Resistors and Ohm's Law - Video Tutorials & Practice Problems

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1

concept

Resistance and Ohm's Law

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3m

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Hey, guys. So in this video, we're going to cover two really important concepts, the resistance of a conductor. And that's really powerful equation that you need to know called OEMs law in order to solve problems. Let's go ahead and check it out. So before when we talked about the charges that flow through a conductor, we've sort of pretended that these electrons or charges could pass through this conductor sort of unobstructed. They were freely able to go from one side to the other. The reality is that they're not that simple, because inside of a conductor there are atoms. And as these electrons are traveling through this conductor, they're bouncing off of the atoms inside of that conductor. And all of those collisions that take place sort of create this internal friction that resists the amount of charges that could go from one side to the other. In other words, this internal friction resists the current. So if you have a conductor and you place some voltage or some potential difference across it, and you have these atoms that air trying to cross from one side to the other, the relationship between the current and the voltage is given by this equation where eyes equal to V. But you have to divide it by this term called Are now this our term is called the resistance and this resistance here is basically the ability for a conductor to resist the amount of charges that are trying to make it across in a certain amount of time. Three units for this resistance are in homes and given by this this letter, uh, this Greek letter capital Omega And what this resistance represents is that the larger the resistance of a conductor, the smaller amounts of current can pass through with smaller amount of charges can pass through in a specific amount of time. So we have this relationship between I, V and R, but more common than you'll see it written in this way. And this is called OEMs law. This is really important. You definitely need to commit this to memory. This is V equals I times are this is kind of like the f equals m A of physics too. So definitely commit V equals ir to memory because you're gonna be using it a lot in the future. Alright, guys, it's basically it. That's the equation that you need to know. Let's go ahead and work out some practice problems right here. Okay, so we've got a conductor and we have a voltage of 10 volts across it. We're told that six micro columns of charge flows through it every 1.5 seconds. Were supposed to figure out what the resistance of this conductor is. So as a variable when other. In other words, we're trying to figure out what our is. And we've only got one equation to you so far. So we're gonna use OEMs. Law V equals I times are, if we rearrange for this, we know that the voltage divided by the current is equal to the resistance. So we have with the voltage is that's just the 10 volts. So how do we figure what the current is because not told without explicitly is So we have to do is we have to say that the currents is going to be the charge divided by the amount of time. Remember, we can use this relationship that I is equal to Delta Q, divided by Delta T from a previous video. So if we have an amount of charge, that's flowing through six times. 10 to the minus six, divided by an amount of time, which is 1.5 seconds on. This is just gonna be four times 10 to the minus six. And that's gonna be amps. Remember that current is gonna be amps. Okay, so now we have to plug this basically back into this equation, and we have that. The resistance is equal to 10 volts, divided by the amps, which is four times 10 to the minus six. And we have a resistance that's going to be 2.5 times 10 to the sixth. And that's gonna be in homes. That's sort of like the simple for that. All right, so that's how you figure out the resistance of a conductor. Alright, guys, Thanks for watching. Let me know if you guys have any questions.

