Thank you for helping me complete my electronics in electric vehicles

Well explained. Thank you😊

thank you so much. This helps me understand a lot more

Thank you . Well explained

Thank you I finally understand this, your a life saver

@KcoolScience

4 жыл бұрын

Glad i could help!

thank you show much, your video made it clear for me to do the problems that i have been doing for weeks recommend this video for everyone. thanks

@KcoolScience

5 жыл бұрын

I'm glad it was helpful!

Thanks, great explanation

@KcoolScience

5 күн бұрын

Glad it was helpful!

got mid term test 2morrow i been lackin big time . this video made me understand this alot more than in class . thanks a lot

@KcoolScience

2 ай бұрын

Yay! Glad I could help!

Really clean expression. Thanks to you :)

@KcoolScience

7 ай бұрын

Awesome- glad it was helpful!

Thanks for the help

Thank you so much for saving my college fees not to go free😊

@KcoolScience

3 ай бұрын

Yay glad I could help!

Very helpful thanks 🙏

@KcoolScience

28 күн бұрын

Glad i could help!

Not sure how you got 25.7 ohms at 10:26... my math provided an Rt of 26.31. Regardless- thank you for your help.

@MrRyanlevi

2 жыл бұрын

1 divided by .038 equals 26.31 but 1 divided by .0388 is 25.7

@clutch-400

4 ай бұрын

I'm with you on that. I was wondering the same thing.

@robbiebettilyon5304

Ай бұрын

@@clutch-400 so... is the answer 26.315? I'm a little upset that answer is 25.7 ! I think that the answer should be 26.315 !

So amazing

@KcoolScience

2 ай бұрын

Thank you!

Thank you I need this for my union classes. Do you have anymore on parallel and series? I couldn’t find anymore? Thank you

@KcoolScience

3 ай бұрын

Here's two other's I have that touch on it- more lecture style, but the conceptual one will help out a lot with remembering how the laws work- kzread.info/dash/bejne/oXxt2ZmjiteXYsY.htmlsi=7i8huLTXmtqGwh1m kzread.info/dash/bejne/pqScx8ucl869ZKg.htmlsi=vsmsMaerYWa0q615 Glad I could help! I'm going to try and make some more just straight practice videos soon!

Hi. your last problem has a bit of an issue. starting at 16:15 you say that "more resistance, less current" and that the 12R will get half as much current as the 6R because it's double the size. cool so far. Your very next calculation then incorrectly shows the opposite happening. Then when you go to solve that parallel resistor section in a different way, you end up saying that the 6R gets one third of an Amp and the 12R gets two thirds, which is also incorrect. It should read that the 6R gets 1/3A and the 12R gets 1/6A. the formula is I1 = (Rt/R1)*It and Rt for that section is 4ohms. so its ((4/6) *.5) = 1/3A and ((4/12)*.5)= 1/6A.

@illi1154

4 ай бұрын

Yup! Great explanation

Thank-you for helping me,,any video for calculating equivalent resistance from the original circuit

@KcoolScience

2 ай бұрын

I'm not sure what you mean, but you can think of the total resistance as the equivalent resistance of all the resistors in the circuit!

How did you come up w .333 and .17 @time 17:43 ? Thx

@KcoolScience

3 ай бұрын

Because the 12-ohm resistor is 2x the resistance of the 6-ohm resistance, it will get half as much current. Since the total current is 0.5A, if we divide it by 3 we can give the 6-ohm half the resistance (0.17 amps) and the 12-ohm twice that (0.33 amps). Basically, using the proportionality to get there. A better way would have been solving for the equivalent resistance (1/R= 1/12 + 1/6; R=4), then multiplying by the current (0.5 A* 4 ohms= 2 V) to get the voltage drop, then using V/R=I to get the individual currents. (2V/6= 0.33A and 2V/12= 0.17 A)

Do you have a video for combination circuits? if not can you do one?

@KcoolScience

5 жыл бұрын

There is a combination circuit at the end, but I can make one with more complex circuits!

What about the the prospective fault current and prospective short current whats the difference between them?

@KcoolScience

2 ай бұрын

I will be honest, I have never heard either of those terms before(i teach at the high school level, degree is in chem not physics), so I don't know!

How u calculate the voltage drop across each rezistor

@KcoolScience

7 ай бұрын

Depends on which type of circuit- in a parallel, typically you know its the same as the voltage of the battery (in a simple parallel), where in a series you typically solve for current first using the total resistance, then use V=IR, with the individual resistances to calculate the voltage drops.

@georgegeorgel7254

6 ай бұрын

@@KcoolScience u can also calculate in paralel per rezistor

@kelvinguevara7540

2 ай бұрын

​@georgegeorgel7254 in a parallel circuit its gonna be the same across each resistors

If the Voltage isn't present how do you find it?

@KcoolScience

5 күн бұрын

Depends, what other information are you given? If you have current and resistance, use V=IR, if it’s a simple parallel, equal to battery voltage, anything more complex would need more steps

how did you get 25.7????

@KcoolScience

6 ай бұрын

As explained in a previous comment, I made a rounding error and used "0.038" instead of "0.038888," 26.31 Ohms would be more precise.

HOW DID YOU GET 25.7? I THINK YOU COMPUTE IT WRONG?

@KcoolScience

6 ай бұрын

As explained in a previous comment, I made a rounding error and used "0.038" instead of "0.038888," 26.31 Ohms would be more precise.

Amoung?

Lol. I can't spell half the time myself...And I used to win spelling bees along time ago!

@KcoolScience

2 ай бұрын

And this is why I don't teach English!

The last step was reversed. It should be 2 volts/6 ohms(.333) plus 2 volts/12 ohms (.166) = 0.5A

@KcoolScience

3 ай бұрын

Oops!

Thanks for the teaching me

@KcoolScience

Ай бұрын

My pleasure!

You are so sweet ❤️❤️❤️

@KcoolScience

5 күн бұрын

You are so kind! 🥰

Iam lost

@KcoolScience

3 ай бұрын

Maybe this video will be a good place to start? It's my attempt to explain circuit conceptually, so the math makes more sense. kzread.info/dash/bejne/oXxt2ZmjiteXYsY.htmlsi=negipWxhGFTP_Ng7

I’m gonna get kicked outa the school

Thank you!!

@KcoolScience

Жыл бұрын

You're welcome!