Capacitive Reactance, Impedance, Power Factor, AC Circuits, Physics

[@TheOrganicChemistryTutor]

Full 1 Hour Video AC Circuits: www.patreon.com/MathScienceTutor Direct Link to The Full Video: bit.ly/3vEFYjI Physics PDF Worksheets: www.video-tutor.net/physics-basic-introduction.html Final Exams and Video Playlists: www.video-tutor.net/

[@icevariable9600]

I wish that most of the educational (electrical) tutorials on YT were this concise and informative. So much packed into so few minutes. Good job.

[@MrKartos1]

I have to say, understanding the math first behind electronics is making me grasp it much faster. Starting out I only had seen tutorials just explaining basic values but nothing like that. Your method is great! Thank you!

[@__-pl3jg]

In case anyone finds this helpful for the first formula... 1.) 2•3.14 = 6.28 2.) 6.28 • 60 = 376.99 3.) 376.99 • 0.00005 = 0.01885 4.) 1 ÷ 0.01885 = 53.05 Summary: Don’t forget to add parenthesis in your calculator or you'll get a different result. Formula: 1÷(2•3.14•60•0.00005) = 53.05Ω Can also be written as 1÷(2•π•60•50•10^(-6) = 53.05Ω Pico farad: 10^(-12) Also written as 0.000000000001 micro farad: 10^(-6) Also written as 0.000001 I'm studying cable properties via the SCTE organization and they weren't explaining the math behind the formulas. I decided to explore KZbin for a better explanation. Unfortunately during the very first formula you skipped right over how you got 53.05ohms....WTF? It would be helpful if in all future videos you show what's being typed into your calculator to achieve the end result. Thanks for taking the time to upload all the other information though :)

[@jamiehawman66]

Don't suppose you know the formula for impedance do you ever time I do it I don't get anywhere near what he got so I must be doing it wrong on calculater :/

[@jamiehawman66]

Says √ of 40 squared + 53.05 squared he gets 66.44 but God knows how

[@Terrobul]

Thanks for adding the formula 👍🏼

[@dre6243]

Im so confused on where you got the 0.0005 im about to jump off a bridge i hate math😕 i know this is a old comment but i hope you see this and can explain

[@__-pl3jg]

@@dre6243 - 0.00005 is the decimal version of 50•10^(-6). And the 50•10^(-6) is part of the formula used. I simply changed it to decimal so I could enter it into a calculator easier.

[@Chimigami]

Currently taking AC Circuits & was considering dropping the class until I found this. Thanks for explaining in a way that just plain makes sense.

[@highgatehandyman6479]

excellent breakdown. i watch this before the exams. and it always serves me well thank you

[@malcolmhodgson7540]

Worth its weight in gold! Just made it crystal clear to me! Thanks. Off to watch your other videos on electronics.

[@wingedvictory8694]

Yooo why this man makes everything so easy to understand like what the hell

[@folkloreafrica27]

Thank you so much am actually having exams but know nothing about these,my mind is at peace now😁😁😁😁

[@titamuncher]

Thank you for this, im finishing a test tomorrow on this, and i literally knew nothing about the calculations

[@nandnijamnadas6811]

how did it go

[@titamuncher]

@@nandnijamnadas6811 got 69%for it👌

[@michaelrogers2597]

Great video, Thank you very much for simplifying Capacitors for me.

[@lxmndas]

A very simple step by step explanation for beginners. Very appealing!

[@tanvi9403]

such a great video! i think the Series RC Circuit video should also be in this playlist, i think that's included in Physics 2 as well!

[@Hossain_MD_Arif]

Good explanation bro. Do similar way RCL circuit please

[@jimchen0714]

It's really helpful. Thank you so much!!!

[@christophermontoya5526]

Thanks for your help. Classes are cancelled, so it's not like I can ask my electronics teacher how to solve a problem in which resistance, capacitance, and Hertz is stated, and then asks for calculated impedance.

[@svwtsvfcb]

When we put a resistor in series with a capacitor, we get the impedance of the circuit equal to Z = R - j*Xc, with Xc = 1/jwC. This means that when w increases, Xc decreases, and Z thus increases since we deduct Xc from R by a lower number. If we decrease the frequency, Xc increases, and Z thus decreases. If Z decreases, the current increases, and vice versa. In this video however, you say that I increases when Xc decreases...

[@sircliff323]

made it understandable and without baggage.

[@muhammadusman5773]

I know it might not be possible now but can you do the same if it is complex circuit analysis

[@CarlosSantos-iu2ei]

Thank you very much for this video. it helped me to understand this type of circuits.

[@ergunguler3981]

Video için teşekkür ederim. Xc ile ESR arasında fark nedir?

[@markmarshall4875]

It would be great to show how you work this out on the calculator 🙏

[@Shoova]

I know it's a little late, man, but if you reach out to me I can show you.

[@princekittonga3884]

I have exams in less than a week and I've just been given a CAT with these and the lecturer didn't even bother teaching them

[@orangaharris640]

Bit hard punching these in my calculatir ,my answers are alot diferent

[@Salamanca-joro]

3:03 why did you bot use this formula and just did it right after? Like 12v divided 40 ohms ?

[@trevontiibarber270]

Hella late but what would be the phase angles of this circuit?

