Just finished the playlist on electronics for beginners. Been learning electronics for a couple of years and still learned a lot. I really worry that sometimes I don't have a good basic grasp of the fundamentals and this playlist helped me a lot.
another cool application for the voltage/current trade-off is high voltage power lines. the higher the voltage, the less current you need to transmit the same power. less current means less voltage drop across the power lines, and lower voltage drop means less power being consumed by the lines themselves. the benefit of this is twofold, it's more efficient, and you don't need as much material to make the cables.
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?
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.
I feel your pain. This stuff can be tricky to understand. There are at least five different videos on UA-cam 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.
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. ;`>
Appreciate the videos. Still find it difficult to make sense of voltage, and keep up with concepts like Power, but will go through the playlist for a few times to keep things in mind.
pls don't tell me i'm the only one who prefers electron flow over conventional current flow. i really want to understand electricity and all those components but because almost all the explanations are in conventional current flow i get confused. i'm a more visualizing person, i want to understand what the electrons actually do.
dude these videos are burning up the blank pages in my notebook! ...I kinda love it when I find a UA-cam channel that makes me fetch a pad and write stuff down...
Fantastic video!! and great series on electricity basics. Easy to understand. Thank you. One small typo in video at 1:06 - "... no engineer is going to covert volts into joules/coulomb and amps into joules/second.. " . You meant amps into coulombs/second?
Leonardo Rodrigues Leonardo Rodrigues 5 years ago (edited) Thanks for the high quality videos. Instructive and amusing. I have been working on a project in which I'll have to use a current limiting circuit to protect my DC power supply. Can you make a video about the different choices for this case? I don't know if it would be popular enough, but is just a suggestion. Cheers
Does V x A = W only apply to DC and linear AC loads? When does power factor come into play? Also, why are transformers rated in kVA and generators rated in W?
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
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....
Power is the rate at which energy gets supplied, or used up and instead of saying joules per second, we have a unit we use for power, called the watt. Also 1 watt is equal to 1 joule being transferred. Voltage and current that leads to power. Voltage and current can be combined in different ways to get the job done.
Thanks for the high quality videos. Instructive and amusing. I have been working on a project in which I'll have to use a current limiting circuit to protect my DC power supply. Can you make a video about the different choices for this case? I don't know if it would be popular enough, but is just a suggestion. Cheers
I'm really grateful. Quick question though; what causes the resistor to go up in smoke?..voltage, current or power? If power, that should mean a low voltage, and a high current could generate such heat
1:11, amps into coulombs per second Real-world example, if you use air conditioning with declared power of 3500 watts, and your city grid is 220 V, that means that your air-con device will pull 15,9 amps through the electric wire to put it to work. (3500/220 =15.9), simply the declared power of an electric device tells you how many amps it is going to need. A 500 watt PC will suck 2.27 amps through the wire. All power consumption in ampers sum up, and if it exceeds the Fuse limit, it will break the circuit for your own protection. You can make a mistake by putting the higher value fuse than it is the capacity of the specific wire. In that case, it will let the higher ampers pass...and the wire will melt, burn and cause a fire.
I'm curious to truly know if an electrical component going up in smoke is tied more to current or power as from what I've learned having too much current while keeping a low voltage but the same wattage is a good way to lose power while frying whatever the power is going through simultaneously
+Afrotechmofs fantastic man! I wanna mention something fail which most of the people don't know about. For example that LED lamp which u showed rated as 40 W, but the issue is, when we gunna power it with some power supply like a transformer (not a battery), we have to calculate how much the transformer will drag. Depending on the bad efficiency of the most of transformers we'll find that the power which is needed to power the LED is about the double, about 80 W and that's horrible! Am I right? Thx
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!
I am very new at all this, so I need to ask: When the resistor burned, how come the LED kept giving light? I would expect that, when the resistor burned it stopped working, which means it stopped "choking the amps", and so too many amps would flow into the LED and it should have burned also. And, yet, the LED kept giving light. Am I getting something wrong?
does low voltage like change the current to be high or is it when the voltage is lower then current, essentially the electrons just siting there with no where to go almost and just start to heat up like pressure building up on a hose?
So let's let's say that a microwave draws 1000w of power per second. What confuses me is that what part of the power is spent. So only some of the energy is turned into microwaves. And we pay bills in watt hours, so it is obviously not drawing 3.6million watts per hour. So my question - is the power drawn different from the power that is dissapated?
