminal voltage, I just look at, oh, they got the little dotted thing that they're trying to say, it's the battery with the internal resistance. Okay? All right. Did they tell me total current anywhere, Miguel? That's going to give us a nice shortcut. This is 1.8 amps. This is 1.8 amps. Oh, you know what they didn't tell me? They didn't tell me the chairlift. But what's the voltage here? If I know 2, I know 3. Could someone go I times R for me? Sorry. 4.5, is that right? What's the voltage here? Sorry? 7.2. So, Gary, how high is this hill? 0.5. How high is this hill? To get from here to here, what's the total height you have to cross through? 11.7, right? With me? So, the total height from here to here is 11.7. If I hear you correctly, you're saying when you got off the chairlift and skied down the bump, the height was what? So how must, what height, what voltage must the mystery battery minus the bump have given us if the ski hills were 11.7 meters volts high? The terminal voltage has got to be equal to these two. This is a very different approach now to finding terminal voltage. We're not using the equation because we don't know the actual battery size. We're looking at the rest of the mountain and saying, well, how high did it have to be for the rest of the mountain to work? The terminal voltage here is 11.7 volts. By the way, what's the current going through this little internal resistor? 1.8 amps. How many volts do I lose going through this resistor? 0.9? Okay. So, here's my question. How high must this chairlift be if I can lose that much, that much, and that much and end up back where I started from? Because I'm willing to bet part E, part B probably asks, what's the EMF? Let me go see. What's the EMF of this cell? Absolutely. What? Irwin, what did you say? It's the two hills and if I add to it the little internal resistor, now let me pause because you're saying, hey, you're adding. The original equation is terminal, actually, let's not use VAB. That's what they used to use, but now they use the terminal. That's the EMF minus IR. Gary, can you see if I got this by itself? I would plus this over to this side. There's my terminal plus it over to that side, right? Same equation except I moved this to there and we already said the terminal was that. The EMF then is going to be 11.7 plus, what was the voltage that we lost going through here? Did I write it on here? 0.9 volts, 0.9 volts, and we get 11.7 plus 0.9, 12.6 volts. That's your maximum. Before you hooked up this battery, it had an EMF, a maximum theoretical voltage of 12.6. Is this helping? I hope? Maybe? Sorry, that kind of. Do you want me to do a few more terminal voltage ones still? Okay. Then I'll say specific questions from the review. Are there any that you're wondering about? Yes, Alyssa? 54. 52 or 54? You're sure now? This one here? 54. Geez, now you got me doing it. Oh, a scholarship question. Okay. I'm not quite sure, but sure. Let me see. Power supply, 6 volts. It delivers a 3 amp current check. When this same supply is connected to two identical bulbs wired in parallel, the current from the supply is 5, check, find the internal resistance. So if these two are identical, what's the current going through here? 2.5, right? What's the current going through here? What's the voltage here? The voltage here is 2.5R. What's the voltage of 2.5R? What's the current going through here? Here's what I think. Looking at this circuit, by the way, my gut is I'm going to get two equations and have to do a little bit of systems plugging one into the, I don't know yet. If I walk this ski hill, if I walk this ski hill, I think how high is the chair lift? And that equals this voltage I times R plus this voltage 5 times little r. Or I could have walked this path as well, but there's the same ski hill. Over here, this is going to be three amps, three amps. If I walk this ski hill, how high is our chair lift? Six volts, and that equals 3R plus 3 little r. How many variables do I have? How many equations do I have? I should be able to solve this. And I think what I would do, I always, when it came to systems, much preferred the substitution method. What I would do is I would go on, divide by three, divide by three, divide by three, and that's going to give me two equals big R plus little r. It's going to give me big R equals two minus little r. Is that okay so far? And I'm going to stick that right there. I'll get six equals, by the way, if you're going, what the, this is a scholarship question involving systems that's fair game on a scholarship question, you're not going to have one this ugly on a test, not even close. But I'm a nerd, I'm going to keep going, six equals 2.5 bracket two minus r plus five r. Oh, this is so much fun, let's keep going. Six equals five minus 2.5 r plus five little r. And little r is what I'm trying to find. By the way, Alyssa, I did some thinking right here. I said, which one do I want to get by itself, because I could have minus big R from both sides and got an expression for a little r. No, I want