 Here we go. The diagram below shows part of an electrical circuit, okay? What are the magnitude and the direction of the current passing through resistor R5? I have no idea, but I'm going to try and hopscotch my way there using our skier rules. For example, I notice I have seven amps going this way. How many amps going this way right here? For me to get seven coming out, what must I have right here? This has to be four amps. I went, okay. So I have four amps going this way. Oh, and I have two amps going this way. So four this way, two this way. How many must be coming in? Six. Oh, six amps. I got the magnitude direction towards the left. Six amps. Oh, I guess I only needed the magnitude anyways. Would have been a little better if it also had six to the right, I guess. But there you go. Kirchhoff's laws for current. Current coming into a junction, and I've used the analogy of skiers equals current coming out of the junction. Number two, find the current through R4. I would love to use the total current. I quickly glanced around. No. Then I would love to use a resistor where they've told me two things. By the way, this is the fancy circuit symbol for a light bulb in case you're wondering. I went, no, they haven't told me two things. I'm going to have to rewrite this as one resistor. How many resistors do I have in parallel here? Three of them. So one over R parallel is going to be one over 18 plus one over nine plus one over three, and then reciprocal afterwards. One, one over 18 plus one over nine plus one over three. Oh, reciprocal is going to be two. I said, okay. Mathematically, this circuit is the same as having a two ohm right there and an eight ohm right there before you hit the 30 volt battery. There's my quick scratch diagram, and I think that's enough for me to figure out the total resistance. What's the total resistance of this circuit? So R total is 10 ohms. Now I can tell you I total because that's going to be B total divided by R total. It's going to be 30 divided by 10, three amps. I said, great. Now I know where I'm kind of headed here. So the current is flowing this way. It's the current splits up between those three, but how many amps are going through here? Three. And that's why I said you always look for it. They've told you two things first because if you know two, you know four. I'll use that. I'm not interested in power. Although I think I am later on in this question. What's the voltage here? We lose 24 volts going through that resistor. So how many volts must we lose going through each one of these steels? Each one's got to be six volts high because I can go this way, or this way, or this way. So six volts, six volts, six volts. Now they want the current through R4. Current equals V over R. So current four is going to be voltage four divided by resistor four. The voltage in, oh, I didn't realize I was done. Three amps. Good gosh, Mr. Dewey. You guys were going, you're done, Mr. Dewey. I4 equals three amps. Fine. I was doing part, there was a part B. Was there not on the next page? Okay. Find the voltage drop across R2. And that's what we kind of had just worked our way towards. And that we said was six volts. And I'll show you how I gave up part marks a little bit later. But right now, if you got three amps, four marks, if you got six volts, two marks, what other questions could they ask? Instead of voltage drop, I could say how many watts are in this bulb, or this bulb, or this bulb. Could have asked for the current in each one. Because as soon as you know two, you know three. Number three, find the power dissipated by R1. Have they told me two anywhere? Nope. Have they told me I total? Nope. So I'm going to need to rewrite this. Now this time there's two separate parallel circuits. There's this one. One over R parallel equals one over R1 plus one over R2. One over 14 plus one over 14. One over 14 plus one over 14. It's going to be seven, I think, when all is said and done. Okay. So this one is the same as a seven ohm resistor. So this one is the same as a seven ohm resistor. What about this one here? One over R parallel equals one over 24 plus one over 24 plus one over 12. One over six. Oh it is six. R parallel equals six. So if I do my little scratch diagram, 12 volts, really it's this. 12 volts, a seven, a three, and a six. What's my total resistance here? What's my total current here? Well it's going to be the voltage divided by that .75 amps. Is that right? Now let's go back to this diagram. They want the power in R1. Power is either going to be VI or I squared R or V squared over R, whichever. Now there is a bit of a shortcut here. Are these two resistors the same? So here's my total current of .75 amps. What can you tell me the current does if the resistors are identical? Splits evenly. If there was three 14s it would split into thirds. If there was four 14s it would split into fourths here. Two 14s. It's going to split in half. Half of .75, this is .375 amps and this is .375 amps. Power is I squared R. It's VI but if you plug V equals I times R into there you use that. I could use VI but then I have to find the voltage here. I already got the equation. It's going to be .375 squared times 14. Clear. .375 squared squared times 14. You get 1.97 watts. Is that right? 