I think you have your units mixed up a bit. It doesn't draw 1000w per second, because watts are already in "per seconds". It just draws 1000w, which means 1000 J per second. So leaving a 1000w microwave running for one hour means that it uses 1000 watt-hours of energy (since it's 1000w of power running for 1 hour).
you said that Ohm's law is a linear relationship between current and voltage provided the resistance is constant. However, in the experiment you are referring to at 0:15, when you set up the voltage to 1, a current of 1.8 Amperes was displayed. But when the voltage was increased to 2, the current was 2 Amperes. This is not a linear increase. Why did that happen?
@@Afrotechmods 4 years after and I still got a reply!!! WOW! Your content is AWESOME!!! Your 3 videos on Current, Voltage and Power are the best I have found on the Internet and better than any textbook. I might be pushing my luck but there are are few aspects of basic electricity that I have not been able to grasp even though I have scoured the Internet for explanations. I would be most grateful if you could shed some light on these. Even electrical engineers seem to contradict themselves on some of them. 1. What is the difference between charge and electrons? 2. Is electricity a flow of charge or of electrons? If it is a flow of charge, and charge is a property of the electron (and not a physical element, as some people seem to put it), how does charge flow? And is the definition that "electricity is a flow of electrons" wrong? 3. When voltage is increased, are more electrons (or charge) being sent from the battery, through the wire and to the load, or is the same number being sent but at a faster rate? One site even claims that more energy is getting transferred to the electrons while their number and speed remain the same. 4. If at a vertical cross-section of the wire, there is x number of electrons/charge, and the battery is sending, say, x+100 of electrons/charge, will the wire not be able to handle this and will consequently burn, just like a resistor or fuse? And if the battery sends x-200 electrons/charge, does this mean that, if the voltage is increased, there is room for about 200 more electrons/charge to be sent? Or is this an erroneous over-simplification? 4. If the battery is providing more energy to the electron/charge, how is the electron/charge storing this energy? Is the electron vibrating more, for example? 5. What causes the pressure (potential difference/voltage) in the battery? Is it the difference between the NUMBER of electrons at the negative terminal and that at the positive terminal, or is it the AMOUNTof energy that the electrons/charge hold at the negative terminal with respect to the positive terminal? Please do continue to make videos. Praying that you are in a generous mood to educate this ignoramus :-) Thanks beforehand for your replies.
@@salimr47181) Charge is a fundamental property that some types of matter have. And electrons are a type of matter. i.pinimg.com/originals/9f/57/7b/9f577b9a840c4c8bd1870523bb44de3a.gif 2) It depends on how you define "electricity". It is technically more accurate to say it is the flow of charge. "Electricity is electrons flowing" is just a simplification to try get people's heads around basic concepts before delving into quantum physics. If you really want to get into more detail, electrical current isn't actually even the flow of charge, it is the propagation of an electric field which results in the movement of charged particles. Sparkfun has a more physics-ey explanation that you might find helpful. ua-cam.com/video/z8qfhFXjsrw/v-deo.html The rest of your questions suggest to me that you are looking for answers in the wrong place - most electrical engineers usually don't look at things at that low a level, you want to be learning from physics and chemistry teachers. That being said, they are excellent questions so definitely keep searching for the answers!
Thank you for your answers and suggestions. Your explanation about charge and electric field helped me a lot. I now understand what you mean, although the aspect of what exactly increases and changes when voltage goes up is still unclear. I will keep on searching for answers and, if I find them (and remember), I will surely post them here. On a side note, this quest for answers has also helped me understand why I (and other people like me) find it hard to understand physics. It comes down to how we approach the subject matter. We need too many answers right at the start and we are serial learners; we find it difficult to move on to the next topic unless we have fully understood the current one. This assumes that there is an answer to every question we raise, which is often not the case in science (and I am discovering that there might not be a single human being on this planet who fully understands all aspects of electricity). This also requires a teacher with in-depth knowledge, excellent teaching skills and tons of patience. This is exceedingly rare.Those who do well are those who can treat what they do not understand as black-boxes, or are satisfied with analogies, and are able to connect the black-boxes together to get an overall picture of the subject. And they move on with their lives! I believe that this is one of the trait of a good engineer. For people like me, that one thing that we do not understand might consume our days and nights (as it is happening with me currently). I wonder if I can train myself to learn otherwise. Thanks again.
For people that don't get it: Think of electricity as water flowing through a pipe. Volts is water pressure. Amps is pipe diameter. If you increase one or the other, you get more "power".
