are you winning son?

What’s your accomplishment(long word ah)

You're very welcome!

195 right

Thank you!

hahaha

exept for those who run on blue smoke

That's magic smoke. FWIW, the LED was still glowing while the resistor was burning. DAMN! And I was willing to take side bets on which component would go open-circuit first. NO FUN! But, since LED's have an internal resistance, they are self-limiting, therefore negating the need for a series resistor, unless the voltage supply is substantially higher than the nominal forward voltage of the LED. But I digress. I vote for the resistor. If the resistor wins the vote, you have to ship me the LED. ;`>

That's not true. When a component gets too hot, it starts to burn, thus producing smoke.

@@31redorange08 you racist

Voltage is literally the hardest thing in EE to understand. By the time I graduated university I'd say only half the people in my class truly understood it. It takes time but once you work with voltage enough everything starts to click.

​@@Afrotechmods Thanks for replying. I had trouble understanding 1V 1.8A is 1.8J. Easy maths, but hard for me to reckon with more than 1J per second. Great videos

David Leef - That describes my motivation pretty well! I would only add that it’s REALLY fun when we understand ANY of it and use that understanding to make fun gizmos.

This is pretty simple for me

I feel your pain. This stuff can be tricky to understand. There are at least five different videos on UKposts explaining watts. Different teachers often explain things in different ways. Watch all five of of them and very likely one of them will make sense to you. A watt is defined as the amount of energy (disapated as heat) of one amp of current flowing through one ohm of resistance in one second.

@@rmorgan453 I feel like the easiest explination would be to think about it in terms of joules used per second. You start with the basic Coulomb, that would be the charge of 6.24 * 10 to the 18 electrons. Then you go to current, that is measured in Amps. 1 Amp would be equal to 1 Coulomb (the charge of 6.24 * 10 to the 18 electorns) passing through a point (that you chose to measure) in one second. Next you have to think about Voltage in terms of Jules per Coulomb. Every Coulomb has the rated voltage in joules. Combining them all you get to power that is simply joules/second. The amount of energy something uses every single second is what power is, and that is called watts for sake of simplicity.

Agree! It’s like Greek.

And yeah, Thomas Edison lost the bet that DC>AC. Shame on him

but doesn't high voltage also cause high current. i thought current was reactor to how much volts you had so even if you dont need the high you'd still get it right? im learning so please forgive my ignorance.

The last time I dealt with transmission lines was underneath a Ford van. I had to throw my coveralls away, because I couldn't get rid of the red stains.

@NoVaKane Yes and no. Whenever we work across a fixed load, then current is directly proportional to voltage. Kirchoff's law. But in practice, higher voltage results in lower current during POWER TRANSMISSION. You see, the higher voltage is stepped down before and during DISTRIBUTION. So, if the end user (subscriber) has a 240 volt service, rated at 100 amps, that would translate to 1 amp at 24 KV, or 100 milliamps 240 KV. Now, if we were to apply 240 KV to the same load at the subscriber, everything plugged in would try to draw 1,000 times the current. All the magic smoke would escape. But since we are dealing with VA (Volt-Amperes), instead of current, the two factors cancel out, and we deliver the same number of VA over the entire system. Since transmission lines run at 100 times or more volts than the distribution grid, and the power load is the same, the current in these HV lines is less, since the voltage hasn't been stepped down yet. I hope this makes sense.

@@gyrgrls Forgive me if I sound like an absolute idiot but I thought that the voltage is what pushes the charge and current is the flow of charge so if the voltage is high, surely the current would be high too?

are you alive

This might end up being the last video I make unless I get corporate sponsorship or donations start pouring in.

I had a Patreon. >99.9% of my subscribers weren't interested, plus I even had some fake pledges with fake credit card numbers.

the problem is you dont make enough videos to atract people. If you ever thought of making money making ideos, you should give Dave's (from EEVBlog) talk a watch. He talked about how he does it and one of the things he said was exactly that. Make videos, make them often, keep the channel alive so that people wont think you are dead and so it will keep the channel in their minds. It's just a matter of not making a video for a couple week and most people will have already forgoten about you and only when you come back will they recall "oh, the guy that made videos!" My recomendation is at least making a video once every two weeks. It doesnt have to be big, but once you start growing and getting money, money will give you the motivation to make more and not get frustrated in the process. As a comparison, dave makes 3 to 4 videos a week and has done so for the last 5 or so years. You have like 75% his subs, and little less than the view rates he has. What you are lacking is a corporative organization to take advantage of it like dave has.

Sad, but understandable.

+Afrotechmods Whatt? I thougt you have over 400$ on patreon?

