In your example for parallel resistance you came up with 0.9 ohms. You then mentioned product over sum and the equation you wrote was (2x3x4)/(2+3+4). That gives you 24/9 which is not 0.9. Anyway, I went ahead and looked it up. Product over sum only works for 2 resistors. If you have more than 2, you have to calculate 2 of them then take that result and use it as 1 of the 2 factors in the calculation with the next resistor. The original formula you showed for parallel resistors can also be written as 1/Rt=1/R1+1/R2+1/R3. It's actually the same formula, but it may be easier for some to visualize than the double fraction.
Yes, parallel resistance is the reciprocal of the sum of all parallel resistive reciprocals.; put simply, find the reciprocal of each parallel component (1/R), sum all of these together, then take the reciprocal of the resultant sum (1/R total). Furthermore, a good way to check whether you've completed your calculation correctly is by comparing your resultant resistance to the resistive element in the parallel arrangement which has the lowest resistance. If your calculated total resistance of the parallel arrangement is lower than the resistance of the lowest resistive element, you have most likely done it correctly.
For those who may have forgotten if your resistor values are the same you can simply take the value of one and divide that by the amount of resistors to get total resistance.
Ive used PIE formula ALOT, especially when an ellement is being changed and you want to use your clamp meter to test the amprage, you can calculate what Amps would be drawn by the ellement, when you test it to make sure is functioning as it should
@@fredmauck6934 While you are correct, in my 18 years of experience I have found that for most circuitry I work with in commercial settings the wire sizes are determined by the engineer who draws up the plans. Those plans go through plan review and we build the project to the print specifications. On small time and material renovation projects, considering that a voltage drop of 3% in the branch circuit and 2% in the feeder is acceptable, I have only needed to use the formula for voltage drop two or three times during my career. The most common math I have used is degree multipliers for bending conduit. I am a member of the IBEW living and working in Washington State. Always have your UGLY'S Sparkies.
Dustin, as a senior electrical engineering student and commercial electrician, I must say…your teaching and knowledge is phenomenal. Although, as electricians we technically don’t need to know the intense calculus, differential equations, and linear algebra behind RLC circuits, I’m glad you mentioned it. Thank you for all of your informative videos!
I completed calc 3 and everything before it and honestly, its 90% of complete waste of time and practice. I suspect its all about the $$ and we need to make education about education and not about profits.
Thank you for this lecture. I am a electronics hobbyist trying to get off the grid at the same time trying to get a job as an electrician apprentice. This will help me get ahead. Can't wait to check out other content!
That is not Ohm’s law. Ohm’s Law states that it takes one volt of pressure to push one ampere of current through tone ohm of resistance. What you presented as Ohm’s Law is actually formulae derived from Ohm’s Law. I missed out on a job opportunity once because I mistakenly explained it as you did, and the master electrician doing the hiring did not like that.
@@JoseVargas-ki5hc yeah, I’m pretty sure that is what I have said. It is not Ohm’s Law. It is a formula derived from Ohm’s Law. Not the same thing. Tell me again how I am wrong, especially since you agree with me?
@@douglasabler3581the title says “5 formulas” okay you got it? And then he proceeded to explain the formula….you got it? A FORMULA …..from Ohms law….FORMULA
@@JoseVargas-ki5hc And in the first part of the video he shows a formula. And claims the the FORMULA IS THE LAW. Its not. It is a formula derived from the law. Got it?
@@douglasabler3581He said Ohms law is the relationship between voltage, amperage, resistance. Okay. So Ohms law is a fundamental principle in electrical circuits that states the relationship between voltage, current (amperage), and resistance. Specifically, it says that the current flowing through a conductor between two points is directly proportional to the voltage across those two points and inversely proportional to the resistance of the conductor. Ohms law has 3 main formulas like he then showed us. Is he wrong?
11:48 that formula is incorrect. It should be: Rt = (R1*R2*R3)/(R2*R3+R1*R3+R1*R2). Also "Joules Law" would more appropriately be called "Watts Law". Joule is a unit of energy, watt is a unit of power (energy over time). 1 watt = 1 joule per second.
@@coldspring22 Not true. Dustin's formula gives (2*3*4)/(2+3+4)=2.67 ohms which is incorrect. As others have noted, his formula only works for 2 resistors.
A&P license years ago, electrician for 25 yrs, contractor for last 6 yrs. Worked with 100s of electricians. That was very well explained for the average electrician!
@@zany5148 Better by a lot? ... Yes, always. However, it's superiority in specific areas. That's one of the joys of working in the field, it's experiencing all the individual co-workers ... and discovering what's their particular strong points are. Some are immediately obvious, others often require time to fully appreciate their attributes. Besides, different types of work, different environments ... have radically varying proficiency needs. Tenant finish ... is entirely different than a pharmaceutical plant, underground duct-bank, of 25, 6" conduit (five high, five wide), over a mile in length, serving multiple buildings, w/several underground vault pulling points. Those are two different disciplines. Electricians can get extremely good, in quite different ways. Devicing out a room is a whole different world from threading 4" rigid with a hog-head, and a porta-pony! How's that for a retired boomer tangent?
The Pythagorean theorem is good to know for figuring out how to roll offsets and helps you calculate degrees of bend to recalibrate old benders. Makes conduit bending easier when smart benders are available.
I am curious as to how it would help calibrate the bender? That seems really useful out on the field! Mind explaining it? And I use pythagorean theorem all that time for rolling offsets as well.
@@travisharrington5819 The Pythagorean theorem is also good at determining phase angles between voltage and current. You can also use trigonometry and find the cosine of the phase angle to get to power factor.
Hey Dustin! Might be worth mentioning the reasons behind the 1.73 with respect to calculating 3 phase power as well with the P=I*E bit. Also, for those asking about E vs V for voltage, it was called Electromotive force before Alessandro Volta put his last name on it!
@@RJFerret beat me to it but yes Dustin has addressed it a few times in the past. Can't recall the exact video but yes he has. I know because I learned the significance of 1.73 from Dustin
One of the most common misconception is that voltage is the same as electromotive force. Voltage is NOT a force. Voltage is the amount of energy per unit charge (1 Volt = 1 Joule/Cuolomb). To resolve this issue, physicists simply say emf (never electromotive force).
Learned all of these great formulas 50 years ago but have not used the vo!tage drop in over 25 years. Got lazy and just go to one of the free web sites to calculate voltage drop. Did an interesting voltage drop calculation 40 years ago. A sparky coworker was arguing g with his sparky dad about what size wire to run to 10 pole mounted driveway luminare that were 100' apart and think each was 75 watt HID. Coworker did not want more then 5% vo!tage drop at furthest luminare. Anyway came up that he had to use #8 copper to first luminare, then #10 copper to forget the exact # but maybe 4 then #12 copper. Just used basic ohms law for DC circuits on this of course AC circuit. Voltage at last luminare was just under 5% total voltage drop & within one volt of what I calculated. His dad told him to just run #12 copper from panel to.last luminare because it was under 10 amp load.
I did hundreds of calculations on parking lot lighting. Your answer seems perfectly correct. At 480 volts, sometimes the conductor needed for VD was smaller than #12. But since 20A circuit, had to minimum at #12.