Hi. On my wall there are sockets hat are able to output 1500W. If I plug a small lamp to the wall it get 10w from the 1500w available. If I plug a bigger lamp it get 100w from the 1500w available. Who is to decide that a lamp can get 10 or 100w? Is there a resistance on every lamp that permit to a small lamp to not be burned by the power of the wall socket?
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? Because I saw a video some time ago about why we use ac current instead of dc current, and the reason was that it is very easy to transform low voltage with high current to high voltage and low current (and viceversa) with ac, and that it was more efficient to deliver electricity with a high voltage and low current. Also some videos here on youtube about contact welders show that they rip open a microwave transformer and mod it to get a low voltage with a massive current from the wall. 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? (the high voltage on power lines and those contact welders that people made that need a lot of current)
_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.
in experiment at 3:26 (140V battery, 3V LED, 6800 ohm resistor), what if the circuit only had resistor (without LED), with 1/4W resistor, the current must be lower than 20mA and the resistor didn't smoke ?
I am not an EE but from what I have read, in AC circuits the voltage and the current are slightly out of phase. So how does one calculate the power factor?
mediaguardian well google,english is not my primary language,but i study EE,so there are 3 equations: 1. P=U*I*cos(phase in °) in Watts 2. Q=U*I*sin(phase in °) in VAR 3. S=U*I in VA
can I fix multiple LEDs (assuming each needs 3V) together WITHOUT a resistor, as long as [amount of LEDs] x 3V = [the V in the battery I choose].......?
I work in I.T but tinker with this stuff. I hate how they size UPS in VA yet almost every PC, server, switch, etc they say it in watts. Like say a 24 port POE switch has a PSU that takes 700 watts. Then you have to find the right UPS for that and those are usually in VA which makes things kind of confusing, frustrating and a pain. I wish the UPS s would just say recommended max load 100 watts. then you know that it can run a 700 watt switch fine, or maybe 2 in some situations.
I always heard NEVER plug anything with a motor or laser printer into a UPS. I heard a story of some tech needed an outlet plugged in a mini vacuum to the UPS and blew the UPS up. Laser printer I think is more an issue of wattage that the fuser draws.
jtech0 The reason they rate these in VA is because of the maximum CURRENT RATING of these devices. Say, you're pulling 1000 VA and using 746 watts. All is well, if your power supply is rated at 1000 VA. You are drawing one horsepower. But your POWER FACTOR is only 0.746. And, in real life, it gets worse. You are lucky to have a 0.65 power factor... not in the UPS device itself, but in the LOAD. Then, again, it all depends on what devices comprise your load... IHTH
Are you kidding? These fuses and breakers are rated in AMPERES. PERIOD. Nothing to do with wattage. Wattage refers to the actual work consumed by the load, and in certain applications, is a fraction of the VA product. You have a lot to learn. Actually, I lied to you. Breakers and fuses DO have a MAXIMUM voltage rating, but it has nothing to do with initial blowing or tripping. Instead, it is a safety specification, in order to ensure RELIABLE circuit interruption on overload. But still: a 20 amp fuse is designed to open at 20 amps, no matter what the OCV. Put a million volts across a blown fuse, however, and it's liable to arc over.
Very very basic question When you say the bulb consume 40W every second, then the energy meter in the house, how will it calculate the consumption? Since it is Wh, so is the calculation 40W x 3600sec? Wouldn't that be too much or incorrect? Because I thought, whatever the power rating of the electrical equipment, that is the consumption per hour, ie, 40Wh. Equation looks similar but the final calculation the meter does would be different for billing.
instead of putting in a stronger resistor, can I simply put multiple resistors in parallel to divide the power onto more resistors? or doesn't power work like that?^^
Blasulz1234 2 equal resistors of, let's say, 5 Ohms in parallel will have an total resistance of 2,5ohms, so you would have to buy stroger resisitor, but if you're going to buy more resistor why don't buy one of 5ohm that can handle the power? But if you already have 2 10ohm resistors, yes, the power will split, the circuit will work and you won't get smoke
Why, when I feed my electrical designs from the car (12-14v, tons of Amp), it produces so much heat in the regulator lm7805, do I need a resistance before the circuit?
Without knowing more about your circuit I can't give a full answer but if you are using it to power a USB outlet for instance (I'm guessing because of the 5v) then I would suggest a buck regulator design rather than a linear regulator, they tend to be more efficient at dropping voltages.