1.97 watts. Let me look at my actual answer key. So in terms of part marks, if you need them for question number two, I gave one mark if I saw that you had done a half mark for finding the two ohms. One mark if I saw any variation of Ohm's law written somewhere. Half mark for getting the three amps. Oh no, one mark for getting the three amps. Half mark for going V over T. Here, there's how I gave out part marks. If I saw the seven ohms somewhere that got you one. If I saw the six ohms somewhere that got you one. If I saw 12 over 16 somewhere that got you one. If I saw .75 that got you one. And then I squared R. Did anybody do it differently? Did anybody actually find the total current, total resistance and then split their way? So you could find the voltage here. And then you could say, I know what the voltage drop is here because this was a six ohm resistor. V equals I times R. .75 times 6. Each one of these is going to be how many volts? You could figure that out. Each one of these is how many volts? And then again, if you know two, you know three. Give yourself a lovely score out of 15, please. And then pass them forward, please. So less than seven. Ready? All we're doing today is making our circuits a bit more accurate. Now what we're not going to deal with this year is wires that have resistance because wires do have resistance, but it's pretty difficult because as the wires heat up, the resistance changes. So we're going to do a little bit more of a calculus to analyze them nicely. But what we are going to look at today is batteries. Batteries actually have an internal resistance. There is no perfect battery. Brandon, when the charges in our circuit go through the battery, we've been fibbing a little bit. We've had a lovely battery, but I got to be honest, each little battery should actually have built in itself a tiny little internal resistor. It loses a bit of energy just going on the chairlift itself. So that's what we're going to be talking about today. We call it terminal voltage. Terminal voltage is the actual measured voltage that you get when you measure between the ends of a battery or cell. And the symbol, Nick, is this. We have our maximum theoretical voltage. The symbol for the maximum theoretical voltage is this curly letter E. We'll talk about what that means in a second. It stands for EMF. Inside the battery, you have an internal resistance. And the way that they show that this is part of the battery is they will often put a little dotted line box around it to say, this is inside the battery. What this resistor is kind of like is when you get off the chairlift at the ski slope, do you ski straight to the nearest hill? Or when you get off the chairlift, is there about a little four foot down slope to get off the chairlift? That's what we're talking about with this internal resistor. It's the tiny hill you ski down getting off the chairlift before you head to the slope. What that means is a six volt battery, for example, you never get six volts completely from it. The chairlift may be six volts high, but you're going to ski down a tiny hill. You're going to lose a tiny bit of voltage just going through the battery. The EMF of a six volt battery, the maximum theoretical chairlift. And the way I remember this, by the way, what analogy have I used for the battery? What have we called it in our ski hill analogy? It looks sort of like a C. It's an EMF symbol, I know, but it's your maximum chairlift height. But then you lose a tiny bit getting off the battery. Let's read. Terminal voltage is the voltage potential difference between the ends of a battery or cell. The ends or posts of the battery are called terminals. A cell contributes some resistance to a circuit. This is called the internal resistance of a cell. Here's your diagram. Here's the circuit diagram. Where I equals the current through the cell. Little R, we use a lowercase R because it's supposed to be, Nick, ideally, a very small resistance. In fact, if we could make it perfect, that'd be even better. We haven't got there in our technology yet. We use a little R for any internal resistance, but it's there. And then we use this symbol here, and this is your chairlift height, voltage. And then they use V with a subscripted T for terminal voltage. And this is your height after skiing off the chairlift. Most of you have skied at least once. When you get to the top of the chairlift, there's a little downhill slanky thing. So in my ski hill analogy, the internal resistance, even though it's sometimes written above the positive like in this circuit, it's sometimes written below the negative like in this circuit. It kind of varies. But it's like that tiny little bump when you get off the chairlift. It's actually not the resistance that's the bump. It's the voltage that you lose going through here. Voltage is what times what? V equals what times what? I times R. So as soon as you have a current flowing, you lose a bit of IR going through here. Oh, now here's where the ski hill analogy breaks down. More current means more IR, somehow this dip gets bigger. So that doesn't really happen on mountains. It doesn't change the height. It's okay. What happens to the cell voltage measured by the voltmeter if we close this switch so that the current drawn by the circuit increases? We're going to do a little scratch diagram over here. So here's our battery, but this battery has an internal resistance. And I'll show that by doing a little dotted rectangular box around the battery. And then the rest of the circuit, we have one resistor and we have now a second resistor. And we're going to label this event with the lowercase R. And we're going to label this battery with that funky little E symbol. And we're going to call this location A. We're going to call this location B. Now this, when you go through a resistor, Kyle, you lose some height. You lose some voltage. And the voltage is I times R. Before the battery is connected, is any current flowing? So what's your I times R through here? Zero. Before the battery is connected, the voltage is the EMF. That's what appears on the