[@Shoova]

Use inverse tangent. So it would be inverse tangent multiplied by negative capacitive reactance divided by resistance. It would look a but like: ϕ tan −1 =Xc/R). Or in this case, ϕ = tan-1(-53.05Ω/40Ω). So phase angle is roughly -52.983°. Roughly, if I didn't screw it up.

[@jakedaniel5821]

no matter how I input that equation for Xc in my calculator I would get a really small decimal. can anyone help me with the right way?

[@Johnmatrixx81]

Hello =) Just calculate the numbers below first: 3,14 x 3,14 x 60 x 0,00005 = 0,01885, then take 1/0,01885= 53,05 Not all calculators follow the same rules, like cellphones and pads. Thats why we need to learn how to think, rather then what to think =)

[@chichikimeli]

@@Johnmatrixx81 i got the same values as jake daniel but you deserve a nobel peace price

[@secretsather]

Super helpful. Thank you

[@bigdividends5487]

I dont really have a great grasp on this subject so if I say something that makes no sense don't judge me. So I have an electromagnet, it generates 90.5 Tesla, B=(permeability)(loops per meter)(amps,current). the permeability is 2.3 x10^-2, loops per meter is 787.4, amps is 5. That's just with direct current, but since AC goes back and forth and deals with frequency could we just multiply the amps(5) times the frequency in the AC circuit? For context, the circuit I'm talking about is a very simple circuit nothing crazy just a power supply and electromagnet.

[@perelium-x]

7:07 I got 2.16 Watts using the V^2/R and I*V formula, but using I^2R i got 1.305, why is that?

[@EducationalLearning-i7l]

Because you found the Power across that Specific Resistor using (I^2R) When you multiply the Electro Motive Force of the circuit by the Current of the circuit, aka (I*V), you take all the Resistances of the circuit into the calculation. As he explained at 6:42, the Capacitor gives Energy back to the Circuit, unlike the Resistor which only takes Energy from the circuit. Therefore the Impedence of the Capacitor is not calculated when finding the Power Disappation in the Circuit, which would be the case if you did (I*V), but only the Resistance of the Resistor. Hence why the right forumla to use is (I^2R)

[@perelium-x]

@@EducationalLearning-i7l Thanks but I don't even remember what I asked. 😂It will help someone for sure.

[@EducationalLearning-i7l]

@@perelium-x 👍

[@EducationalLearning-i7l]

@@perelium-x i edited and added a bit more clarification to my explination if you would like to reread it

[@hamzehalzoubi8447]

excuse me can you say why didnt multiplying (I*Z) , and just multiplying (I*R) ??? in 7:18

[@allegrahudson2550]

I actually have been finding P by taking Vr x Irms

[@leviakrman9032]

ابه نيو خوش يشرح والله

[@brandenom4059]

How is capacitor used to block both AC and DC currents???

[@Shoova]

It depends on the frequency.

[@niobeeyt4889]

Thank you.

[@roaebrahim1437]

thank you soooo much

[@edwardquintana8609]

Can someone please help me with this question. Which of the following should be included in a calculation of impeadance for an A.C circuit that contains resistance inductance and capacitance. A) the resistance only B) the value of none of these circuit elements. C)The value of all 3 impeadant circuit elements . D)the value of the inductance and capacitance only.

[@kenmichaels1842]

I've not seen a reply and not sure you got your answer elsewhere since it's been a year since you've posted but...Since inductance is expected to be the larger number per the equation for impedance, most electrical engineers are only going to calculate with known resistance and inductance. Capacitance is often so small of a value that's it's nearly insignificant. I'm sure there are some engineering brainiacs that will dispute it but, as a power engineer who has worked with electrical utilities....resistance and inductance is the overriding concern.

[@geedestin3389]

It’s C

[@edwardquintana8609]

@@geedestin3389 thanks gee you are correct it is confirmed

[@edwardquintana8609]

@@kenmichaels1842 thanks Ken yea the answer is actually c

[@geedestin3389]

Edward Quintana anytime

[@successthruknowledge]

I think that the video was okay in terms of showing how the calculations work and how each result helps to solve the next problem; however he did a poor job explaining exactly what is capacitive reactance itself. The way that I understand the term it means to oppose changes in voltage (or current?) in the circuit

[@beginsaingenieriaelectrica9701]

hy! every one could you said how you represent the 10 -6

[@yok-m2w]

0.000001 (called micro)

[@sltiktok6952]

Vrms=irms×z ?

[@coolkid9770]

is Vsource = Vpeak ? if yes then why is it Vsource = Irms * Impedance why not Vrms = Irms * Impedance can someone explain this to me? thanks in advance

[@khaileng3020]

V source =VAC =Vrms Vpeak =Vsine wave peak =Vm Vsource=Vrms =VAC=Irms × impedance Vpeak=Vm=Ipeak × impedance

[@farmlandtiger1377]

low pass filter and high pass filter circuit drawings are wrong. Anyone noticed?

[@yok-m2w]

Left one is low pass filter (allows higher frequencies), Right one is high pass filter (allows lower frequencies), It seems there is no mistake in his explanations. You may want to listen again more carefully.

[@LiyuansLife]

as a 11 year old, doing algebra 2, i can't understand a single thing 😭

[@JohnVKaravitis]

I thought that half the energy was lost by charging up a capacitor.

[@hirushapatabendige]

sound low

[@DriniDuma-lk8dr]

lum nona qe te ka

[@mikaa4279]

This is not helpful if you are trying to teach it's confusing