Hello, thanks for the suggestion, but it's not for usb. The small part of the regulator circuit of 5v is to feed a PIC , The regulator circuit is basic, with a capacitor of 470uf or more at the input of the regulator, and at the output one capacitor of 10uf and another of 100pf. Basically only food to the pic, and some transistor if the case with the 5V, if I have other peripherals like leds or sensors, the food direct from the 12-14v of the car with their respective resistances, the LM35 It is perfectly capable of feeding my circuit with its 500mA and these work well when I install it from the source that I have for tests, but when I install it in the car everything starts to go wrong, It gets very hot sometimes it burns. I thought of putting a resistance of about 100 ohm at the beginning of the whole circuit
Midian Alberto Khaiyow Ok, i'm too tired at this time to read that properly so I'll skip to trouble shooting. You say it works on your supply but gets hot in the car. I assume you are testing at 12v? The car when running will be around 14.5v(ish) so try testing at that voltage and see if that replicated the problem?
Hey man, so I got a suggestion to translate this video to Swedish (Since Im Swedish I guess) and I was a bit curious to do that. But I wonder, do you get control over that translation or is it just Google that will use it for their future AI? ;)
Why does it not use power from what it gets, based on voltage? This phenomena has baffled me - the load dictates or the supply dictates the power used? E.g. putting a "large" fan on a "small" circuit.
Great video! Hey, also by any chance, could you give me advice for my channel? I also do Electric things, and any criticism would be very much appreciated :D
Keystone Science It's already good (better than mine, nyahaha (painful laugh)), just keep going and improve your vid quality. I too, just started making videos a few months ago, I hope it goes well..
UA-cam views are proportional to the amount of cats in a video. Something about your audio doesn't quite sound right, I can hear clipping/distortion sometimes. It sounds like you are having to talk really loudly into your mic just to get things at a normal audio level. Might be worth investing in a new external mic to plug into your camera. Otherwise I think your content is pretty good.
+Keystone Science u have a great channel, subscribed ;) Actually, try to work more with audio and the cam plz, and plzzzz try to talk a little slowly, English isn't my native language lol
Please brief us regarding the 3 formulae: P=VI P=(I^2)R P=(V^2)/R Is there any difference between their usage? How do we know which one is to be used in a given problem?
I explained this to someone below but here we go: 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.
Just finished the playlist on electronics for beginners. Been learning electronics for a couple of years and still learned a lot. I really worry that sometimes I don't have a good basic grasp of the fundamentals and this playlist helped me a lot.
These videos have simplified a concept that can and have been overcomplicated. Thank you for your amazing explanations!
another cool application for the voltage/current trade-off is high voltage power lines. the higher the voltage, the less current you need to transmit the same power. less current means less voltage drop across the power lines, and lower voltage drop means less power being consumed by the lines themselves. the benefit of this is twofold, it's more efficient, and you don't need as much material to make the cables.
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?
I learned more from these videos than my formal education. Thank you! :)
Excellent series of lectures for beginners in electronics. Highly recommended. I hope there will be more to come.
I feel like i'm never gonna understand this stuff, and I so want to!
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 UA-cam 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.
Man these electricity basics videos are soo useful to me. Thank you
You've forgot to mention a crucial thing, which is that everything runs on white smoke, once the white smoke gets out it doesn't want to work anymore.
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
its hard to overstate how helpful these videos are. you are a hero
Thank you!
You are the best teacher one needs to have,for explaining these stuffs. Thankyou
Appreciate the videos. Still find it difficult to make sense of voltage, and keep up with concepts like Power, but will go through the playlist for a few times to keep things in mind.
pls don't tell me i'm the only one who prefers electron flow over conventional current flow. i really want to understand electricity and all those components but because almost all the explanations are in conventional current flow i get confused. i'm a more visualizing person, i want to understand what the electrons actually do.
atif i too, we can rock. 😀
If you are getting confused then you should probably learn the conventional way and go from there.
Same here
That's like being confused cuz you don't know how many bricks are in a kilometer 🤷♂️
You just neet to imagine that in convertional current positive charges are flowing instead of electrons
You cleared almost every previous doubt I had, Thank You 💪
dude these videos are burning up the blank pages in my notebook! ...I kinda love it when I find a UA-cam channel that makes me fetch a pad and write stuff down...
Thanks for another great video.
Fantastic video!! and great series on electricity basics. Easy to understand. Thank you. One small typo in video at 1:06 - "... no engineer is going to covert volts into joules/coulomb and amps into joules/second.. " . You meant amps into coulombs/second?