A year later lol. Did you just subtract the 'voltage drop' from your nominal voltage at each pole or just at the wire size changes? or am I wrong all together? was your goal to keep the VD% at a certain threshold at a % of the way.. (2.5%VD MAX, at half the distance)
Dang, just had my fourth year (final year) mid term tonight! Wish this was out a WEEK ago! You da man, Dustin, imma follow up on this come finals weeks and Journeyman exam
The ending was great 😂 "we're all dummies" my time has come and really helps to remember all of this formula memories back in my tech trading school so wish me luck in my electrician career soon i hope 👍🤞
Get taping it back together I'm hvac tech lot of my side charts are falling apart can go on line but sometimes u get conflicting information my books and charts don't conflict if that makes sense 😅
Sorry not sorry I am necromancing this comment but I'm a homeowner DIY with BS/MS in electrical engineering. I understand so much of the theory behind electrical and even a lot of the practical applications but the appropriate Ugly's manual is an absolute game changer. I have probably learned as much reading that book (I skipped a lot of the motor sections because I didn't even like that when I was in school 😂) than I did at least in grad school. Electricians are my heroes - thank you all for what you do especially when you do really neat and tidy installations. 😊
Glad to here u using practical applications maybe I won't cus engineers as mutch 😅 ur sience is there but some times u get in the field practical calculations are needed engineering with practicality is always appreciated
Thank you for all you content. Ive been an electrician for about 3 years now recently went back to become an official apprentice. I am taking a class to challenge up to a year 2 apprentice. Thank you!!! ive been watching you for about 2 years now. thx!!!!
Not so. You can do the replacement resistance for 2 resistors. Then again for the replacement resistance and the third. Not that it's a convenient method.
@Major General Suppose you have 3 resistors of 3 ohms in parallel. work out for 3 and 3 : you get 1.5 as replacement resistance. then work out 1.5 and 3, you get 1 ohm. Thatś the final answer.
You can use product over the sum if you complete them in twos. solve r1 and r2, which will give you R12, then solve R12 and R3. It is doing it out in long form, but if you do not have the inverse function available on your calculator, it is a proper method to solve for resistance in a parallel circuit.
Can also use the conductance formula. Conductance being the inverse of resistance or G=1/R. The formula is Gp= G1+G2+G3. So, Gp = 1/2+1/3+1/4 or Gp=0.5 + 0.3 + 0.25 Gp=1.083 R=1/G R=1/1.0833 R=0.92
@Nitrodamous At the very least invest in some CDs at the bank. They're almost stupid proof. Not the greatest. They're like an old timey savings account back when those payed anything.
@2:43 Amperage and Current are the same thing. An Ampre is a derived unit using the SI unit Coulomb for charge. 1 Ampre is 1 Coulomb per second. Using the water analogy, current would be amount of water per second, like gallons per second. Voltage is analagous to pressure.
Man I’m sooooo grateful to run across a knowledgeable craftsman like yourself I went to electrical school before, and just watching and learning is such a refresher it’s like being in school all over again this is like something I will never forget no matter how old I get I can do some electrical work and throw some pipe on the rack u feel me
Hi Dustin As an Electrician from the UK… I am very interested in how Electrical installations are done over the pond in the USA. However I see loads of videos on installation techniques and the differences on the wiring , voltages and the Codes you all use in the States. I’ve never seen a video on Testing an American Electrical installation. We have to Dead test our circuits in a few different ways. Mainly to prove Earth (cpc) continuity, this also gives us R1+R2 ohms reading between the phase and circuit protective conductor (cpc) you call this “Ground”. Its Part of the formula Zs=Ze+(R1+R2) Ze is a live test and is achieved at the distribution board and gives you an impedance value of the external earth path of the Phase and ground. Zs is the total phase and ground impeach reading at the furthest point of any single circuit in a dwelling or business. Polarity of the phase to make sure Fused and circuit breakers are only on the phase conductor not the neutral. An insulation resistant test to make sure there is no short circuit between, Phase to cpc, phase to neutral and cpc to neutral (in single phase installation). Where its a 3phase installation we need to do a phase to phase test too. Then we do a series of live tests. The main one being an Earth ( ground) Fault LoopImpedance test. This would determine the max ZS value. This will also show if the circuit breaker (mcb) will disconnect the supply within given time allocation for a breaker size. There are other tests like RCD tripping Tests its like a Ground fault breaker. It Would be very interesting to me to see your complete Electrical testing procedure in a video or if already have one please link it to a reply.. thank you in advance ❤️👍
As an EL-01 in Washington State with 18 years experience I tell you this, Keep a current NEC, have an Ugly's, and know your conduit bend multipliers. It is really that easy. Just do the work.
as many have pointed out, the "product over sum" formula only works if there are 2 resistors. If there are 3 resistors, say A, B, C, then the formula is: R = (ABC) / (AB + AC + BC). Other than that, great videos.
@@nhitc6832 You're better off just using this in every scenario. Avoids confusion by having to memorise various different formulas for unique arrangements/instances; whereas 1/Req = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn works for every scenario.
@@tomctutorfor any arbitrary number of resistors it would be (product of all resistances) / {sum of(product of all resistances except this resistor) for all resistors)} Which makes a lot more sense in actual math notation with the big sigma and big pi and index notation. But for 4 resistors, that works out to: R = ABCD / (BCD + ACD + ABD + ABC) There is no magic in this formula, if you just rearrange the more basic equation with algebra you can derive this one. But you have to multiply so many numbers now that you're better off just using the other equation.
The rule of "product over sum " works only when one have 2 resistors in parallel. Regardless this slip your work teaching is awesome. I am a Telecomminications and Electronics Engeneer and watching your videos I remember my class of Electrotecnia long time ago. Congratulations. Greetings from Mexico.
"@nhitc6832 2 months ago as many have pointed out, the "product over sum" formula only works if there are 2 resistors. If there are 3 resistors, say A, B, C, then the formula is: R = (ABC) / (AB + AC + BC). Other than that, great videos." Yo I stole this from a dude further down the line. The formula in the video don't work for this exacting application. Just use the formula written in blue at the capacitance part. I actually got it to work after crying and coming back a week later.
What you call “Joule’s Law” should be called Watt’s Law, because power P is in units of watts, not joules. As you probably know, 1 joule of energy per second is 1 watt.
According to statistics, 98% of the world's population has an IQ that is considered to be 'below normal'. As a tutor in a past life (and in my opinion), I found that the best approach to teaching anyone the 'percent circle' is to simply place a multiplication symbol immediately between the two lower variables (in place of that vertical line you have in yours). This eliminates any lingering confusion that may remain in those 'select few'. Once they learn that Percent Circle, they can apply it to anything, including the financial market, distance/velocity, etc.
This doesn't even make sense since "normal" implies most prevalent and common. So no 98% would be normal by your statistics, which I already know is complete wrong without even looking it up.
helpful; but the r1*r2 / r1+r2 (or product/sum) just works for "2" resistance in parallel and not more. the actual and correct resistance value when 2,3,4 are in parallel is 0.9 ohms, where the (wrong) product/sum offers 2.6 ohms
You can only use the product over sum method on 2 resistors at a time. When working with 3 like the case in the video (2ohm, 3ohm and 4ohm) you would do the 2 and the 3 ohm resistors first which would come out to 1.2ohms. 1.2ohms is now the equivalent resistor to the first two and would be the new value for both of those. So now you take that and do the whole method again with the third resistor (3ohms). 1.2 x 3 / 1.2 + 3, which comes out to roughly .857ohms. You can keep using this method with more resistors just remember the resistance total will keep getting smaller the more you add.
