Hey, guys. So in this video, we're going to cover two really important concepts: the resistance of a conductor and this really powerful equation that you need to know called Ohm's law in order to solve problems. Let's go ahead and check it out. Before, when we talked about charges that flow through a conductor, we 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 can 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 are trying to cross from one side to the other, the relationship between the current and the voltage is given by this equation where I is equal to V, but you have to divide it by this term called R. Now this R 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. The units for this resistance are in ohms, and it's given by this Greek letter, capital Omega. And what this resistance represents is that the larger the resistance of a conductor, the smaller amount of current can pass through it. The 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 commonly you'll see it written in this way. And this is called Ohm's law. This is really important. You definitely need to commit this to memory. This is V equals I times R. This is kind of like the F equals MA of physics 2. So definitely commit V equals IR to memory because you're going to be using it a lot in the future. Alright, guys. That'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 6 microcoulombs of charge flows through it every 1.5 seconds. We're supposed to figure out what the resistance of this conductor is. So as a variable, we're not in other words, we're trying to figure out what R is. And we have only got one equation to use so far. So we're going to use Ohm's law. V equals I times R. If we rearrange for this, we know that the voltage divided by the current is equal to the resistance. So we have what the voltage is. That's just the 10 volts. So how do we figure out what the current is? Because we're not told what that explicitly is. So what we have to do is we have to say that the current is going to be the charge divided by the amount of time. So remember we can use this I = ΔQ Δt from a previous video. So if we have an amount of charge that's flowing through, 6 times 10 to the minus 6, divided by an amount of time, which is 1.5 seconds, then this is just going to be 4 times 10 to the minus 6, and that's going to be amps. Remember that current is going to 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 4 times 10 to the minus 6. And we have a resistance that's going to be 2.5 times 10 to the 6, and that's going to be in ohms. That's sort of like the symbol for that. Alright? 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.

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# Resistors and Ohm's Law - Online Tutor, Practice Problems & Exam Prep

Understanding the resistance of a conductor is crucial in electrical circuits. Resistance, denoted as R, opposes the flow of electric current, represented by I, when a voltage V is applied. Ohm's Law, expressed as V = IR, is fundamental for calculating these relationships. Additionally, resistivity ρ quantifies a material's resistance, influenced by its length and cross-sectional area, impacting circuit design and efficiency.

### Resistance and Ohm's Law

#### Video transcript

### Resistivity & Resistors in Circuits

#### Video transcript

All right, guys. So we talked about resistances in conductors. We want to talk about two related concepts in this video, which are the resistivity of a material and also how we deal with resistors in circuits. Alright? Let's go ahead and check it out. So any material has a property called the resistivity. So this resistivity 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 rho and the units for that are in ohm 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 or this material, I need to know the resistivity, 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 ρ·ℓ/A, where I just want to reiterate that this resistivity depends only on the material that it's made of. So basically, it's just a constant that's gonna be given to you. For instance, rho lead has a resistivity of this number right here. But if 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 homeworks. But this is basically the relationship between resistances and resistivity, 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 too, 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 resistances 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, we're always gonna consider or assume that wires have zero resistance. And really, they have some, you know, nonzero. It's like 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, these little squiggly lines, we're gonna say that this resistor here R 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 a second. So we've got these wires here have zero resistance. Okay? So let's go ahead and check out an example problem right here. You've got a wire that's 25.1 meters long and 6 millimeters in diameter. It's got a resistance of 15 milliohms. Ohms. This number is actually pretty small already and we can see that a wire that's 25 meters 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 a potential difference right here. We're supposed to figure out what the resistivity of this wire material is. So in other words, we're supposed to be figuring out what rho is equal to. So let's go ahead and set up our equation. The relationship between resistance, rho, the length over the area is R = rhoL / A. So we have what 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 πr2. Where I just want to reiterate, this r 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've got L that's gonna come down. And that means that R * A / L is gonna be equal to rho. So in other words, the resistance right here, which is 15 milliohms, 15 times 10 to the minus 3, now we have to figure out what the area is. The area is just gonna be pi times 0.003 because we're given 6 millimeters 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. And now we have to divide it 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. Okay? So if you go ahead and work this out in your calculators, plug everything carefully, you should get a rho, a resistivity that is 1.69 times 10 to the minus 8, and that is gonna be ohm meters. Now this corresponds to a material that's copper, which is usually what most wires are made of. So this is copper. So we need a wire that's 25 meters long. That's like 75 feet long just to get a resistance that's 15 milliohms, which is very very small. Okay? So again, this is sort of reiterating that wires have zero resistance in a circuit. Alright. So now what we're supposed to do is we're supposed to figure out what the current in the wire is. Now how do we do that? Well, we're told specifically that the voltage across this wire is 23 volts, and now we have what the resistance is we can figure out the current using Ohm's law. So if we need to figure out I, we just have to relate it back to V = I * R. Now, I know, again, we're supposed to assume that this has 0 resistance, 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 wouldn't have to worry about it. Okay? So we've got, V / R is gonna equal to I. So we've got that 23 divided by 15 times 10 to the minus 3 is gonna give us the current, and that's.equal to 1.53 times 10 to the 3rd, and that's gonna be in amps. Alright. So that is the current due to the resistance and the voltage. Let me know if you guys have any questions.