You are correct of course. RIP UA-cam's annotations feature :*(
Leonardo Rodrigues
Leonardo Rodrigues
5 years ago (edited)
Thanks for the high quality videos. Instructive and amusing. I have been working on a project in which I'll have to use a current limiting circuit to protect my DC power supply. Can you make a video about the different choices for this case? I don't know if it would be popular enough, but is just a suggestion. Cheers
Does V x A = W only apply to DC and linear AC loads? When does power factor come into play?
Also, why are transformers rated in kVA and generators rated in W?
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
These videos are very good at explaining this, thanks!
Power is the rate at which energy gets supplied, or used up and instead of saying joules per second, we have a unit we use for power, called the watt. Also 1 watt is equal to 1 joule being transferred. Voltage and current that leads to power. Voltage and current can be combined in different ways to get the job done.
Thanks for the high quality videos. Instructive and amusing. I have been working on a project in which I'll have to use a current limiting circuit to protect my DC power supply. Can you make a video about the different choices for this case? I don't know if it would be popular enough, but is just a suggestion. Cheers
are you alive
I'm really grateful. Quick question though; what causes the resistor to go up in smoke?..voltage, current or power?
If power, that should mean a low voltage, and a high current could generate such heat
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 😅
So what you are saying is the resistor in that setup can take a maximum of 42.23V reduction in that circuit to max out at .25Watts
1:11, amps into coulombs per second
Real-world example, if you use air conditioning with declared power of 3500 watts, and your city grid is 220 V, that means that your air-con device will pull 15,9 amps through the electric wire to put it to work. (3500/220 =15.9), simply the declared power of an electric device tells you how many amps it is going to need. A 500 watt PC will suck 2.27 amps through the wire. All power consumption in ampers sum up, and if it exceeds the Fuse limit, it will break the circuit for your own protection. You can make a mistake by putting the higher value fuse than it is the capacity of the specific wire. In that case, it will let the higher ampers pass...and the wire will melt, burn and cause a fire.
so using those style of resistors is basically bleeding off energy as heat to 'dilute' the currents overall energy?
Good im found all of your video im now gonna learn some electric Thank you very much
Great video as always
I'm curious to truly know if an electrical component going up in smoke is tied more to current or power as from what I've learned having too much current while keeping a low voltage but the same wattage is a good way to lose power while frying whatever the power is going through simultaneously
Thanks for sharing its helping with my latest projects
dude you are very good explainer :)
I was having a similar case but thank God It has been fixed
so much enjoyable and informative. thank you.
very Good informetion of e.c Unit given.
Thanks for these videos!
love your videos please recommend a good app for the resistors calculator thanks
Thank you
This is straightforward.
+Afrotechmofs fantastic man!
I wanna mention something fail which most of the people don't know about. For example that LED lamp which u showed rated as 40 W, but the issue is, when we gunna power it with some power supply like a transformer (not a battery), we have to calculate how much the transformer will drag. Depending on the bad efficiency of the most of transformers we'll find that the power which is needed to power the LED is about the double, about 80 W and that's horrible!
Am I right? Thx
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% !!!
I am very new at all this, so I need to ask:
When the resistor burned, how come the LED kept giving light?
I would expect that, when the resistor burned it stopped working, which means it stopped "choking the amps", and so too many amps would flow into the LED and it should have burned also. And, yet, the LED kept giving light.
Am I getting something wrong?
does low voltage like change the current to be high or is it when the voltage is lower then current, essentially the electrons just siting there with no where to go almost and just start to heat up like pressure building up on a hose?
thanks bro.. the best channel !
Great job explaining.
well sir i appreciate of your teaching method nice
So let's let's say that a microwave draws 1000w of power per second. What confuses me is that what part of the power is spent. So only some of the energy is turned into microwaves. And we pay bills in watt hours, so it is obviously not drawing 3.6million watts per hour. So my question - is the power drawn different from the power that is dissapated?
I think you have your units mixed up a bit. It doesn't draw 1000w per second, because watts are already in "per seconds". It just draws 1000w, which means 1000 J per second. So leaving a 1000w microwave running for one hour means that it uses 1000 watt-hours of energy (since it's 1000w of power running for 1 hour).
i've loved your vidio it help me a lot thanks master you r great
love your videos!
thanks a lot!!
you said that Ohm's law is a linear relationship between current and voltage provided the resistance is constant. However, in the experiment you are referring to at 0:15, when you set up the voltage to 1, a current of 1.8 Amperes was displayed. But when the voltage was increased to 2, the current was 2 Amperes. This is not a linear increase. Why did that happen?