side of the battery, the 1.5 volts, the 6 volts, whatever kind of battery you're looking at. What they're actually telling you is the EMF. That's the chairlift height. As soon as you connect it, you have some current flowing through here, you lose some voltage. This little voltage loss is going to be I times little R. So if you close this switch, which makes the total current get bigger, what's going to happen to your voltage loss? What's going to happen to this term here if that gets bigger? Yeah, so instead of maybe getting, well, okay, when it's disconnected, let's suppose it's a 6 volt battery. When you first hook it up, you might be getting 5.8 volts. When you close the switch, increase the current. Now you might be measuring 5.6 volts across the terminal. So decreases. We're going to write this. Voltage loss equals I total times little R. Remember your internal resistor. As I increases, voltage loss decreases. Why write decreases when you use an up arrow for increases, Mr. Dewick? As I increases, let's try this again. As I gets bigger, the amount of voltage that you lose through here gets bigger, because I times R. And the terminal voltage is your chairlift height minus what you lose skiing off the chairlift. You guys stay here, right? Your terminal voltage, which is this height here, it's your chairlift height minus this height right here. So the terminal gets smaller because you're subtracting a bigger number. Example two says find an expression for the terminal voltage. This is the actual voltage once you're running a current through it. I think I just did. It's your chairlift minus whatever you lose going through the battery. For which diagram? This one here. We said that you can either have the resistor, sometimes they'll put it above the positive, like in this diagram. Sometimes they'll put it below the negative, like in this diagram. They'll always put a box around it, or they'll label it with a lower case R. In all honesty, I suspect in real life it depends on the design of the battery. So back to my question. It says find an expression right there. The terminal voltage is how high you are when you get off the chairlift, minus whatever tiny amount you lose skiing off the chairlift to the first hill. Unfortunately, this amount depends on the current, which means my ski hill analogy breaks down because depending on how much current is flowing, that changes how much you lose skiing off to the next hill, which doesn't happen in real life. Sorry, not perfect. So here's a good question. Yo, I'm getting there. Hang on. That's recharging a battery. Patience, young grasshopper. Have I done a big box yet with the formula inside the box yet? No, no. Example three, find the terminal voltage. And Troy, that phrase there is what's telling you what's going on over there. The cell is supplying the energy. Okay. What's the total resistance in this cell? Not 30. What's the total resistance in this cell? 32. What's the total voltage in this cell? And now I'm going to be fussy. Instead of using V-total, I'm going to use my EMF symbol. What's my chairlift height? 20. Is this okay so far, Aaron? Right? What's the total current flowing through this? Well, it's going to be total voltage divided by total resistance. Remember, this is your V-total, your chairlift height. It's going to be 20 volts divided by 32. How many amps are flowing through this particular circuit? 0.625. So how many amps are flowing through this 30 ohm resistor? 0.625. How many volts do we lose going through this 30 ohm resistor? Can someone go I times R, please? 18.75. 18.75 volts. How many amps are flowing through the battery itself? 0.625. Oh, I called this volts up here, didn't I? Let's call that amps, Mr. Dewick. Sorry, amps. 0.625 amps. How many volts do we lose crossing the battery itself? Go I times R. 1.25. So here's what this means. Miguel, the battery itself is labeled 20 volts. But if you hook this up and you put a volt meter right there and right there, you would not measure 20 volts. You know what you would measure? The terminal voltage is the EMF minus I times R. It's the 20 minus, what did we get when we went I times little R? I'll put the numbers here. 0.625 times 2. You would measure 18.75 volts, 18.8 if I go to 2 or 3 sig figs. That's the actual voltage you're getting from this circuit. Cherlift, actual. Example 4. It says find the terminal voltage and find the internal resistance. Okay. So we have an ammeter and we know we're getting 1.5 amps. So how many amps are flowing through here? 1.5 amps. How many volts do we lose going through this resistor? Troy? 45 volts. But how high is my chairlift? How high is my EMF? 50. So what must I lose going through the battery getting off the chairlift from that internal resistor? I must lose 5 volts. Oh, how many amps are going through here? Oh, first of all, the terminal voltage, what would the terminal voltage be measuring? 45 volts. 50 minus what you lose going through the battery. Oh, and what would the internal resistance be? Well, resistance equals V over R. Over R, Mr. DeWitt, V over I. How many volts is there inside this resistor? 5 divided by what's the total current? By the way, do you see why if they tell you the total current? 1.5. What must the internal resistance of this battery be? 3.3 repeating? 