No, that formula applies to any current form, but when you have a current that changes throughout the time you need advance math to solve it. And we often use different units because of convenience and habit, when you measure how much energy you used in your house in a month for example you mesure it in kWh instead of Joule, it is easier that way. Thats why

You're wrong. You can't measure power in AC as in DC. If you just simply do P=VxA will only show you the Apparent Power, not the real power. In order to explain why transformers are rated in KVA, you need to know what POWER means. There are 3 types of power: 1. Real Power (W), Reactive power (VAr) AND Apparent Power (KVA). In order to measure the power in AC, you need to know this: u = U (sin omega t + fi 0) i - I(sim omega t + fi 0) Note that "u"/"i" and "U"/"I" (in capital) are not the same. U/I - The voltage/intensity displayed on the multimeter (Peak voltage/intensity) u - the real voltage. If you get U x A = The peak power (apparent power, also rated in KVA) - Now you know why the transformer is rated in KVA. That's the peak power they can support. If you put a load bigger than the maximum KVA, you will most likely burn it. If you get u x a = The real power (KW or W). You need to know the real power on generators so you know what you expect the load will do. If the load is bigger than the real power, it will burn the generator. Basically generators are not rated in KVA because you don't need to know the peak power. The peak power is there for only 50Hz/60Hz (50/60 times/s) but the load you put on the generator it is a constant load. "fi 0" is the resistive inductance or the capacitive inductance. This is a better mugshot (pun intended) about this explanation: www.sgs-engineering.com/wp/wp-content/uploads/2015/09/beerkva.jpg

Forgot to say: Any variations between u x A or U x a you will get the Reactive Power

Also, sorry for any mistakes if there any but English is not my main language and is not that easy to translate physics terms from Romanian to English and also, I only slept 2 hours in the past 2 days....

Riddim Dubstep your the man. That was explained very well. Helped me out a lot. Thanks

You are correct of course. RIP UKposts's annotations feature :*(

Tom Tommy transformers are just about the most efficient possible way to convert electricity. That being said, do you mean voltage droop on the transformer, where unloaded, it's x volts, but with a load, it .5x volts?

+Ethan M thx for helping first! Then, no I didn't mean the voltage drop. I meant the efficiency. As I know η=(Vo*Io)/(Vin*Iin). I have a LED strip which is powered by a cheap 12V 10A transformer, so I did a few of measurement and I got theses numbers, for the input I=28.8m.A and V=220v so P=6.3W. And for the LED strip which is the output I got 244m.A and of course V=12v so P=2.9W................... so......... η= 2.9W/6.3W = 44% !!! Is it fail? Did I forget considering anything? THX again!

The Traveler 10% okay, but in my case it's 56% !!!

Power. And yes low voltage high current can do it if the resistance is low enough.

@@Afrotechmods thank you, sir. I'm a big fan, though a medical student 😅

_This is something i've been confused about for some time: Is the amount of heat a lightbulb or any cable with resistance gives off just measured with VxI?_ Yes, resistive loads convert almost 100% of electric power to heat. _The point is, if it doesn't matter if the current or the voltage is high as long as both multiplied yield the same watts, then why is that?_ Power lines have a certain resistance and according to Ohm's law there will be a voltage drop if current flows through a resistor. So the lower the current, the lower the voltage drop. V=RxI Let's say the power line has a resistance of 10 Ohms and you want to deliver 1MW. At 10kV you'd have to pass 100A trough the power line, so the voltage drop across the conductor would be V=10Ohms x 100A=1kV and therefor the power loss would be P=1kV x 100A=100kW. If you use a higher voltage for the same task, lets say 500kV, you'd only have to pass 2A trough the power line. So the voltage drop would be V=10Ohms x 2A=20V. That gives you a power loss of P=20V x 2A=40W. For spot welders made from MOTs the opposite applies. Normally these transformers have a primary winding with let's say 1000 turns which sees mains voltage and a secondary coil with roughly 10 times the number of turns so they step up the voltage by a factor of 10. If you want high current at low voltage, you replace the secondary winding with a low number of turns of very thick and conductive wire, lets say 10 turns. That would step down the voltage by a factor of 100 and on 220V AC you'd get 2.2V on the secondary. Now the primary winding is typically rated for something like 800W, so you draw 3.6A from the socket while drawing roughly 360A from your secondary winding because P(in)=P(out)-P(heat). I hope this cleared things up a bit for you. If you still have questions, feel free to ask.

swiss Oh now I understand. Thank you! That was very clear. I completely missed that the heat given off by the resistor is given by the voltage drop times the amps, and that the voltage drop increases with the current passing through it. I think you cleared all my doubts about it! :D Just out of curiosity, are you an electrical engineer, or studying something related to electronics?

Ciroluiro You're welcome. I'm not an electrical engineer by any means, just kind of a jack of all trades who doesn't mind sharing his knowledge. I've also been called a smart-ass many times and I hold that title with pride.