by Patrick Ford

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

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