The third resistor would be 4ohms not 3ohms, 3ohms was the second resistor. So it would be 1.2 x 4 / 1.2 + 4, which would come to 0.923. I'm still learning, Your comment helped me to understand the math behind what Dustin was illustrating, just saw that you put 3ohms as the third resistor instead of 4ohms 🤙
@@weavercattlecompanyhow is he wrong id like to learn more about this, im a student for Harley and I’m currently in an electrical class so I’m trying to absorb everything I can
One that I love that I'm actually somewhat surprised wasn't covered is the formula to calculate the distance multiplier for offsets. I've had to do some obscure degree offsets to get a pipe run to work and look good with how obstructions were positioned. If anyone doesn't know it, it's 1/(sine of the degree). Every apprentice I've had, I give that formula to and they all have clean work when running pipe because of that
Aluminum is generally more conductive than most types of covers. Aluminum has a high electrical conductivity, which means it can easily transmit electric current. This property makes it a popular choice for various applications that require good electrical conductivity, such as power transmission lines, electrical wiring, and heat sinks.
Another great video Dustin. I did many many calculations to limit my voltage drop on branch circuits to 3% as suggested in NEC. Since I’m really looking conductor size needed to achieve the goal, I exchanged Vd and cm but did 3% of the volts of the system. One calculation and I knew how many cm was needed to be at or below 3%. Go to the table and pick the conductor size. As. Cm=2xKIL/3% of V. I also did 3 phase slightly different. For 3ph 3w balanced load same as single phase and multiply by .66. For 3ph 4W balanced load multiply by .5. In the thousands of designed circuits over 35 years, I never had a low voltage or breaker tripping issue.
I can't imagine how helpful a resource like this can be for those learning the trade today ... it's just very cool! I went through my apprenticeship 40 yrs ago ... a resource like this would've been priceless. Keep up the good work.
I am currently studying these formulas to apply for the electricians apprenticeship. These videos are very helpful, along with some others here on TH-cam. I want to make sure I can pass the aptitude test.
I'm trying to learn some electrical stuff to get a head start before I start taking classes or doing an apprenticeship, this is very helpful and informative. I appreciate the time and effort you took to make these videos! I subbed👍
I started in Australia as an apprentice electrical fitter in the mid 1960s. My experience was different from this . Much of our basic formulas were the same, but, we didn't get into voltage drop. We just had a different approach, and much of our work was fairly standardised. We also did not refer to Joules' laws, they were power formula. Earthing in an installation was important, and knowing the national standard wiring rules. I have to say I was gratified that the math was not complicated.
2:46 P is for power which is a unit of work/energy per unit time. It is the rate a which work is being done or energy is being used or absorbed. Joules per second or watts Amperage is the unit of current which is in fact a rate but it is a rate analogous to the flow of something like a gas pump. Imagine charge to be analogous to fuel and the rate which fuel is being pumped into your car then it can be kind of a good way to get a better understanding of amperage as the flow of charge. This rate of charge does not already account for the rate which energy is being transferred, not independently anyways. The product of current and voltage accounts for the energy per unit time because current is Coulombs per second and voltage is energy per unit charge or joules per second. The multiplication of current and voltage allows us to cancel out the units of charge so the product is joules per second. This doesn’t take anything away from his fantastic explanation of the formulas but I want to caution viewers who are in a classroom setting, units and dimensional analysis are very important. As I tell my students, be aware of the details of the units which are being cancelled out and which are left remaining, they should provide a secondary confirmation of the correct units of your equation.
Could you do a video in the future on symbols you’d find in the field like for plugs switches contrators on schematics etc.? I can’t find any good videos on that
Not even an electrician, but a computer scientist/Software engineer and I just like this kind of stuff. If you can understand all this electrician stuff you can probably be a pretty dang good programmer.
A couple interesting observations comparing my training (electronics technician) to this (electrician). Line voltage drop isn't nearly as important in small circuits or inside wiring panels, so that formula was more of "Here it is, but you don't need to memorize it. Just know where to find it if you need it again, but you probably won't." And I don't think we even touched on horsepower. Though I did need to learn how to count and do math in binary. ;-) Also, it may have just been the teacher, but I learned my ohm's law and power formula in triangles, not circles. Side note... Because of the power formula, this video should have been released yesterday... on Pi day. ;-) (Yeah, power formula is PIE, and 3.14 is Pi. But they are pronounced the same.)
Hello Christopher, You bring up interesting points for low voltage in considerably smaller circuits. They are incredibly more complex and efficient. As an electrician though, especially in servicing, these formulas are important. We also don’t really deal with digital circuits and needing to know binary code. So, this video is very informative for the average apprentice electrician going for their journeyman license.
Liked & commented for YT analytics. Great detail at 7:50... Appreciate the obvious time you spent putting this specific video together because the info and specifications are so important. It's very clear you spent a lot of time writing out the slides and notes Glad you didn't just "wing it". Carry on!
Cool video, but I'd like to point out I've been an electrician in an industrial setting for 25 years and only know ohms law and how to calculate wattages- and very rarely ever even need to use those.
Ohm's law in a single phase AC circuit is more of a rough rule of thumb result. In a 3 phase AC circuit it can become a ballpark result. Ohm's law is accurate in DC circuits while in AC circuits it's more of a rough calculation as several instructors over the years put it.
Not trying to be a stickler, maybe I’m wrong, but I believe the “product over sum” method only works if there are only “two” resistances. 3 or more have to be the inverse equation. At least that’s how I learned Jt and I can’t get the math from your equation to equal 0.9 . I get 2.6 . Let me know your thoughts. Great video either way though 👌🏼
I wish when i took up electronics in high school that ohms law was explained to me this simple,instead of making it seem like you had to have a degree in physics to understand it.
@@mikecafe5364 well I'm an actual engineer and EMF is NOT the same as voltage. That's a fail in exams. EMF, is denoted as E or greek epsilon and is an energy transfer. Voltage V is neither a force nor an energy, as it would be measured in newtons otherwise. (Voltage is a potential potential difference). The video poster clearly doesn't know ohm's law properly or really know what he's talking about, because voltage is never denoted as E. V is the correct denoted term for voltage (potential difference) for reasons that ohms law is a simplified formula derived from a complex mathematical equation, derived from energies, forces etc. The video poster is also an idiot by stating electricians need to know deep maths - complete nonsense. He fails to state that ohms law does not apply to DC or diode based electrical circuits. I could go on - but he's a TH-camr trying to earn money rather than working in the real world.
@@StuartCarolyn I agree. But I have seen E used in examples using batteries and internal resistance, where E is the battery terminal voltage when there is no load current being drawn. V = E-IR Which says: the battery terminal voltage when the battery is in circuit and a current being drawn, V, is equal to the unloaded battery voltage minus the voltage drop caused by the internal resistance of the battery which is IR, the current times that resistance
Joules Law is a law stating that the heat produced by an electric current i flowing through a resistance R for a time t is proportional to i 2Rt. joules law is about the heat produced in a electrical flow. P in the p=is is watts which is the amount of power consumption or worked done.