2

concept

Resistivity & Resistors in Circuits

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5m

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Alright guys. So we talk about resistance is in conductors. We want to talk about two related concepts in this video which is the rest festivity of the material and also how we deal with resistors in circuits. All right, let's go ahead and check it out. So any material has a property called the resistive ity. So this resistive, it is just a measure of how effective this particular material is at resisting charges and currents that flow through it. So it's given by this Greek letter row and the units for that are in omega meters. So if I have this conductor right here and I have some length and some cross sectional area, then in order to figure out the resistance of this conductor of this material, I need to know the resistive iti, which is basically just a constant and it just depends on what it's made of. And the resistance of this whole entire conductor is gonna be row times l over a where I just want to reiterate that this resistive ity depends on Lee on the material that it's made of. So, basically it's just a constant that's gonna be given to you for instance, row lead has a resistive ity of this number right here. But we're dealing with copper or gold or silver. Those are gonna have all different numbers. You're not gonna be expected to memorize any of them. They're gonna be given to you on tests or homework's. But this is basically the relationship between resistance is and resistive ITI, which is a property of that material. Okay, so it just depends on how effective it is at resisting charges and also how long it is toe divided by how wide it is. Okay. So similar to how we talked about circuits with capacitors. First we talked about capacitance, and then we talked about what a capacitor is in a circuit. It's the same thing here we've talked about. Resistance is, and a resistor is just a circuit element that has some resistance. And we're gonna hook it up to a battery to form a simple circuit, just like we did with capacitors. But in circuits were always gonna consider or assume that wires have zero resistance and really they have some, you know, non zero. It's like very, very, very small. What that means is that when we have a circuit connected to this resistor, but which, by the way, is given by this symbol right here. This little squiggly lines were going to say that this resistor here are has some resistance, but that the wires that hook up this resistor to the battery have little to no resistance. We're just gonna go ahead and assume that these things have zero resistance. Let me go ahead and write that out. Second. So you've got these wires here have zero resistance. Okay, so let's go ahead and check out a new example Problem right here. Who? You got a wire that's 25.1 m long and six millimeters in diameter. It's got a resistance off 15 millones. This number is actually pretty small already on. We can see that a wire that's 25 m is required for a very, very small amount of resistance. This is why in circuit problems, we assume wires to have almost zero resistance. Okay, so we're told there's potential difference right here. We're supposed to figure out what the resistive ity of this wire material is. So in other words, we're supposed to be figuring out what row is equal to. So let's go ahead and set up our equation. The relationship between resistance row the length over the area is row is R equals row L over A. So we have with the resistance of the material is I know how long this wire is. And if I can figure out if I'm assuming that this wire is cylindrical, then I can figure out the area by pi times r squared where I just want to reiterate this are right here is the radius of the wire and not the resistance just so you don't get those two things confused. Okay, so let's go ahead and manipulate this equation. I've got a that's gonna go over. We got l. That's gonna come down. And that means that our times a over l is gonna be equal to row. So, in other words, the resistance right here, which is 15 million homes 15 times 10 to the minus three. Now we have to figure out what the area is. The area is just gonna be pi times 0.3 because we're giving it 6 million m in diameter. But we need the radius. So that means we need half of this number right here. And then we need to put it in the right units. Now we have to square that. Now we have two divided by the length of the wire, which is gonna be 25.1. And this is in meters, so we don't have to change anything about that. OK, so you go ahead and work this out in your calculators, plugging everything carefully. You should get a row. A resistive ity. That is 1.69 times 10 to the minus seven. And that is gonna be ohm meters. Now, this corresponds to a material that's copper, which is usually what MOCs wires are made of. So this is copper. So we need a wire that's 25 m long. It's like 75 ft long just to get a resistance. That's 15 millones, which is very, very small. Okay, so again, this is sort of reiterating that wires have zero resistance in the circuit. All right, so now we're supposed to do is we're supposed to figure out what the current in the wire is. How do we do that? Well, we're told specifically that the voltage across this wire is volts. And now we have with the resistance is so we can figure out the current using homes law. So if we need to figure out, I we just have to relate it back to V equals I times are now I know again, we're supposed to assume that this has zero assistance, but we're told specifically that this thing does have some resistance, so we have to plug that in. Okay, If this was a circuit problem, we didn't have to worry about it. Okay, so we've got V over R is gonna equal toe I So we've got that 23 divided by 15 times. 10 to the minus three is going to give us the currents, and that's equal to 1.53 times 10 to the third. And that's gonna be it's not cool arms. That's gonna be an apse. Alright, so that is the current due to the resistance and the voltage. Let me know if you guys have any questions and I'll see you guys the next one

3

Problem

Problem

A resistor has a current through it of 5 A. If the EMF across the resistor is 10 V, what is the resistance of this resistor?

A

2 Ω

B

50 Ω

C

0.2 Ω

D

5 Ω

4

example

Current Through Unknown Resistor

Video duration:

2m

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Alright, guys, look, we're gonna work this one out together, so we have a cylindrical resistor and we're told a bunch of information about it. What's the resistive ity? The dimensions of that resistor. And also how much E m f vultures across it. We're supposed to figure out what is the current across the resistor right here. So if we wanted to figure out current and its relationship to the resistance and voltage first we have to go and related to OEMs law. So if we wanted to figure out the current is we have to relate it to V equals I r. So if we just divide over the are so then V over r just becomes the current. So I have the voltage. I'm told that the e m f or the voltage is five volts. Now what I need to do is figure out the resistance. I'm told what the resistive ITI is and some of the properties of this resistor, but I don't have the resistance, so I have to go and use another equation. This resistance, remember for any resistor of some resistive ity is gonna be row times l over a and we assume that it's cylindrical. We have that. It's gonna be row times the length divided by the area of the cylinder, which is going to be, by the way, this is the cross sectional area. So if we have, like a cylinder like this, so we have a cylinder, then this area right here is actually the cross sectional area of the specific cylinder, not the actual surface area. That's actually ah, common pitfall for that Students run into, so it's gonna be the cross sectional area. So I'm gonna write that cross section area. So that's what that A is equal to. Okay, so that means that the cross sectional area is just gonna be the area of a circle, which is pi times R squared, which is the radius. So this is the radius. Okay, cool. So let's go ahead and calculate with the resistance is it's gonna be row, which is 2.2 times 10 to the minus seven. Now we have the length of the this particular resistor, which is two centimeters of 20.2 Now we have to divide it by the area, which is gonna be pi times. We have a radius of five millimeters. So we're gonna do 50.5 and then square. That's and then all that stuff that's gonna go in the denominator, right? So we get a resistance of 5.6 times 10 to the minus five, and that's gonna be an alms. So now we can stick it back into homes long to figure out the current is So that means that the voltage, which is five volts to buy the resistance 5.6 times 10 to the minus five is gonna be equal to the current. In other words, we have a current of 8.9 times 10 to the fourth, and that's and that's actually a lot of currents right there. So that is the currents. That's how we figure out these kinds of things. We use our resistance relativity equations and using homes laws. Okay, let me know if you guys have any questions

5

Problem

Problem

Two resistors are made of the same material, one twice as long as the other. If the current through the shorter resistor is 5 A, what is the current through the longer resistor if they both have the same potential difference and cross-sectional area?

A

2.5 Α

B

5.0 Α

C

10 Α

D

Not enough information given

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