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?

### Current Through Unknown Resistor

#### Video transcript

Alright, guys. Look at it. We're going to work this one out together. So we have a cylindrical resistor and we're told a bunch of information about it. What's the resistivity, the dimensions of that resistor, and also how much EMF or voltage is across it. We're supposed to figure out what is the current across this resistor right here. So if we wanted to figure out the current and its relationship to the resistance and voltage, first, we have to go ahead and relate it to Ohm's law. So if we wanted to figure out what the current is, we have to relate it to V=IR. So if we just divide over the R, then V/R just becomes the current. I have the voltage. I'm told that the EMF or the voltage is 5 volts. Now what I need to do is figure out the resistance. I'm told what the resistivity 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 resistivity is going to be ρ×l/A. And we assume that it's cylindrical. We have that it's going to be ρ 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 a cylinder like this, so if 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 a common pitfall that students run into, so it's going to be the cross-sectional area. So I'm going to write that, cross-section area. So that means that the cross-sectional area is just going to be the area of a circle, which is π×r2, which is the radius. Okay. Cool. So let's go ahead and calculate where the resistance is. It's going to be ρ, which is 2.2×10-7. Now we have the length of this particular resistor, which is 2 centimeters or 0.02. Now we have to divide it by the area, which is going to be π×0.0052, and then all that stuff that's going to go in the denominator. Right? So we get a resistance of 5.6×10-5, and that's going to be in ohms. So now we can stick it back into Ohm's law and figure out what the current is. So that means that the voltage, which is 5 volts, divided by the resistance, 5.6×10-5, is going to be equal to the current. In other words, we have a current of 8.9×104 and it's amps. So actually a lot of current right there. So that is the current. That's how we figure out these kinds of things. We use our resistance, resistivity equations, and use Ohm's laws. Okay? Let me know if you guys have any questions.

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?

2.5 Α

5.0 Α

10 Α

Not enough information given

## Do you want more practice?

More sets### Here’s what students ask on this topic:

What is Ohm's Law and how is it used in electrical circuits?

Ohm's Law is a fundamental principle in electrical circuits, expressed as $V=IR$. It states that the voltage (V) across a conductor is directly proportional to the current (I) flowing through it, with the proportionality constant being the resistance (R). This law is crucial for calculating the relationships between voltage, current, and resistance in circuits. For example, if you know the voltage and resistance, you can determine the current using $I=\frac{V}{R}$. Understanding Ohm's Law helps in designing and analyzing electrical circuits effectively.

How do you calculate the resistance of a conductor using its resistivity?

The resistance (R) of a conductor can be calculated using its resistivity (ρ) with the formula $R=\frac{\rho L}{A}$, where L is the length of the conductor and A is its cross-sectional area. Resistivity is a material-specific property that quantifies how strongly a material opposes the flow of electric current. For example, if you have a copper wire with a length of 2 meters and a cross-sectional area of 0.0001 square meters, and the resistivity of copper is $1.68\times 10\u207b8\Omega m$, you can calculate the resistance as $0.336\Omega $.

What is the difference between resistance and resistivity?

Resistance (R) and resistivity (ρ) are related but distinct concepts in electrical circuits. Resistance is a measure of how much a specific object opposes the flow of electric current, and it depends on the object's dimensions and material. It is given by the formula $R=\frac{\rho L}{A}$, where L is the length and A is the cross-sectional area. Resistivity, on the other hand, is an intrinsic property of a material that quantifies how strongly the material itself opposes current flow, regardless of its shape or size. It is measured in ohm-meters (Ω·m). While resistivity is a material constant, resistance varies with the dimensions of the conductor.

How do you calculate current using Ohm's Law?

To calculate the current (I) using Ohm's Law, you can rearrange the formula $V=IR$ to solve for I: $I=\frac{V}{R}$. This means that the current is equal to the voltage (V) divided by the resistance (R). For example, if you have a circuit with a voltage of 12 volts and a resistance of 4 ohms, the current would be $3A$ (amperes). This calculation is essential for understanding how much current flows through a circuit given a specific voltage and resistance.

What are the units of resistance and resistivity?

The unit of resistance (R) is the ohm (Ω), which quantifies how much a conductor opposes the flow of electric current. One ohm is defined as the resistance between two points of a conductor when a constant potential difference of one volt applied between these points produces a current of one ampere. The unit of resistivity (ρ) is the ohm-meter (Ω·m). Resistivity measures how strongly a material opposes the flow of electric current per unit length and cross-sectional area. These units are fundamental in electrical engineering and physics for analyzing and designing circuits.

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