Hi, my name is Charles. I am one of the maths teachers from the Maxim Workshop. I am now going to teach you how to do some math. In this video, I am going to show how to calculate voltage drop. Now the voltage drop, commonly known as pd or potential difference is more commonly related to physics and of course, electricity. Now, we are going to a particular circuit, you might find, you have a power supply. That's normally denoted by this, let's say battery or something like that, and we give this a rating of 10 volts. Now, voltage again is like the electrical force that provides the electrons with the electrical energy to move around, okay. Now, we might have a few resistors within our circuit. And, you might think of this, I mean a good analogy would be to think of the electrons that whiz around, move at an extremely fast pace, it's actually the energy that goes around at the speed of light. But the electrons that whiz around, you might think of them as being like water along a stream. And these as your resistors, you might think of them as being logs of wood within say, your street park. Now, any canoeist that goes around will find some sort of resistance at these two points here. Now, in order for him to actually move fast, those two points at the same speed, will obviously have to try harder, I apply a force. Now, this is the electrical force that you see here. Now, over these two points we find that force is used. Electrical force is used to move the current through these points at the same speed, okay. So within a circuit like this, the current does flow at the same speed. I mean, even though you find the resistors here, the current moves through these points at the same speed. This force maintains the speed of the current, okay. So, if we get these two resistor values, you might call this 2 ohms, and we might say that this one has a resistance of 3 ohms. Now, that's the standard measurement for resistance. Within electricity, you have relationship like this, V=IR, so the voltage equals I multiplied by R. So the first one, what you want to do is add these two resistances up and in the configuration that we have them here, we can do them. So, 2+3 is 5, so your total resistance is 5 and your total voltage is 10, okay. So we have a manipulation of this, provides us with a current which is just basically v. We take the R over to the side of the V, now the R is multiplying with this side. So when it gets over to this side, it's going to do opposite action which is division. So, we got V divided by R. Now, again the voltage is 10 and the resistance is 5. So, all we have here is 2 amps as our current, okay. Now, we go back to these two points here and these two points here. Now just a recap, we want to find the voltage that would be used across these two points to maintain the current speed. And of course, these two points to maintain the current speed. So, the equation here, we have used this equation again and again. And, so we know the amps and we know the voltage, okay. So, what we don't know though is the voltage across these two points individually. So, we know in total but not assigned to these two points. Now, to calculate the voltage drops across this resistance and this resistance, we are going to need this equation. So, you come out there and we say the voltage drop across these two points is equal to the current that flows across here by the resistance. So again, V=IR. So the current is 2 amps and the resistance is 2 ohms. So we got 2 amps multiplied by 2 ohms which is just equal to 4 volts, okay. Now, that is the particular voltage drop that occurs across here. Now, that gives you two ways of calculating the voltage drop across here. The first way is the simplest, you have to actually look at the whole circuit and say that, whatever voltage we start of with as the source and all the other parts within here must use it up. So obviously, if we have 4 volts ov

**万花筒**

- Appraisal: French Bru Doll, ca. 1895
- Appraisal: 1919 Rockwell "The Little Model" Oil Painting
- Appraisal: Jessie Willcox Smith Oil Painting with...
- September 4, 2013 - "Impressionism, Fashion, and Modernity"
- ANTIQUES ROADSHOW - Appraisal: 1909 Charles R. Knight Oil
- Appraisal: Chinese Enameled Porcelain Bowl, ca. 1920

#### Discuss

0 comments characters remainingSubmit