the little r to stay and the big r to cancel because they want the internal resistance, which is little r. Excellent. I think I'm going to get this. Six equals five plus 2.5 little r. Is that right? Gathering like terms. Minus five from both sides. You'll get one equals 2.5 little r. R is going to be one divided by 2.5. Whatever the heck that is, I don't know. What do you get? 0.4? Two sig figs or three? Is that the answer? I hope, I think, I hope, I think, I hope. That was 10 marks? Wow. Either I've gotten smarter or it had gotten easier. I don't know if you can get there, Adam, using complete skier without having to do some kind of two variables, two variables, maybe. I'm not scared of systems. I like them in nap 11, actually. I won't do the one from yours, but I'm pretty sure there's one in here. I'm using principles of physics right to explain question. I know there was one in here somewhere. Yeah, I totally care. Bear with me. Let me go hunting. Of course, it was number, oh, 25. Okay. Oh, I did that one already. Let's find another one, Mr. Duk. Let's try and find another one. Let's go the older one. Pardon me? Yeah, these are old. Okay. This is sort of what you're asking, but it's phrased differently. It says use the following diagram to answer question 22 should be 21. I had to change the numbers. Yes? We're good. Great. Closing this switch is the same as adding an extra resistor. Okay. With this open, this resistor does not exist. What's your total resistance right now? So I'm going to say our total is 6. I total originally is 6 divided by 6. Yes, it's 1 amp, and I'm not worried about sig figs. This is rough work. If I close this, when you, I'm going to generalize, add a resistor in parallel. First of all, did I have a bunch of stuff to vanish? What did that go to, Mr. Duk? Okay. So we've already said originally our total is 6 ohms. I total is 1 amp originally. What's the resistance of this in parallel? Because these two are now in parallel if I close the switch. What do you get? What's the smallest resistor in here? What was the resistor originally to? Adding a resistor in parallel actually lowered. What did you say our parallel was? 1.2? So what's our total? 5.2, lower than 6. So if our total goes down, what's I total going to be? 6 divided by 6 volts divided by 5.2. 1.5. Okay. Thank you. So here's the key, folks. If you add resistors in series, that just increases your overall resistance. If you add them in parallel, it lowers your overall resistance, even though you're adding a resistor. And if you lower your overall resistance, if this gets smaller, what's going to happen to your overall current? It gets bigger. Now let's answer the question. Is overall current going through here? So will it read a smaller, the same, or bigger? Maybe, no. Maybe, no. Increases, increases. So far, so good. Now, if the current gets bigger, what will the voltage drop across here be if the current just got bigger? Bigger, smaller, or the same? Why? V equals I times R. If that number gets bigger, this number gets bigger. So the voltage drop here is going to get bigger. Originally, it's 4 volts I times R. Under the new system, it's 4.8 volts with the switch closed. Makes sense so far? Okay. If I lose 4, well, first of all, if I lose 4 volts, what must I lose going through the 2 ohm resistor? 6, 4, 2 volts. If I have 4.8 volts once the switch is closed, what must I lose going through either one of these? So instead of the original value, which you said was, now I'm reading, does it got bigger, smaller, or stay the same? Voltmeter decreases. This is wrong. Okay. That's a good example of using principles of physics, except often they'll bring in the notion of power, except a set of power, they'll say brightness of a level. VI is your brighter bulb, or weaker bulb. Again, this is kind of counterintuitive because Jacob, you'd be saying, look, you add an extra bulb and this one got brighter. Yep. Yep. This one got way, way, way, way, way different. Did I answer your question? Okay. Next. Yeah. 8 as in the number of sides on an octagon, as in 2 more than the number of sides on a hexagon, as in 1 more than the number of sides on a septagon, or sexagon depending on which part of the world you're from. But usually in North America we like the word sept because nobody wants to say the S-E-X for it anymore. Go figure. 