Because the motor is not a linear load. Not all loads are linear and therefore not all loads follow ohm's law.
@@Afrotechmods 4 years after and I still got a reply!!! WOW!
Your content is AWESOME!!! Your 3 videos on Current, Voltage and Power are the best I have found on the Internet and better than any textbook.
I might be pushing my luck but there are are few aspects of basic electricity that I have not been able to grasp even though I have scoured the Internet for explanations. I would be most grateful if you could shed some light on these. Even electrical engineers seem to contradict themselves on some of them.
1. What is the difference between charge and electrons?
2. Is electricity a flow of charge or of electrons? If it is a flow of charge, and charge is a property of the electron (and not a physical element, as some people seem to put it), how does charge flow? And is the definition that "electricity is a flow of electrons" wrong?
3. When voltage is increased, are more electrons (or charge) being sent from the battery, through the wire and to the load, or is the same number being sent but at a faster rate? One site even claims that more energy is getting transferred to the electrons while their number and speed remain the same.
4. If at a vertical cross-section of the wire, there is x number of electrons/charge, and the battery is sending, say, x+100 of electrons/charge, will the wire not be able to handle this and will consequently burn, just like a resistor or fuse? And if the battery sends x-200 electrons/charge, does this mean that, if the voltage is increased, there is room for about 200 more electrons/charge to be sent? Or is this an erroneous over-simplification?
4. If the battery is providing more energy to the electron/charge, how is the electron/charge storing this energy? Is the electron vibrating more, for example?
5. What causes the pressure (potential difference/voltage) in the battery? Is it the difference between the NUMBER of electrons at the negative terminal and that at the positive terminal, or is it the AMOUNTof energy that the electrons/charge hold at the negative terminal with respect to the positive terminal?
Please do continue to make videos.
Praying that you are in a generous mood to educate this ignoramus :-)
Thanks beforehand for your replies.
@@salimr47181) Charge is a fundamental property that some types of matter have. And electrons are a type of matter.
i.pinimg.com/originals/9f/57/7b/9f577b9a840c4c8bd1870523bb44de3a.gif
2) It depends on how you define "electricity". It is technically more accurate to say it is the flow of charge. "Electricity is electrons flowing" is just a simplification to try get people's heads around basic concepts before delving into quantum physics. If you really want to get into more detail, electrical current isn't actually even the flow of charge, it is the propagation of an electric field which results in the movement of charged particles.
Sparkfun has a more physics-ey explanation that you might find helpful. ua-cam.com/video/z8qfhFXjsrw/v-deo.html
The rest of your questions suggest to me that you are looking for answers in the wrong place - most electrical engineers usually don't look at things at that low a level, you want to be learning from physics and chemistry teachers. That being said, they are excellent questions so definitely keep searching for the answers!
Thank you for your answers and suggestions. Your explanation about charge and electric field helped me a lot. I now understand what you mean, although the aspect of what exactly increases and changes when voltage goes up is still unclear. I will keep on searching for answers and, if I find them (and remember), I will surely post them here.
On a side note, this quest for answers has also helped me understand why I (and other people like me) find it hard to understand physics. It comes down to how we approach the subject matter. We need too many answers right at the start and we are serial learners; we find it difficult to move on to the next topic unless we have fully understood the current one. This assumes that there is an answer to every question we raise, which is often not the case in science (and I am discovering that there might not be a single human being on this planet who fully understands all aspects of electricity). This also requires a teacher with in-depth knowledge, excellent teaching skills and tons of patience. This is exceedingly rare.Those who do well are those who can treat what they do not understand as black-boxes, or are satisfied with analogies, and are able to connect the black-boxes together to get an overall picture of the subject. And they move on with their lives! I believe that this is one of the trait of a good engineer. For people like me, that one thing that we do not understand might consume our days and nights (as it is happening with me currently).
I wonder if I can train myself to learn otherwise.
Thanks again.
Thanks a lot for all the tutorials :)
I like your videos.
For people that don't get it: Think of electricity as water flowing through a pipe. Volts is water pressure. Amps is pipe diameter. If you increase one or the other, you get more "power".
Hi. On my wall there are sockets hat are able to output 1500W. If I plug a small lamp to the wall it get 10w from the 1500w available. If I plug a bigger lamp it get 100w from the 1500w available. Who is to decide that a lamp can get 10 or 100w? Is there a resistance on every lamp that permit to a small lamp to not be burned by the power of the wall socket?