I believe that’s the point right he did in that way faking a mistake to see if we the expetators notice that mistake🧐 Thank you for sharing your knowledge 🙏🙏💐🌷🌺🌷🌺💐🌷💐🌺
You could also add in the Voltage Drop formula for sizing wire using a voltage drop percentage too. I tend to use that variation of the formula more often. Oh and rEsistance. E!!! LOL
CONTENT IDEA VIDEO: Can you put different size guage wires into conduit. For example, for a washer and dryer inside a 3/4 inch emt could you run #10 wire and #12 wire inside same 3/4 inch emt, two separate circuits, same conduit. charts only show how many wires you can fit inside conduit, but the same awg, not mixed. Interesting question. Thanks in Advance Dustin Miami, FL
Of course you can, why would it matter? You may have to think about derating the maximum current rating for the cable to accommodate the fact it is in an enclosed space and cooling is not as effective as if it were clipped to a wall.
I self taught harmonics ,tested well pump soft starts for harmonics 200 horse well pump motors had pipe that looked like termites had eaten it lost many pump 250 feet down the wells
For parallel resistors it’s usually easier on a calculator to set it up like this: RTotal^-1 = R1^-1 +R2^-1+ R3^-1 There is usually a little ^-1 key on the calculator that’s very useful
V I R Ohms Law and P I V joules Law is how we do it the V is voltage the P is simply power. Voltage drop, wow! I often wondered how you in the US do it (missionary😂) We take a value from our Regulations known as mV/A/m ( milli Volt / Amp / per metre) then the length of wiring in metres finally the I b design current. Divide by 1000 = Voltage Drop. Maximum V D for lighting is 3% other 5%. Resistance exactly the same formula we use. While we may talk a different language it still the same science that will stay consistent as our respected Codes and Regs change. Presently in the UK we are on our 18th Edition Admedment 2 2022. Edition are every 10 years with the allowance of 3 Admedments in between. To finish I have been enjoying your presentations.
I appreciate what you do for our trade. All of this basic knowledge is being lost with the younger apprentices that are coming into the trade. They rely on apps for everything......if they even know to do any calculations.
For resistors in parallel, the "product/ sum rule" can only be applicable to only two sets of resistors connected together in parallel. Thank you, Sir!
Hi Dustin, thanks for the interesting lecture! Q. In the last part about putting resistors in parallel, you’re talking about the current is moving in a different way, BUT I’ve learned there is actually no movement of electrons in a wire and that the energy is actually moving outside the wire(s)/circuit. If my assumption is right, how and what is current (Amp’s) actually really, if the electrons barely move (especially in a alternating current)? Q2. If my assumption is right, how does a resistor in a circuit even work if the electrons barely move and it’s the energy that’s moving outside the wire(s). I mean, if the energy surpass (outside) the wire, how can a resistor even function?
Thanks for this man. Studied for my masters and failed. It had been a long time since school. Had I refreshed here I probably would have been alright but froze.
Great video... and I like the presentation.. the digital board, the readability of the material on the board, etc. Also, you explain well, and you look at the camera in a professional manner, etc... overall a great job... easy to follow the material.
feels like ive done tons of 30 degree offsets but rarely do I get to take time to practice other variations. Can we get a video with like 45 or 60 degree bends ??
The US is still stuck with things circular mils ending up with cm meaning circular mils . In the rest of the world cm means centi meter . Cable area in metric is mm^2 .Change over to metric , it is easier
In your example about Power, this is adaptable to resistive load but if the load is inductive you need to assume a Power Factor or see a table with assume power factor for example an induction motor and multiply it to get the real power.
Hello, from Dominican Republic
I'm an electrician. But I am watching this video to improve my english.
Brother donde aprendiste? Yo quisiera prender estoy en Estados Unidos
Klok mi hermano. I’m a electrician from Washington Heightz 😎🇩🇴🦾
Same here, I want to improve my engliwh as well.
@@GARY-zo5fonecesitas trabajar para una empresa eléctrica y ellos te envian a la escuela o puedes entrar a un sindicato de electricistas
@@GARY-zo5fo your English is pretty good
In your example for parallel resistance you came up with 0.9 ohms. You then mentioned product over sum and the equation you wrote was (2x3x4)/(2+3+4). That gives you 24/9 which is not 0.9.
Anyway, I went ahead and looked it up. Product over sum only works for 2 resistors. If you have more than 2, you have to calculate 2 of them then take that result and use it as 1 of the 2 factors in the calculation with the next resistor.
The original formula you showed for parallel resistors can also be written as 1/Rt=1/R1+1/R2+1/R3. It's actually the same formula, but it may be easier for some to visualize than the double fraction.
Yes, parallel resistance is the reciprocal of the sum of all parallel resistive reciprocals.; put simply, find the reciprocal of each parallel component (1/R), sum all of these together, then take the reciprocal of the resultant sum (1/R total). Furthermore, a good way to check whether you've completed your calculation correctly is by comparing your resultant resistance to the resistive element in the parallel arrangement which has the lowest resistance. If your calculated total resistance of the parallel arrangement is lower than the resistance of the lowest resistive element, you have most likely done it correctly.
I knew that thanks to my mike holts exam prep book
Thx I just seen this and was like wait that’s not right I had a feeling something was off
For those who may have forgotten if your resistor values are the same you can simply take the value of one and divide that by the amount of resistors to get total resistance.
It would be really cool if you could show us some situations where these formulas could be useful in your day to day work as an electrician.
The Voltage drop formula would used to calculate wire size when the feeder length become s excessive, generaly over 100 feet.
Ive used PIE formula ALOT, especially when an ellement is being changed and you want to use your clamp meter to test the amprage, you can calculate what Amps would be drawn by the ellement, when you test it to make sure is functioning as it should
@@fredmauck6934
While you are correct, in my 18 years of experience I have found that for most circuitry I work with in commercial settings the wire sizes are determined by the engineer who draws up the plans. Those plans go through plan review and we build the project to the print specifications.
On small time and material renovation projects, considering that a voltage drop of 3% in the branch circuit and 2% in the feeder is acceptable, I have only needed to use the formula for voltage drop two or three times during my career.
The most common math I have used is degree multipliers for bending conduit. I am a member of the IBEW living and working in Washington State.
Always have your UGLY'S Sparkies.
Already existing building, adding receptacles or lights.
I took electrical theory in high school. 30 years ago. Yes, it's time to brush up. Thanks. Good job.
Do it on your own....
Dustin, as a senior electrical engineering student and commercial electrician, I must say…your teaching and knowledge is phenomenal. Although, as electricians we technically don’t need to know the intense calculus, differential equations, and linear algebra behind RLC circuits, I’m glad you mentioned it. Thank you for all of your informative videos!
Bobby, as a person who went to vocational school for electrical. How the hell do I get a job... spent money for this schooling and certificate.
If you only install wires, you don't. If you are an engineer you have to otherwise what's the point to become engineer in the first place...
I completed calc 3 and everything before it and honestly, its 90% of complete waste of time and practice. I suspect its all about the $$ and we need to make education about education and not about profits.
You're a journeyman and now an electrical engineer? Whoa, good job man
@@zhumusic-ng9tr hey buddy, did you ever find a job?
Thank you for this lecture. I am a electronics hobbyist trying to get off the grid at the same time trying to get a job as an electrician apprentice. This will help me get ahead. Can't wait to check out other content!
That is not Ohm’s law. Ohm’s Law states that it takes one volt of pressure to push one ampere of current through tone ohm of resistance. What you presented as Ohm’s Law is actually formulae derived from Ohm’s Law. I missed out on a job opportunity once because I mistakenly explained it as you did, and the master electrician doing the hiring did not like that.
