Listrik Dinamis • Part 1: Hukum Ohm, Hukum Coulomb, dan Resistor

Hello everyone, welcome back to the General Science channel. This channel is for those of you who want to understand Math, Physics, and Chemistry for high school. In this video, we will discuss the 12th grade physical education, namely Dynamics or Direction Linear Linear. And in this first part, we will discuss the Law of Ohm, Law of Coulomb, and Resistor.

For the next part, you can click the link in the upper right corner or you can also click the link in the description to watch the video. To get a complete understanding, make sure to watch this video from beginning to end. Before we start, don't forget to click subscribe by pressing the button on the bottom right and don't forget to press the bell so you don't miss our latest videos. Okay, let's get started.

Let's discuss about Coulomb law, Ohm law, and resistor law. The first formula is Coulomb law, which you already know from the SMP, which is I = Ki/T. I is the power of the electric current in Ampere, Ki is the power of the electric current in Coulomb, or C, and T is the time in Seconds. Okay, next is Ohm law. Ohm law is V = I x R. V is the difference in potential or voltage in Volt.

I is the force of current in amperes and R is the resistance in ohms. Okay, let's go to the third formula, which is the resistor and resistance. R is equal to Rho L/A. R is the resistance in ohms. Rho is not a mass of a type, Rho here is the resistance of a type. The unit is ohm meter. L is the length of the wire in meters and A is the width of the wire in meters per square.

So this is the formula to calculate a wire-shaped resistor that has a resistance of Rho with length L and with a width of the transducer as A. Let's go to the last formula. Resistor and temperature. So if the resistor is heated, the resistance will also increase. Here, for example, the resistance of the original resistor is R0. Heated with a temperature change of delta T.

then the resistance after heating or RT is R0 times 1 + alpha delta T where alpha is the resistance coefficient in per degree Celsius or per Kelvin later known in the question for easier explanation, let's go straight to the topic

A flash light is charged with a current of 0.75A by a battery that produces a difference of potential of 6V. Determine the A, the voltage of the flash light, the B, the amount of charge and the amount of electrons that pass through the flash light in 1 minute. So here we first identify what is known as a current. This is a current, this means I.

then here the potential difference, this is V we look for the first one is the wavelength of the lamp, what is the wavelength? the wavelength is R, we know V and I, we ask to find R, so we use the law of Ohm, okay? so V is equal to I times R, the law of Ohm, which means 6 is equal to 0.75 times R

or R = 6 / 0.75 the result is R = 8 Ohm okay, let's go to question B the amount of content and the amount of electrons what is this content? the symbol of this content is Ki what is this electron? the electron is N what is N? I will explain later let's calculate the amount of content first it means we have to use Coulomb's law okay, what is Coulomb's law? I = Ki per T

i is 0.75, the key is found. What is t? t is time. Time is in seconds. Because it's still one minute, we have to make it a second times 60, which means 60 seconds. Okay, so here we put 60, which means the key is 0.75 times 60, or the key is 45. Now we will calculate the amount of electrons. Now,

This is what you need to pay attention to. 1 electron is 1.6 x 10^-19 coulombs. 1 electron is 1.6 x 10^-19 coulombs. This means that if the content here is 45 coulombs, there are several electrons, which means that 45 coulombs is divided by 1.6 x 10^-19.

So if 1.6 times 10 to the power of minus 19 is labeled as E, then we can make a formula N equals Ki per E. So it's equal to 45 times 1.6 times 10 to the power of minus 19. If we calculate it, 45 divided by 1.6 is 28.125. 10 to the power of minus 19 is increased, so 10 to the power of minus 19.

or we make it into a 2.8125 x 10^20 electron so there are 2.8125 x 10^20 electrons that flow through the light bulb for 1 minute Let's go to the next question. The wire A has a resistance of 40 ohms. If the wire B has a resistance of 3 x 4A

while the diameter is 2 times the wire A and the length is half times the wire A, then determine the width of the wire B. Okay, here is the type of the width. What is the type of the width? The type of the width is Rho, the diameter is D, and the length is L. So which formula do we use? We use the formula R = Rho L/A. Do you still remember the formula? We use it here.

So what is this diameter for? This diameter is to calculate the width of the wire. The width of the wire is considered a circle. Because of the circle, the width of the wire, A, is equal to PR squared. Or if we use a diameter, A is equal to 1/4 PD squared. Before we work, let's do the data first. Here we compare the thickness of the wire between wire A and wire B. The first thing we know is the thickness of wire A, 40 ohm. What does that mean? RA.

Ra = 40 Ohm. Then, the wire B has a 3xA wire. What does that mean? Rho = 3 Rho . Okay? The diameter is 2xA. So, the diameter of the wire B is 2xA. dB = 2 dA. Lastly, LB = 1/2 LA.

To work on a comparison method like this, I have explained in other videos, namely by dividing the formula. What is the formula for dividing? So, what is the formula for RA? RA is equal to Rho A La per A A. What is the formula for RB? RB is equal to Rho B Lb per Ab. I divide these two formulas, the one on the left is divided into RA per RB, and the one on the right is also divided, so it's like this.

then the same sign is removed, we put one in the middle, so like this then we enter the data that is known in the question, how much is the RA? 40, yes, here 40 per RB, the RB is searched, yes, you are told to determine the range of the wire B, so here is RB, then here the Rho A, leave it in the form of Rho A, yes, the LA also leave it, well, let's make AA a quarter

pi dA squared, right? a = 1.4 pi d squared, so if a is the width of the radius of the wire, a means it is equal to 1.4 pi dA squared, dA is the diameter of the wire a. The root B, we immediately replace it, replace it with what? The root B is replaced with 3 root a, okay? Then Lb, half La, half La, so here half La per, well, it should be 1.4 pi db squared,

1/4P db^2, we replace the db with 2 dA. So, this will be 1/4P^2. Then we subtract the same, what is the same? Rho A, then LA, then 1/4P. Then we calculate, here it means 40/RB.

What's left on the right? 1/da^2. The bottom one is still 3 and a half here. So write it here. 3 times 1/2 times... The bottom one is 2da^2, so it's 4da^2. So here we calculate 40/Rb equals...

this is per dA^2, right? I'll just go back to the bottom. So here it becomes, multiplied by, 3 becomes the bottom, half becomes 2, 2 multiplied by 4, 8 dA^2. dA^2 is subtracted, then multiplied by the cross, it becomes 8Rb = 120. Rb = 15 Ohm. Okay, we found the answer. The distance from the cord B is 15 Ohm. Let's go to the next question.

For example, about resistors and temperature. A heating element has a resistance of 150 ohm at 200 degrees Celsius. If the resistance coefficient of a wire is 0.005 per degree Celsius, determine the resistance of the heating element at 800 degrees Celsius. So here we have the data first, what is known, what is known is the initial resistance at 200 degrees, we consider it initial.

So that's R0, R0 is 150 ohm. Okay, then here is alpha, the coefficient of the voltage gap is alpha, alpha equals 0.005 per degree Celsius. Then we need the data of delta T, what is the temperature change? From 200 to 800, that means delta T is 800 minus 200 is 600.

Okay, we are looking for the limit of the heat element at 800 degrees Celsius or RT. What is the formula for RT? RT is equal to R0 in the bracket 1 + αΔT. Just enter 150 times 1 plus 0.005 times 600. So 150 times 1 plus 0.005 times 600, the result is 3.

So 150 times 1 plus 3 is 150 times 4, 600. That's all for this video. Thank you for watching. If you like it, please like and share this video. If you have any suggestions, criticisms, and suggestions, you can write them in the comments section. See you in the next video.