8, total power. Okay. I'm going to, I notice my target right away, I see a resistor with two things on it. I'm going to flog that to death because if I know 2, I know 4 things. But first I'm going to copy this question. You see the resistor I was talking about again? Okay. So if I know 2, I know 3, and they gave me power. Now I know power equals VI. The problem is they didn't give me I here. What else is power equal to? Well, if V equals I squared R, power is equal to, sorry, let's try that again. If V equals IR, power is equal to, if I put an IR right there, power is equal to I squared R, which means that I is equal to P over R square root. Is that okay? Now let's fill in that gap, P 9.2 divided by 33 square root. What do you get? Yes, you, all of you are. You're still at playland. I know. You get 0.528. Anybody else? Divide by R square root, yes? Power, divide by resistance squared. You get 0.528. Yes? Okay. So I have 0.528. What's the voltage drop through here? I times R times 33, 17.42, I'll carry if you have to digit, 17.42. How high is the chair lift? Can I go like this and end up back where I started from? So 24 volts minus 17.42. How high must this mountain hill resistor voltage be? 24 minus answer, 6.5. How many volts must be over here? Yes. Oh, over here. Ian, do I know two? I know three. What's the current which happens to be the total current, which is one of the things that I always wanted to find anyhow? What's the total current here? V equals IR. So R, not R, Mr. Dewick. V equals V over R is this number divided by 10, which I think I could have done in my head, 0.658 amps. What's the current here? What's the current flowing into this junction? I think you thought it right and said it wrong. 0.658, right? Yeah, okay. That's my writing. I'll never make fun of you for not being able to read my writing. Although for Pete's, no, never mind. How many amps are right here, Ian? How many went this way? How many went this way? Yeah, I know. Now do it. 0.658 minus 0.52. Oh, I think I could do that in my head. 0.13 amps. I'm just filling in. Oh, wait, wait. If I know two, I know three. Except I don't care about the voltage. What is this question asking me for? Find the total. So can you go VI? Tell me the power here. How many watts? 2.26 watts. Oh, Ian, how many watts? V times I, 4.33. Yes, watts. What's this question want me to find? Power, scalar or vector? Add them up. 4.33, 2.26, and 9.2. 4.33 plus 2.26 plus 9.2. 92, Mr. Dewey, 9.2. 15.8. Is there an answer like 15.8 or 16 or something like that? There is. Power is nice. It's a scalar. Remember, I initially started saying in this unit, Ian, if you know 2, you know 3, really it's 2, you know 4 because we kind of forget about power is also one of the things that we're sometimes interested in. Does that make sense? One over R parallel is one over all of the other ones. You're saying that's the only, in circuitry right now at this level of your schooling, yes. But I can't answer that yes or no because if I answered, yeah, you're right. Actually, you're not right. There's other places where that occurs as well. But for now, Ian, sure. Next, good question. Any others from the review? Okay. On the provincial, on every single provincial exam, the second last question, they're going to give you some data and they're going to ask you to interpret a graph because they want to know can you do like a line of this bit or whatever. So this was taken from one of those. Okay. In this particular one, it says plot a graph of V versus I. So they want voltage on the y-axis, current on the x-axis. The biggest current I get is one. I would probably go 1, 2, 3, 4. I would probably go 0.25, 0.5, 0.75, 1. 1.25 is my scale here. I can see some people saying graph is a silly thing, Mr. Dewey. What number was this? Melissa, because I chopped the number off. Sorry. 44. So I would go 0.25, 0.5, 0.75, 1, 1.25. The highest voltage I get is 12. 1, 2, 3, 4. I'll probably go 3, 6, 9, 12. And a graph. The 0, 0 change colors so it stands out, Mr. Dewey. 0.25, 3. 0.5, 6. I'm getting a lovely line of best fit here, I'll bet you. 0.2, yeah, this is bang on. And I would connect it with the ruler. This might be accurate. Currents they can be pretty accurate with, which may be why they chose this data. Moving mechanical things, you know, bumps on the floor completely change your data. Current, we can be pretty accurate with current. Why our whole system is based so much on electricity. You weren't wondering about this. You're wondering about B, I suspect. Okay. So here's how I handle this. I told you, because there's a gazillion graphs out there, I think I went on a big grant. I said, some of you in Physics 11 try and memorize every graph. Done. Done. Done. First of all, slope is rise over run. 12 over 1 would be the simplest possible way to go rise over run. They would want to see that bit of work, so I will. 12 over 1, units. It's volts over amps. So the slope is 12. Okay. 12, what? What is voltage divided by amperage? How are these two related in the same equation? Yeah. I would say ohms. That's what they wanted. That's one that says express your answer in appropriate units. Now, what does the slope of the line represent? Let me read if I can just say resistance or maybe they want a specific type. Let me read exactly what they did. Oh, I chopped that part off quite cleverly. That was silly of me. Okay. A student connects a power supply to a circuit and measures the potential difference at its terminals and the current delivered to the circuit. Is he changing the circuit? No. I think it sounds like what he's changing is the potential difference and you're getting a different current. I think this is the overall resistance of the circuit. I