Well the title of the video is “formulas” you should know so you just proved he is right, it’s a formula from ohms law buddy.
@@JoseVargas-ki5hc yeah, I’m pretty sure that is what I have said. It is not Ohm’s Law. It is a formula derived from Ohm’s Law. Not the same thing. Tell me again how I am wrong, especially since you agree with me?
@@douglasabler3581the title says “5 formulas” okay you got it? And then he proceeded to explain the formula….you got it? A FORMULA …..from Ohms law….FORMULA
@@JoseVargas-ki5hc And in the first part of the video he shows a formula. And claims the the FORMULA IS THE LAW. Its not. It is a formula derived from the law. Got it?
@@douglasabler3581He said Ohms law is the relationship between voltage, amperage, resistance. Okay. So Ohms law is a fundamental principle in electrical circuits that states the relationship between voltage, current (amperage), and resistance. Specifically, it says that the current flowing through a conductor between two points is directly proportional to the voltage across those two points and inversely proportional to the resistance of the conductor. Ohms law has 3 main formulas like he then showed us. Is he wrong?
Starting my apprenticeship over after leavening during my 1-2 year. These videos are amazingly helpful for review.
Hi Dustin as an engineering student I appreciate what you are doing Keep it up its just another tool we need in our field.
11:48 that formula is incorrect. It should be: Rt = (R1*R2*R3)/(R2*R3+R1*R3+R1*R2).
Also "Joules Law" would more appropriately be called "Watts Law". Joule is a unit of energy, watt is a unit of power (energy over time). 1 watt = 1 joule per second.
Watts over Volts = Amps...
The easiest way I remember Ohms and Joules' law is WAVE to EIRIE. Watts=amps*Volts & Volts=amps*resistance. After that, it's just basic algebra
No, formula shown by Dustin is correct. Both your and Dustin's formula is same equation shown in slightly different way.
@@coldspring22 Not true. Dustin's formula gives (2*3*4)/(2+3+4)=2.67 ohms which is incorrect. As others have noted, his formula only works for 2 resistors.
A&P license years ago, electrician for 25 yrs, contractor for last 6 yrs. Worked with 100s of electricians. That was very well explained for the average electrician!
I’m curious , was there ever a electrician that was better than the others by a lot
@@zany5148 Yeah, me!
@@zany5148
Better by a lot? ... Yes, always.
However, it's superiority in specific areas.
That's one of the joys of working in the field, it's experiencing all the individual co-workers ... and discovering what's their particular strong points are.
Some are immediately obvious, others often require time to fully appreciate their attributes.
Besides, different types of work, different environments ... have radically varying proficiency needs.
Tenant finish ... is entirely different than a pharmaceutical plant, underground duct-bank, of 25, 6" conduit (five high, five wide), over a mile in length, serving multiple buildings, w/several underground vault pulling points.
Those are two different disciplines.
Electricians can get extremely good, in quite different ways.
Devicing out a room is a whole different world from threading 4" rigid with a hog-head, and a porta-pony!
How's that for a retired boomer tangent?
The Pythagorean theorem is good to know for figuring out how to roll offsets and helps you calculate degrees of bend to recalibrate old benders. Makes conduit bending easier when smart benders are available.
I am curious as to how it would help calibrate the bender? That seems really useful out on the field! Mind explaining it? And I use pythagorean theorem all that time for rolling offsets as well.
@@travisharrington5819 The Pythagorean theorem is also good at determining phase angles between voltage and current. You can also use trigonometry and find the cosine of the phase angle to get to power factor.
Hey Dustin! Might be worth mentioning the reasons behind the 1.73 with respect to calculating 3 phase power as well with the P=I*E bit. Also, for those asking about E vs V for voltage, it was called Electromotive force before Alessandro Volta put his last name on it!
1.73 for three phase was explained in the prior vid on voltage drop FYI.
Wasn't that Lord Voldemort for volts?🙃
@@RJFerret beat me to it but yes Dustin has addressed it a few times in the past. Can't recall the exact video but yes he has. I know because I learned the significance of 1.73 from Dustin
One of the most common misconception is that voltage is the same as electromotive force. Voltage is NOT a force. Voltage is the amount of energy per unit charge (1 Volt = 1 Joule/Cuolomb). To resolve this issue, physicists simply say emf (never electromotive force).
James Watts says high
Learned all of these great formulas 50 years ago but have not used the vo!tage drop in over 25 years. Got lazy and just go to one of the free web sites to calculate voltage drop. Did an interesting voltage drop calculation 40 years ago. A sparky coworker was arguing g with his sparky dad about what size wire to run to 10 pole mounted driveway luminare that were 100' apart and think each was 75 watt HID. Coworker did not want more then 5% vo!tage drop at furthest luminare. Anyway came up that he had to use #8 copper to first luminare, then #10 copper to forget the exact # but maybe 4 then #12 copper. Just used basic ohms law for DC circuits on this of course AC circuit. Voltage at last luminare was just under 5% total voltage drop & within one volt of what I calculated. His dad told him to just run #12 copper from panel to.last luminare because it was under 10 amp load.
I did hundreds of calculations on parking lot lighting. Your answer seems perfectly correct. At 480 volts, sometimes the conductor needed for VD was smaller than #12. But since 20A circuit, had to minimum at #12.
A year later lol. Did you just subtract the 'voltage drop' from your nominal voltage at each pole or just at the wire size changes? or am I wrong all together? was your goal to keep the VD% at a certain threshold at a % of the way.. (2.5%VD MAX, at half the distance)
Dang, just had my fourth year (final year) mid term tonight! Wish this was out a WEEK ago! You da man, Dustin, imma follow up on this come finals weeks and Journeyman exam
Hope you smashed it mate!
The ending was great 😂 "we're all dummies" my time has come and really helps to remember all of this formula memories back in my tech trading school so wish me luck in my electrician career soon i hope 👍🤞
Keeping a current copy of an Ugly's book in your tool bag at all times will solve this and many more needs.
Thank u was about to post whole reason I bought a uglys reference book
Had to tape mine back together.
Get taping it back together I'm hvac tech lot of my side charts are falling apart can go on line but sometimes u get conflicting information my books and charts don't conflict if that makes sense 😅
Sorry not sorry I am necromancing this comment but I'm a homeowner DIY with BS/MS in electrical engineering. I understand so much of the theory behind electrical and even a lot of the practical applications but the appropriate Ugly's manual is an absolute game changer. I have probably learned as much reading that book (I skipped a lot of the motor sections because I didn't even like that when I was in school 😂) than I did at least in grad school.
Electricians are my heroes - thank you all for what you do especially when you do really neat and tidy installations. 😊
Glad to here u using practical applications maybe I won't cus engineers as mutch 😅 ur sience is there but some times u get in the field practical calculations are needed engineering with practicality is always appreciated
Thank you for all you content. Ive been an electrician for about 3 years now recently went back to become an official apprentice. I am taking a class to challenge up to a year 2 apprentice. Thank you!!! ive been watching you for about 2 years now. thx!!!!
Hi Dustin - Just a note, the Product/Sum method only works with two resisters. Three or more you have to do the reciprocal formula.
2*3*4/2+3+4 does not equal .9
Have to agree too
It’s great that you picked up on this mate. It’s always best to confirm your stuff while learning because even the bests make mistake.
Not so. You can do the replacement resistance for 2 resistors. Then again for the replacement resistance and the third. Not that it's a convenient method.