3.33 ohms. Now that internal resistance it's built in, it doesn't change. However, now that I know this, if I hooked it up to a different resistor, I could predict what the current would be and what the terminal voltage would be. Because I would do a new V total or R total, whatever my new resistor was. I know V total is in theory 50. What's it really going to be? What's the terminal? So far so good. Now, earlier I cut Troy off. Troy said, Mr. DeWitt, I'm looking at the formula sheet. And the formula sheet, Mr. DeWitt, that little equation that you gave us, it has a plus or minus right there, doesn't it? Plus is when you're recharging the battery, going against the grain. This is when you're running a regular circuit. And it's this tiny little bit of plus that actually recharges your battery, which is why we call them trickle chargers. It's why they take a long time to recharge the battery because most of the voltage cancels itself out. It's that little tiny bit of extra that trickles in and gradually resets the chemical reaction back to zero. So speaking of recharging, it says find an expression for the terminal voltage if the cell is being charged. So here's my charger. Here is my, excuse me, my terminal voltage here is going to be my chairlift. But this time we have the current going in the opposite direction. Instead of going downhill through the resistor, we're going back uphill. So instead of losing voltage, there's the math of recharging a battery. And like I just finished saying, it's this tiny little bit right here that actually does the recharging, which actually means we don't want perfect batteries for a battery to recharge. It must have an internal resistance. And this is also why the newer nice lithium ones work pretty good. But this is it's because the resistance is a physical chemical issue. This is what's gradually over two or three years starts to wear out and your batteries hold less and less and less and eventually are not able to recharge at all. That's why batteries go bad. Is that a question? No, the chemical, so Matt asked are ones that are not rechargeable, no internal resistance. No, the chemical reaction inside is not reversible. Yep, they're usually carbon batteries or carbon zinc as opposed to the lithium ones. So it says find the terminal voltage. The cell is being charged. Okay, I don't think we actually, I don't think we need to use the terminal voltage equation. I think we can figure this out by using Ohm's law and resistance. I have 30 volts going this way, 9 volts going this way. What's my net voltage? What's my total voltage of this circuit? Like we said last day, 21 volts. What's my total overall resistance of this circuit? They're all in series, which is nice. What's my total overall resistance of this circuit? 11. What's my total overall current? It's going to be the total voltage divided by the total resistance, 21 over 11. My total current is 21 over 11. Folks, there's a lot of chatter going on back there. Can we stop, please? Thank you. 1.91 amps. So how many amps are leaving right here? 1.91. How many amps through here? 1.91 amps. How many amps through here? 1.91 amps. Here's the key. How many amps through here? 1.91 amps. Now, we're going against the grain. Normally, this cell would have a 9-volt voltage drop. We're going against the grain. So the terminal voltage, I'm using V term because here I'm using V total. For terminal voltage, I like to use voltage and a subscript of A and B to say I'm measuring it across location A and location B. Is that what I do on the formula sheet tripe? Can you look? Or is it V term? Yeah, okay. So I'm going to use terminal because we have V total T and V terminal. The terminal voltage is my 9 volts plus my I times little r because we're recharging. It's 9 plus 1.91 times 1. Oh, and I can go 1.91 times 1 in my head. It's going to be 9 plus 1.91, 10.91 volts, and it's that extra 1.91 volts, the extra 1.9 volts. Let's recharging it. Is that okay? In summary, if the cell is supplying energy to the circuit, you have the current going that way. Oh, sorry. If the cell is supplying energy to the circuit, you have the current going this way. This is not a rechargeable battery, my bad. And you have V terminal equals your maximum chairlift minus whatever you lose going through the battery because you're going downhill through the resistor. And next page. If we're recharging so the current is going this way, the current is being forced in the opposite direction, then it's this plus whatever voltage there is in this resistor. Terminal voltage. A little more accurate, a little more realistic. Let's do a few more, I think it's a few more examples. No, exercises and review. Holy smokes. So we are boys and girls officially done the unit, right? Maximum minus whatever you lose because you're going in the correct direction, which means if you're going through resistor, you're losing voltage. Maximum plus because another cell is overpowering and forcing the current in the opposite direction. Okay? Test a week from today. Problem, folks. I don't think I'm going to be able to fit in an after-school tutorial, but I'm also sensing most of you are doing pretty good with the material. I got a meeting Thursday after school. I will be around Friday after school and I will be around Wednesday after school. I just don't know if it's going to be a formal fancy schmancy tutorial. What's your homework? Deliberately a short lesson. Most of your homework is, hey, why don't you start the review? I'm just going to give you a couple to try from here. Try number one. Try number three. Try number four. The rest of those are all lovely, but I'm going to say do one, three and four and your homework is start working on the big circuitry review. Everything is fair game. Now this one in particular, before you ask me questions from the big review, go look at my online answer key because most of the time it'll be one thing that you can't find and I tried to show my work as much as possible. Oh, I forgot that. Okay? The class is yours. Only one quiz this unit. Short unit.