was wondering if maybe it was the internal resistance of the battery. Is that what I said as my answers? I don't think I did. I don't think it's the internal resistance of the battery. I think it's just the overall resistance of the circuit. Is that what I said as my answer? I hope. Sorry. I just said resistance. Yeah. Okay. So what's the slope? Resistance. I've seen some where the line didn't go through 00. It maybe went through here and went like that and your Y intercept was the bump at the end of the ski slope. That was your internal resistance which is why it didn't quite balance off between your maximum theoretical voltage and your measured voltage. That was a nice graph as well. What's the Y intercept? Why doesn't this go through? Oh, it's a little bump afterwards. That's why I didn't start at 00. Is that okay? Is there going to be one like that on this test? No. Will you have at the end of the year, which we're getting closer to, on your Mach Finals at least one graphing class? Yeah. They'll look very similar. That's why I gave you four provincial exams. I made up two Mach Finals identical in layout and outline. One written question per unit. A graphing question using principles of physics right to explain at the very, very end. Oh, and probably somewhere in the middle using principles of physics right to explain question two. They have changed the format a bit since then. It's now more multiple choice and less written. I think it's much better to do more written and less multiple choice. I just know they don't want to pay teachers to mark. Pardon me? I know. I also understand in the current economic climate why they don't want to pay teachers to mark. Although I think it's some of the best bang for their buck that they can do because we don't just mark. We exchange ideas at some of the best professional development I ever did. That's where I can ask any question about anything that's bugging me and I'll get a great explanation at a high school level. Tangent. We're back. Next. Yes. And you are going to have to use that at least once. How will you know? Holy smokes. They're giving me voltage and resistance, but I got no current. Oh, but they're telling me something about Coulombs and there's a how many seconds the batteries on for like you'll clearly be able to see. Okay. So every once in a while they won't give you current, but they'll give you the charge and how long it's running for either directly or indirectly. Oh heck. For you guys, I think the first question. The reason I want to say I think is I'm giving you a different version of the test. I found an older, shorter one because I was initially going to do this over the late start Friday short class. I know it's number one on my original longer version. A couple of more hints. I typed some of these questions, Irwin, and I have to be honest. Initially, I was going to do lovely graphic circuit diagrams for each one I gave up. There are some where I just describe the circuit. A battery is connected to two resistors in parallel. One resistor is three ohms. One resistor is five ohms. Draw the circuit. Draw it. No, because actually one of the IRPs says something like and sketch resistance circuits from this script, something like that. I'm not going to, I didn't give you ones like what the, there, it's pretty much draw this, then draw this, then draw this, split here because it says parallel then join here again and go to your battery kind of a thing. And I only did it on a few. Some I found old provincial exam questions that you haven't seen or I did the diagram, but this unit more than any other one would be a nightmare for me to have to try and type up everything. That's why I'm so glad I bought that book. Any others? Melissa's been trying these, good for her. Greg, let's go back to your question. This is from June 86, the year before I graduated. 23 here. Says use the following circuit diagram. Here's our circuit diagram. I apologize. It's been photocopied over and over and over. That's fine. Says in the circuit illustrated in the above diagram, which of the following changes would cause the greatest increase in the current through the six ohm resistor? Now the current going through there is I total. So really what they're saying is which of these would increase the overall current? Removing the two ohm, removing the 12 ohm, connecting another two ohm between X and Y, connecting another 12 ohm between X and Y. My battery isn't going to change. If I want current to go up, what must go down? Total resistance. How do I lower total resistance? Add another resistor in parallel. So the real question is which of those two would give you a lower overall resistance? And you could do it algebraically. Or yes, Irwin, very quickly you could go one over two plus one over two plus one over four plus one over 12 reciprocal and compare that to one over 12 plus one over two plus one over two plus one over 12 reciprocal. Whichever one of those gave you the lowest resistance would give you the biggest current, which would actually