@Major General Suppose you have 3 resistors of 3 ohms in parallel. work out for 3 and 3 : you get 1.5 as replacement resistance. then work out 1.5 and 3, you get 1 ohm. Thatś the final answer.
Product over sum only works that way with two resistors. (2*3*4)/(2+3+4) is 2.7 not 0.9
I was looking for this comment. I just kept thinking I wasn't doing it right. Thank you
Can you really do the product/sum formula with more than 2 resistors? If you do not use product/sum and use 1/RT ----- I come out with 1.083 Ohms.
yeah for 3 resistors product over sum is (r1*r2*r3)/(r1*r2+r1*r3+r2*r3)
You can use product over the sum if you complete them in twos. solve r1 and r2, which will give you R12, then solve R12 and R3. It is doing it out in long form, but if you do not have the inverse function available on your calculator, it is a proper method to solve for resistance in a parallel circuit.
Can also use the conductance formula. Conductance being the inverse of resistance or G=1/R. The formula is Gp= G1+G2+G3.
So, Gp = 1/2+1/3+1/4 or
Gp=0.5 + 0.3 + 0.25
Gp=1.083
R=1/G
R=1/1.0833
R=0.92
The more I watch the more I want to become an electrician. Love your style of teaching!!
@Nitrodamous At the very least invest in some CDs at the bank. They're almost stupid proof. Not the greatest. They're like an old timey savings account back when those payed anything.
@2:43 Amperage and Current are the same thing. An Ampre is a derived unit using the SI unit Coulomb for charge. 1 Ampre is 1 Coulomb per second. Using the water analogy, current would be amount of water per second, like gallons per second. Voltage is analagous to pressure.
The unit ampere is a SI base unit. One of seven. The unit coulomb is derived.
Man I’m sooooo grateful to run across a knowledgeable craftsman like yourself I went to electrical school before, and just watching and learning is such a refresher it’s like being in school all over again this is like something I will never forget no matter how old I get I can do some electrical work and throw some pipe on the rack u feel me
Hi Dustin
As an Electrician from the UK… I am very interested in how Electrical installations are done over the pond in the USA.
However I see loads of videos on installation techniques and the differences on the wiring , voltages and the Codes you all use in the States.
I’ve never seen a video on Testing an American Electrical installation.
We have to Dead test our circuits in a few different ways. Mainly to prove Earth (cpc) continuity, this also gives us R1+R2 ohms reading between the phase and circuit protective conductor (cpc) you call this “Ground”.
Its Part of the formula Zs=Ze+(R1+R2)
Ze is a live test and is achieved at the distribution board and gives you an impedance value of the external earth path of the Phase and ground.
Zs is the total phase and ground impeach reading at the furthest point of any single circuit in a dwelling or business.
Polarity of the phase to make sure Fused and circuit breakers are only on the phase conductor not the neutral.
An insulation resistant test to make sure there is no short circuit between, Phase to cpc, phase to neutral and cpc to neutral (in single phase installation).
Where its a 3phase installation we need to do a phase to phase test too.
Then we do a series of live tests. The main one being an Earth ( ground) Fault LoopImpedance test.
This would determine the max ZS value. This will also show if the circuit breaker (mcb) will disconnect the supply within given time allocation for a breaker size.
There are other tests like RCD tripping Tests its like a Ground fault breaker.
It Would be very interesting to me to see your complete Electrical testing procedure in a video or if already have one please link it to a reply.. thank you in advance ❤️👍
Lots of theory to learn in the field of Electrical installation,as they say we live and learn every day.
L.E.D.
Learn
Every
Day
😉😉
This is 101@@martf1061
at time 12:42 if u use product over sum for 3 resistors in parallel u will get 2.6 ohm not 9ohms or 0.9ohm
As an EL-01 in Washington State with 18 years experience I tell you this, Keep a current NEC, have an Ugly's, and know your conduit bend multipliers.
It is really that easy. Just do the work.
Where at in wa state?
as many have pointed out, the "product over sum" formula only works if there are 2 resistors. If there are 3 resistors, say A, B, C, then the formula is:
R = (ABC) / (AB + AC + BC).
Other than that, great videos.
Hmm! what about 4 llL resistors?
@@tomctutor when it comes to 4 resistors more, you're better off using 1/Req = 1/R1 + 1/R2 + 1/R3 + 1/R4 + ...
@@nhitc6832 You're better off just using this in every scenario. Avoids confusion by having to memorise various different formulas for unique arrangements/instances; whereas 1/Req = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn works for every scenario.
thanks so much for pointing this out and giving the formula!! this saved me so much time :)
@@tomctutorfor any arbitrary number of resistors it would be
(product of all resistances) / {sum of(product of all resistances except this resistor) for all resistors)}
Which makes a lot more sense in actual math notation with the big sigma and big pi and index notation.
But for 4 resistors, that works out to:
R = ABCD / (BCD + ACD + ABD + ABC)
There is no magic in this formula, if you just rearrange the more basic equation with algebra you can derive this one. But you have to multiply so many numbers now that you're better off just using the other equation.
The rule of "product over sum " works only when one have 2 resistors in parallel. Regardless this slip your work teaching is awesome. I am a Telecomminications and Electronics Engeneer and watching your videos I remember my class of Electrotecnia long time ago. Congratulations. Greetings from Mexico.
Really ... is that what you think?
Explain why this works "Engineer" ...
1/Rt = 1/R1 + 1/R2 + 1/R3 = ( R1 · R2 · R3) / ( R1·R2 + R1·R3 + R2·R3)
.... ( 2Ω·3Ω·4Ω ) / ( 2Ω·3Ω + 2Ω·4Ω + 3Ω·4Ω) = (24Ω / 26Ω) = 0.923Ω
Just simple Algebra 😉
"@nhitc6832
2 months ago
as many have pointed out, the "product over sum" formula only works if there are 2 resistors. If there are 3 resistors, say A, B, C, then the formula is:
R = (ABC) / (AB + AC + BC).
Other than that, great videos."
Yo I stole this from a dude further down the line.
The formula in the video don't work for this exacting application. Just use the formula written in blue at the capacitance part. I actually got it to work after crying and coming back a week later.
Passed my first two tests. One more to go. This video came at a great time. Needed the refresher!!
Thank you for taking back to school. Learned these formulas and more in my electrical engineering degree
Excellent formula review boss! I love the way you teach. Much appreciation
What you call “Joule’s Law” should be called Watt’s Law, because power P is in units of watts, not joules. As you probably know, 1 joule of energy per second is 1 watt.
According to statistics, 98% of the world's population has an IQ that is considered to be 'below normal'. As a tutor in a past life (and in my opinion), I found that the best approach to teaching anyone the 'percent circle' is to simply place a multiplication symbol immediately between the two lower variables (in place of that vertical line you have in yours). This eliminates any lingering confusion that may remain in those 'select few'.
Once they learn that Percent Circle, they can apply it to anything, including the financial market, distance/velocity, etc.
Let me guess, you think you’re in that 2% 😂. It’s hilarious when people post a statistic thinking they’re exempt.
@@joaquinsanchez1251bro thinks he's a main character🤣
This doesn't even make sense since "normal" implies most prevalent and common. So no 98% would be normal by your statistics, which I already know is complete wrong without even looking it up.
Thank you for making this video. I'm taking my Journeyman exam next week and this helps alot.
So, how did the exam go?