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? Because I saw a video some time ago about why we use ac current instead of dc current, and the reason was that it is very easy to transform low voltage with high current to high voltage and low current (and viceversa) with ac, and that it was more efficient to deliver electricity with a high voltage and low current. Also some videos here on youtube about contact welders show that they rip open a microwave transformer and mod it to get a low voltage with a massive current from the wall.
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? (the high voltage on power lines and those contact welders that people made that need a lot of current)
_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.
How was the lightbulb able to pull such low amperage from the wall? Is it because it is an AC power supply?
Yeah it's because it's 120VAC
Awesome Video!
i wanna know why and how did you take 6800 ohms resistor??
in experiment at 3:26 (140V battery, 3V LED, 6800 ohm resistor), what if the circuit only had resistor (without LED), with 1/4W resistor, the current must be lower than 20mA and the resistor didn't smoke ?
Good information. Thank you.
This is an awesome tutorial 😉😉😉😉😉😉
The arrow doesn't indicate the right formula at 2:57It should instead be the one saying I = V/R
I am not an EE but from what I have read, in AC circuits the voltage and the current are slightly out of phase. So how does one calculate the power factor?
mediaguardian well google,english is not my primary language,but i study EE,so there are 3 equations:
1. P=U*I*cos(phase in °) in Watts
2. Q=U*I*sin(phase in °)
in VAR
3. S=U*I in VA
Good video, well educated indeed
Good work
can I fix multiple LEDs (assuming each needs 3V) together WITHOUT a resistor, as long as [amount of LEDs] x 3V = [the V in the battery I choose].......?
I work in I.T but tinker with this stuff. I hate how they size UPS in VA yet almost every PC, server, switch, etc they say it in watts.
Like say a 24 port POE switch has a PSU that takes 700 watts. Then you have to find the right UPS for that and those are usually in VA which makes things kind of confusing, frustrating and a pain. I wish the UPS s would just say recommended max load 100 watts. then you know that it can run a 700 watt switch fine, or maybe 2 in some situations.
I always heard NEVER plug anything with a motor or laser printer into a UPS. I heard a story of some tech needed an outlet plugged in a mini vacuum to the UPS and blew the UPS up. Laser printer I think is more an issue of wattage that the fuser draws.
jtech0 The reason they rate these in VA is because of the maximum CURRENT RATING of these devices. Say, you're pulling 1000 VA and using 746 watts. All is well, if your power supply is rated at 1000 VA. You are drawing one horsepower. But your POWER FACTOR is only 0.746. And, in real life, it gets worse. You are lucky to have a 0.65 power factor... not in the UPS device itself, but in the LOAD. Then, again, it all depends on what devices comprise your load... IHTH
Are you kidding? These fuses and breakers are rated in AMPERES. PERIOD. Nothing to do with wattage. Wattage refers to the actual work consumed by the load, and in certain applications, is a fraction of the VA product. You have a lot to learn.
Actually, I lied to you. Breakers and fuses DO have a MAXIMUM voltage rating, but it has nothing to do with initial blowing or tripping. Instead, it is a safety specification, in order to ensure RELIABLE circuit interruption on overload. But still: a 20 amp fuse is designed to open at 20 amps, no matter what the OCV. Put a million volts across a blown fuse, however, and it's liable to arc over.
thanks for the explanation
Very very basic question
When you say the bulb consume 40W every second, then the energy meter in the house, how will it calculate the consumption?
Since it is Wh, so is the calculation 40W x 3600sec? Wouldn't that be too much or incorrect?
Because I thought, whatever the power rating of the electrical equipment, that is the consumption per hour, ie, 40Wh.
Equation looks similar but the final calculation the meter does would be different for billing.
Arjun Menon a 40W bulb running for an hour would cost you 40W x 1h = 40Wh.
Arjun Menon 40W * 3600s. This would give you the amount of energy in J (Joules)
miningmanna mc Wh is too a unit of energy, just not S.I (system International)
pit fermi I know, I was pointing out, that he may have confused Watthours with Joules
yes. my bad. confused basic units with S.I :D
instead of putting in a stronger resistor, can I simply put multiple resistors in parallel to divide the power onto more resistors? or doesn't power work like that?^^
Blasulz1234 2 equal resistors of, let's say, 5 Ohms in parallel will have an total resistance of 2,5ohms, so you would have to buy stroger resisitor, but if you're going to buy more resistor why don't buy one of 5ohm that can handle the power? But if you already have 2 10ohm resistors, yes, the power will split, the circuit will work and you won't get smoke
yea I'm aware of that, but thanks!
wow, i like the explanation,,,, if i have chance and not bc, i try to watch of ur some video,
Can u please guide me to calculate the time when 48v 2kw motor is connected to 48v 100ah battery.
