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.
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- czcams.com/video/oH9w7jZt12c/video.htmlsi=7i8huLTXmtqGwh1m czcams.com/video/tpheicgkW4E/video.htmlsi=vsmsMaerYWa0q615 Glad I could help! I'm going to try and make some more just straight practice videos soon!
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)
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.
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
Maybe this video will be a good place to start? It's my attempt to explain circuit conceptually, so the math makes more sense. czcams.com/video/oH9w7jZt12c/video.htmlsi=negipWxhGFTP_Ng7
Thank you for helping me complete my electronics in electric vehicles
Thank you I finally understand this, your a life saver
Glad i could help!
Well explained. Thank you😊
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
I'm glad it was helpful!
thank you so much. This helps me understand a lot more
Thank you so much for saving my college fees not to go free😊
Yay glad I could help!
got mid term test 2morrow i been lackin big time . this video made me understand this alot more than in class . thanks a lot
Yay! Glad I could help!
Thank you . Well explained
Thanks, great explanation
Glad it was helpful!
Really clean expression. Thanks to you :)
Awesome- glad it was helpful!
Very helpful thanks 🙏
Glad i could help!
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.
Yup! Great explanation
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.
1 divided by .038 equals 26.31 but 1 divided by .0388 is 25.7
I'm with you on that. I was wondering the same thing.
@@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 !
Thanks for the help
Thank you I need this for my union classes. Do you have anymore on parallel and series? I couldn’t find anymore? Thank you
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- czcams.com/video/oH9w7jZt12c/video.htmlsi=7i8huLTXmtqGwh1m czcams.com/video/tpheicgkW4E/video.htmlsi=vsmsMaerYWa0q615 Glad I could help! I'm going to try and make some more just straight practice videos soon!
Thank-you for helping me,,any video for calculating equivalent resistance from the original circuit
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!
What about the the prospective fault current and prospective short current whats the difference between them?
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!
Do you have a video for combination circuits? if not can you do one?
There is a combination circuit at the end, but I can make one with more complex circuits!
So amazing
Thank you!
How did you come up w .333 and .17 @time 17:43 ? Thx
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)
How u calculate the voltage drop across each rezistor
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.
@@KcoolScience u can also calculate in paralel per rezistor
@georgegeorgel7254 in a parallel circuit its gonna be the same across each resistors
If the Voltage isn't present how do you find it?
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????
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?
HOW DID YOU GET 25.7? I THINK YOU COMPUTE IT WRONG?
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.
Lol. I can't spell half the time myself...And I used to win spelling bees along time ago!
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
Oops!
You are so sweet ❤️❤️❤️
You are so kind! 🥰
Thanks for the teaching me
My pleasure!
Iam lost
Maybe this video will be a good place to start? It's my attempt to explain circuit conceptually, so the math makes more sense. czcams.com/video/oH9w7jZt12c/video.htmlsi=negipWxhGFTP_Ng7
I’m gonna get kicked outa the school
Thank you!!
You're welcome!