Can you do service calculations? I need a good step by step and you do the best explanations on every video
These are theory. Real world needs to factor in many variables if you want to be spot on. Many times, good enough is not good enough.
helpful; but the r1*r2 / r1+r2 (or product/sum) just works for "2" resistance in parallel and not more. the actual and correct resistance value when 2,3,4 are in parallel is 0.9 ohms, where the (wrong) product/sum offers 2.6 ohms
I was like why am I doing exactly what he said but not getting 0.9 ohms
You can only use the product over sum method on 2 resistors at a time. When working with 3 like the case in the video (2ohm, 3ohm and 4ohm) you would do the 2 and the 3 ohm resistors first which would come out to 1.2ohms. 1.2ohms is now the equivalent resistor to the first two and would be the new value for both of those. So now you take that and do the whole method again with the third resistor (3ohms). 1.2 x 3 / 1.2 + 3, which comes out to roughly .857ohms. You can keep using this method with more resistors just remember the resistance total will keep getting smaller the more you add.
Thank you
The third resistor would be 4ohms not 3ohms, 3ohms was the second resistor. So it would be 1.2 x 4 / 1.2 + 4, which would come to 0.923. I'm still learning, Your comment helped me to understand the math behind what Dustin was illustrating, just saw that you put 3ohms as the third resistor instead of 4ohms 🤙
This is a good video. You did a good job. But, the Product/Sum method only works if there are only 2 resistors in parallel.
Wrong!
@@weavercattlecompanyhow is he wrong id like to learn more about this, im a student for Harley and I’m currently in an electrical class so I’m trying to absorb everything I can
@roscoe4092 he did it for 3 in the example
One that I love that I'm actually somewhat surprised wasn't covered is the formula to calculate the distance multiplier for offsets. I've had to do some obscure degree offsets to get a pipe run to work and look good with how obstructions were positioned. If anyone doesn't know it, it's 1/(sine of the degree). Every apprentice I've had, I give that formula to and they all have clean work when running pipe because of that
Really appreciate the videos!! Takes me back to apprenticeship school
Aluminum is generally more conductive than most types of covers. Aluminum has a high electrical conductivity, which means it can easily transmit electric current. This property makes it a popular choice for various applications that require good electrical conductivity, such as power transmission lines, electrical wiring, and heat sinks.
Yeah I came up with the same thing. Product over sum doesn't work unless we are missing something.
Another great video Dustin. I did many many calculations to limit my voltage drop on branch circuits to 3% as suggested in NEC. Since I’m really looking conductor size needed to achieve the goal, I exchanged Vd and cm but did 3% of the volts of the system. One calculation and I knew how many cm was needed to be at or below 3%. Go to the table and pick the conductor size. As. Cm=2xKIL/3% of V.
I also did 3 phase slightly different. For 3ph 3w balanced load same as single phase and multiply by .66. For 3ph 4W balanced load multiply by .5. In the thousands of designed circuits over 35 years, I never had a low voltage or breaker tripping issue.
That's definitely something that I should remember.
I can't imagine how helpful a resource like this can be for those learning the trade today ...
it's just very cool!
I went through my apprenticeship 40 yrs ago ... a resource like this would've been priceless.
Keep up the good work.
I am currently studying these formulas to apply for the electricians apprenticeship. These videos are very helpful, along with some others here on TH-cam. I want to make sure I can pass the aptitude test.
I'm trying to learn some electrical stuff to get a head start before I start taking classes or doing an apprenticeship, this is very helpful and informative. I appreciate the time and effort you took to make these videos! I subbed👍
I started in Australia as an apprentice electrical fitter in the mid 1960s.
My experience was different from this . Much of our basic formulas were the same, but, we didn't get into voltage drop. We just had a different approach, and much of our work was fairly standardised. We also did not refer to Joules' laws, they were power formula. Earthing in an installation was important, and knowing the national standard wiring rules. I have to say I was gratified that the math was not complicated.
At 9:48 check spelling of resistance.
🤣🤣🤣
You are teaching the right thing
2:46 P is for power which is a unit of work/energy per unit time. It is the rate a which work is being done or energy is being used or absorbed. Joules per second or watts
Amperage is the unit of current which is in fact a rate but it is a rate analogous to the flow of something like a gas pump. Imagine charge to be analogous to fuel and the rate which fuel is being pumped into your car then it can be kind of a good way to get a better understanding of amperage as the flow of charge.
This rate of charge does not already account for the rate which energy is being transferred, not independently anyways. The product of current and voltage accounts for the energy per unit time because current is Coulombs per second and voltage is energy per unit charge or joules per second.
The multiplication of current and voltage allows us to cancel out the units of charge so the product is joules per second.
This doesn’t take anything away from his fantastic explanation of the formulas but I want to caution viewers who are in a classroom setting, units and dimensional analysis are very important. As I tell my students, be aware of the details of the units which are being cancelled out and which are left remaining, they should provide a secondary confirmation of the correct units of your equation.
Perfect video to hear while making money.
Make money make money thanks again for the video !!!
Could you do a video in the future on symbols you’d find in the field like for plugs switches contrators on schematics etc.? I can’t find any good videos on that
Not even an electrician, but a computer scientist/Software engineer and I just like this kind of stuff. If you can understand all this electrician stuff you can probably be a pretty dang good programmer.
Just refreshed my memory back in Technical college over 26 years ago. Is a good learning. video.
A couple interesting observations comparing my training (electronics technician) to this (electrician). Line voltage drop isn't nearly as important in small circuits or inside wiring panels, so that formula was more of "Here it is, but you don't need to memorize it. Just know where to find it if you need it again, but you probably won't." And I don't think we even touched on horsepower. Though I did need to learn how to count and do math in binary. ;-) Also, it may have just been the teacher, but I learned my ohm's law and power formula in triangles, not circles.
Side note... Because of the power formula, this video should have been released yesterday... on Pi day. ;-) (Yeah, power formula is PIE, and 3.14 is Pi. But they are pronounced the same.)
Hello Christopher,
You bring up interesting points for low voltage in considerably smaller circuits. They are incredibly more complex and efficient. As an electrician though, especially in servicing, these formulas are important. We also don’t really deal with digital circuits and needing to know binary code. So, this video is very informative for the average apprentice electrician going for their journeyman license.
I've watched your video's in the past, simple and to the point. Looking forward going down the TH-cam rabbit hole viewing the others!!! Thanks!
Good teaching here as a field tech in the phone company I used a vom every day when troubleshooting t-carrier lines.
What about Pythagorean theorem..? I use that on the field alot more than any of those that you mentioned... Helps alot with rolling offsets..
Interesting, informative and worthwhile video. A very worthwhile, simplified approach.
Liked & commented for YT analytics.
Great detail at 7:50...
Appreciate the obvious time you spent putting this specific video together because the info and specifications are so important. It's very clear you spent a lot of time writing out the slides and notes
Glad you didn't just "wing it".
Carry on!
Cool video, but I'd like to point out I've been an electrician in an industrial setting for 25 years and only know ohms law and how to calculate wattages- and very rarely ever even need to use those.
Product over sums only works for 2 resistors,
Ex. 1 || 2 || 3 = 6/11 = 0.545
That's not 1*2*3/(1+2+3) = 6/6 = 1.
Yes indeed, VERY well done indeed. I need to watch, intend to, at least several more times 2 fully digest it. Ths again bro!