Is it possible to boost 100watt to 250 watts like boosting voltage
Why, when I feed my electrical designs from the car (12-14v, tons of Amp), it produces so much heat in the regulator lm7805, do I need a resistance before the circuit?
Without knowing more about your circuit I can't give a full answer but if you are using it to power a USB outlet for instance (I'm guessing because of the 5v) then I would suggest a buck regulator design rather than a linear regulator, they tend to be more efficient at dropping voltages.
Hello, thanks for the suggestion, but it's not for usb. The small part of the regulator circuit of 5v is to feed a PIC , The regulator circuit is basic, with a capacitor of 470uf or more at the input of the regulator, and at the output one capacitor of 10uf and another of 100pf. Basically only food to the pic, and some transistor if the case with the 5V, if I have other peripherals like leds or sensors, the food direct from the 12-14v of the car with their respective resistances, the LM35 It is perfectly capable of feeding my circuit with its 500mA and these work well when I install it from the source that I have for tests, but when I install it in the car everything starts to go wrong, It gets very hot sometimes it burns. I thought of putting a resistance of about 100 ohm at the beginning of the whole circuit
Midian Alberto Khaiyow Ok, i'm too tired at this time to read that properly so I'll skip to trouble shooting.
You say it works on your supply but gets hot in the car. I assume you are testing at 12v? The car when running will be around 14.5v(ish) so try testing at that voltage and see if that replicated the problem?
Hey man, so I got a suggestion to translate this video to Swedish (Since Im Swedish I guess) and I was a bit curious to do that. But I wonder, do you get control over that translation or is it just Google that will use it for their future AI? ;)
your awesome great video!
I think you need clearer explanation for the 120V AC (I believe this is RMS voltage) driving light bulb compared to LED requiring DC.
Why does it not use power from what it gets, based on voltage? This phenomena has baffled me - the load dictates or the supply dictates the power used? E.g. putting a "large" fan on a "small" circuit.
because watching things catch fire is fun
Nice video :-)
Amps
not joule per second
coulomb per second*
(at 1min 12 secs)
at 1:10 you said amps into joules per second, instead of coulombs per second...
Doh. Thanks. Added an annotation.
@@Afrotechmodsannotation...
I've learnt alot. 🤖
Nice video
You are awesome thankksss alot
Great video! Hey, also by any chance, could you give me advice for my channel? I also do Electric things, and any criticism would be very much appreciated :D
Keystone Science
It's already good (better than mine, nyahaha (painful laugh)), just keep going and improve your vid quality.
I too, just started making videos a few months ago, I hope it goes well..
UA-cam views are proportional to the amount of cats in a video. Something about your audio doesn't quite sound right, I can hear clipping/distortion sometimes. It sounds like you are having to talk really loudly into your mic just to get things at a normal audio level. Might be worth investing in a new external mic to plug into your camera. Otherwise I think your content is pretty good.
I just subscribed!
+Afrotechmods thanks xD yeah, my microphone was like a 75 cent one off eBay, so, in other words it's probably terrible quality
+Keystone Science u have a great channel, subscribed ;)
Actually, try to work more with audio and the cam plz, and plzzzz try to talk a little slowly, English isn't my native language lol
why electricity is transmitted at high voltage ?
Woaaah Thank You Dude..
thx you very much
If one watt is equal to 1 joule per second, why call it a watt? What was wrong with calling it a joule?
5:00 I need that.
Please brief us regarding the 3 formulae:
P=VI
P=(I^2)R
P=(V^2)/R
Is there any difference between their usage? How do we know which one is to be used in a given problem?
thanks alot
thank u sir :)
Works for dc, what about va? And power factor...
www.sgs-engineering.com/wp/wp-content/uploads/2015/09/beerkva.jpg
I explained this to someone below but here we go:
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.
bennett
nd ho to caaalcooolyyyyt temperature...?
Wonderful!
I have 9v battery how can I calculate power of it using multimeter.
P=V×i works only for dc not ac
For ac it was p=v×i×cos0
5:02, u are lit
This video was released 1 month too late for IGCSE candidates :(
My 12v killbots might be hugely inefficient but thats primarily a result of them being programmed to feel pain..
Stardust LEDs
turn down for "watt"