make this bulb go brighter. Pardon me, would be going dimmer because you've added one more bulb for the current to have to split up. They're all going to have to share some of their current with this one. So the I squared R, which is the other way I think of this power, R is not going to change. The voltage is going to drop and the current is going to drop. That would be a good using principles of physics. Those are the kinds of ones that they have. And I'm fairly certain, I'm going from memory here, but I'm fairly certain the one that I gave you. I photocopied the shorter version. I don't mind giving you a shorter version for the full length. I just didn't want you to run out of time on the Friday. I think it was something to do with mucking around with the resistors. Here you go, Alyssa. Another graph question from way back when. You have the resistance versus current, actually versus one over current. This was a fairly tricky one, but whatever, I'm not going to give you one that yucky. Another terminal voltage question. Here, we don't care. That looks like a newer one. In fact, that looks like the one I gave you on your quiz. Oh, that's where I got it from. It's exactly the same. I know one time I lifted one of these and I used it on one of my final exam, some diagram similar to this. And I just changed the battery. I crossed it out and I wrote something above it. It turned out it wasn't a strong enough battery to power the circuit. Mr. Duke, why am I getting a negative voltage in one of my resistors? I have to change that one in a hurry. Sure, absolutely. And easy. You wouldn't find the power. What you would find is either the voltage or the current. Probably the voltage by the time was said and done, because you would have had the voltage here, the voltage here. You would go 12, minus this, minus this. All three of these are the same, right? Once you have the voltage, if you know two, you know three, you have the current and you can go VI. Or if you want to go directly, I think power is V squared over R. I think if I substitute the voltage, there's three power equations, only one of which is on your sheet. That's where I got that one from. Now I know. Find the potential difference across the 4 ohm resistor. So find the voltage drop going through the 4 ohm resistor. Did they tell you the total current? Okay, so you'd say those two are in series. That's a 12. 1 over 6 plus 1 over 12 is 1 over R parallel. And you would do your little scratch diagram over here. This is totally a fair game question. In fact, the odds are pretty good on your written section. I'm not going to give you total current. And I'm not going to give you two, so that if I know two, I know three and it all breaks down. Probably give you at least one where, hey, you got the voltage. You got your resistors. Rewrite it using the resistor rules. Okay. Oh, there's your little internal voltage. So once you had total current, terminal voltage would be 120 minus 1 times whatever the current was minus IR. I don't think I have a recharging battery question on my test. It's fair game on the provincial, but I'll be honest. I haven't seen it in years. I almost think they put it in here because it's a neat concept. Oh, that's what's happening in my iPod. That's why I can't recharge my battery in five minutes. That's why it takes longer to recharge than to drain it sometimes depending on how the system works. Although they're getting better now and I suspect they've come up with better ways to, well, the lithium batteries, I think have a much higher internal resistance. So you can send a much greater difference in voltage through it. But my gut is that's why it's on the formula sheet and I've covered it. I haven't seen it. And again, the one or two times I've seen it, it's like been staggeringly obvious. I think even the question said this is recharging them or something like that. Or it might have had the current clearly going labeled with an arrow or something like that. That make sense? Okay. Oh, here you go. Here is an example of a recharging battery. See it? Well, what's your net voltage? Five? Oh, except actually no. I would lose a little bit through there and lose a little bit through there because this is a tougher question. First of all, the power dissipated by the 18 ohm resistor. I'm pretty sure my net overall voltage, including all of these is, let's see, five volts. My total resistance is one plus one plus 18, 20. I total is five divided by 20. Now I've got I total. Now I can walk through each one of these right the current, right the current, right. And you know the voltage drop and the power drop and whatever you need. This would be an example of a recharging one though. And can I see a date on here? Well, you can see even from the font, I'm pretty sure this was probably around 92 or 93. And I just finished the end of this. That helped, I hope, answer questions. So who would like a copy of the ones that we did? One, two, three, four, five. Five of you? Okay, let me first of all hit pause on the video.