Thank you Dustin, it is really a must to know these formulas. ❤
Ohm's law in a single phase AC circuit is more of a rough rule of thumb result. In a 3 phase AC circuit it can become a ballpark result. Ohm's law is accurate in DC circuits while in AC circuits it's more of a rough calculation as several instructors over the years put it.
Nothing ballpark, rule of thumb, or rough calculation about it, it's accurate.
It's reactive.
Not trying to be a stickler, maybe I’m wrong, but I believe the “product over sum” method only works if there are only “two” resistances. 3 or more have to be the inverse equation. At least that’s how I learned Jt and I can’t get the math from your equation to equal 0.9 . I get 2.6 . Let me know your thoughts. Great video either way though 👌🏼
it indeed only works if there are 2 resistors. When you get 3 or more resistors, the formula will change. Hopefully, he'll make a correction.
This is the first time I see E for Voltage instead of V. E is usually an energy in physics. Or Error in control theory.
QUICK-WITTED Electrician U
👍
From Nick Ayivor from London England UK 🇬🇧 ⏰️ 21:34pm
I wish when i took up electronics in high school that ohms law was explained to me this simple,instead of making it seem like you had to have a degree in physics to understand it.
I went to engineering and not electrician school but have never seen voltage referred to as E before. Very interesting
Well I am an electrical engineering student and E means EMF or ELECTROMOTIVE FORCE wich is the same as voltage.
@@mikecafe5364 well I'm an actual engineer and EMF is NOT the same as voltage. That's a fail in exams. EMF, is denoted as E or greek epsilon and is an energy transfer. Voltage V is neither a force nor an energy, as it would be measured in newtons otherwise. (Voltage is a potential potential difference). The video poster clearly doesn't know ohm's law properly or really know what he's talking about, because voltage is never denoted as E. V is the correct denoted term for voltage (potential difference) for reasons that ohms law is a simplified formula derived from a complex mathematical equation, derived from energies, forces etc. The video poster is also an idiot by stating electricians need to know deep maths - complete nonsense. He fails to state that ohms law does not apply to DC or diode based electrical circuits. I could go on - but he's a TH-camr trying to earn money rather than working in the real world.
@@StuartCarolyn I agree.
But I have seen E used in examples using batteries and internal resistance, where E is the battery terminal voltage when there is no load current being drawn.
V = E-IR
Which says: the battery terminal voltage when the battery is in circuit and a current being drawn, V, is equal to the unloaded battery voltage minus the voltage drop caused by the internal resistance of the battery which is IR, the current times that resistance
Joules Law is a law stating that the heat produced by an electric current i flowing through a resistance R for a time t is proportional to i 2Rt.
joules law is about the heat produced in a electrical flow. P in the p=is is watts which is the amount of power consumption or worked done.
So you confirm that what i learned 20 years ago, that : P=Ei and E=Ri . Are both Holm's Law.
I believe that’s the point right he did in that way faking a mistake to see if we the expetators notice that mistake🧐
Thank you for sharing your knowledge 🙏🙏💐🌷🌺🌷🌺💐🌷💐🌺
You could also add in the Voltage Drop formula for sizing wire using a voltage drop percentage too. I tend to use that variation of the formula more often. Oh and rEsistance. E!!! LOL
Please teach at the IEC in Austin. You would be a great instructor!
CONTENT IDEA VIDEO:
Can you put different size guage wires into conduit. For example, for a washer and dryer inside a 3/4 inch emt could you run #10 wire and #12 wire inside same 3/4 inch emt, two separate circuits, same conduit.
charts only show how many wires you can fit inside conduit, but the same awg, not mixed. Interesting question.
Thanks in Advance Dustin
Miami, FL
Of course you can, why would it matter?
You may have to think about derating the maximum current rating for the cable to accommodate the fact it is in an enclosed space and cooling is not as effective as if it were clipped to a wall.
I self taught harmonics ,tested well pump soft starts for harmonics 200 horse well pump motors had pipe that looked like termites had eaten it lost many pump 250 feet down the wells
this has been my week in class for my apprenticeship
Make a video about Motor Drives, and how they work.
Variable Frequency Drives.
Best way to figure horse power is 4 amps running and 6 amps start up with out a load put into the 746 watts
This video took me back to my EET years. Thanks for the review.
For parallel resistors it’s usually easier on a calculator to set it up like this:
RTotal^-1 = R1^-1 +R2^-1+ R3^-1
There is usually a little ^-1 key on the calculator that’s very useful
keep it up
V I R Ohms Law and P I V joules Law is how we do it the V is voltage the P is simply power.
Voltage drop, wow! I often wondered how you in the US do it (missionary😂) We take a value from our Regulations known as mV/A/m ( milli Volt / Amp / per metre) then the length of wiring in metres finally the I b design current. Divide by 1000 = Voltage Drop. Maximum V D for lighting is 3% other 5%. Resistance exactly the same formula we use. While we may talk a different language it still the same science that will stay consistent as our respected Codes and Regs change. Presently in the UK we are on our 18th Edition Admedment 2 2022. Edition are every 10 years with the allowance of 3 Admedments in between. To finish I have been enjoying your presentations.
Great video. Clear - Consistent - Common sense
Great intro to these formulas! One suggestions...add some units so it's easier to see what the numbers stand for.
I appreciate what you do for our trade. All of this basic knowledge is being lost with the younger apprentices that are coming into the trade. They rely on apps for everything......if they even know to do any calculations.
For resistors in parallel, the "product/ sum rule" can only be applicable to only two sets of resistors connected together in parallel.
Thank you, Sir!
Hi Dustin, thanks for the interesting lecture! Q. In the last part about putting resistors in parallel, you’re talking about the current is moving in a different way, BUT I’ve learned there is actually no movement of electrons in a wire and that the energy is actually moving outside the wire(s)/circuit. If my assumption is right, how and what is current (Amp’s) actually really, if the electrons barely move (especially in a alternating current)?
Q2. If my assumption is right, how does a resistor in a circuit even work if the electrons barely move and it’s the energy that’s moving outside the wire(s). I mean, if the energy surpass (outside) the wire, how can a resistor even function?
Those pie charts. My apprenticeship teacher never mentioned just covering the letter. Now it makes sense why it's drawn like that.
2*3*4/(2+3+4)=2.7 which is not .9. The product/sum only works for two resistors. For more than two the formula is more complicated.
Yes yes yes thank you!!
Great to have them in one spot.
🎉
Thanks for this man. Studied for my masters and failed. It had been a long time since school. Had I refreshed here I probably would have been alright but froze.
Great video... and I like the presentation.. the digital board, the readability of the material on the board, etc. Also, you explain well, and you look at the camera in a professional manner, etc... overall a great job... easy to follow the material.
feels like ive done tons of 30 degree offsets but rarely do I get to take time to practice other variations. Can we get a video with like 45 or 60 degree bends ??
Dustin thank you very much for your great work.
Well said, teach! Can you put the dimensions of the numbers in the equations next time? Thanks!
The US is still stuck with things circular mils ending up with cm meaning circular mils . In the rest of the world cm means centi meter . Cable area in metric is mm^2 .Change over to metric , it is easier
Absolutely is a lot easier.
One year of electrical engineering classes was enough for a major in Mechanical.
In your example about Power, this is adaptable to resistive load but if the load is inductive you need to assume a Power Factor or see a table with assume power factor for example an induction motor and multiply it to get the real power.
Real power is not affected by power factor; apparent power is.
Very nice and very useful video. Many thanks.