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here are five formulas every  electrician should know [Music]   number one is Ohm's Law most electricians know  this like they know nothing else with math it's   Ohm's law and Ohm's law is specifically  the relationship between voltage amperage   and resistance only does not put into any uh  Power efficiency anything like that once you   start getting into Power you're talking about  the transferring of power from one thing to   another thing and that is a process of joules  so uh the the next one we will talk about is   once we introduce power into this but Ohm's log  and voltage amperage resistance now if you want   to solve for one of these you just cover the one  you're trying to solve for and it tells you what   equation to to use so with Ohm's loss there's  three different ways that you can structure it   so you could do the first one E equals I times  R if you're trying to solve for voltage E equals   I times R just remember they're right next to  each other I times R if you wanted to solve for   amperage you could sit and try to like divide out  the r and move everything over or you just look   at this and be like I'm trying to solve for I so  cover I it's e over R so super simple and lastly   as you would expect R equals e over I so that  tool that little chart is really good because   then otherwise you'd have to like do all of this  math and try to multiply by things and like move   things around but if we have an amperage that's  20 amps times a 6 Ohm resistor it's going to be   120 volts if you're trying to figure out how many  amps is it I got 120 volt circuit and I know this   resistor is six ohms you just divide and you get  20 amps but if you don't know what the resistance   is of a resistor but you know that this thing  is drawing 20 amps and it's 120 volt circuit   you can just divide that and figure out six ohms  so this is a really really helpful thing to use   next is going to be Jules law so a lot of people  don't call this Jewel's law but that's what it is   Jewels is the amount of energy that's produced by  something or that's transferred from one system   to another system so when we're talking about  electrical energy going through a light bulb we're   talking about the amount of energy that is coming  from the electrical energy in the electrical   circuit and then is being produced or transferred  into another kind of energy like light and heat   in the case of like an incandescent bulb so Jules  law is usually a function of time so usually in a   formula they're going to have time as a function  but amperage includes time so amperage is a rate   so it's current that's flowing per second so  it already factors that in so we can call this   joules law and use it as such so very similar  kind of thing instead of uh E equals I times R   now we have P equals I times E I just think of Pi  Yami pi all right p-i-e Pi that's how I think of   it so the one thing if we're trying to solve for  how many uh watts of like light bulbs that we have   um What's the total wattage of a circuit once we  turn all these bulbs on well we know it's 120 volt   circuit and uh we know that it's drawing 20 amps  so how much total power is that it's going to be   2400 so we could take all of these values and  the same thing all of them are true if we're   trying to figure out how much current draws in  a circuit well we know we have 2400 watts in a   circuit and say we're trying to figure out like  a 20 amp breaker you know 20 amp circuit well   if it's 2 uh 2400 Watts that we have to deal  with divided by 120 then that's going to be   20 total amps for that circuit you could also do  the inverse so say that we're trying to figure out   um figure things out for what the total wattage  of a circuit is if we have a 20 amp breaker 120   amps then we have 2400 watts to play with in a 20  amp circuit that's another like helpful thing out   in the field so so anyways lastly if we're trying  to solve for voltage which you're never going to   do you're going to take the power whatever the  the wattage is divide that by 20 amps if you're   testing 20 amps and it should be 120 volts so  they all organize nice and neatly again rather   than doing the math to try to figure out and like  divide by E to isolate I and move e over here   you'll have to remember all that just remember the  pie chart next up is voltage drop so voltage drop   is something we're very frequently going to come  across in the field you can actually use Ohm's   law to do a voltage drop calculation without all  the extra variables it's just slightly off it's   not going to be as accurate but kind of like  gets you relatively close but a better way to   do voltage drop is to actually use voltage drop  formulas you know I have two here one of them is   VD voltage drop the other one is vd3 for three  phase so anytime you do three phase you have to   introduce 1.732 into the formula it's also the  square root of three so the square root of 3   equals 1.732 so you're going to notice the  formulas then are very similar one of them   is two times a bunch of craziness the other one's  1.732 times a bunch of craziness which that means   this is a smaller number you're multiplying  by so that typically in a three-phase circuit   there's going to be slightly less voltage drop  voltage drop is more expressed in a single phase   circuit all right so voltage drop Let's do an  example so the two I I think of it there's two   conductors right you have there and back we have  length which is L so if you have a hundred feet   you're not going to measure 100 feet one way and  100 feet back the other way that's what that 2   is for so the two two conductors at 100 feet  that's how I remember the two and the L the K   is a constant that we use depending on the type  of material that we're using so if we're using a   copper conductor the conductivity or resistivity  of a certain conductor is going to be different   so aluminum is a little bit less conductive than  copper is so copper you're not going to experience   as much of a voltage drop as you are for the same  size conductor in aluminum just because they don't   conduct as well there's just more resistance with  the material so K the coefficient for k for copper   is 12.9 so every single one of these that you  ever do you're going to use 12.9 for copper if   it's going to be aluminum K is going to be 21.2 so  just memorize those two numbers and never forget   them the rest of your life and you'll be fine  so anytime you're doing aluminum stuff use that   anytime I'm using copper use that so that covers  the two the k i is your amperage how like if we   had a motor or something that's a 20 amp motor out  in the field you're gonna plug in whatever that   is going to be so that takes care of everything  on the top the bottom is not centimeters it is   circular Mills so every conductor has a certain  diameter which means it also has an area and the   area of that conductor or the cross-sectional area  if you slice the thing in half the cross-sectional   area of it is measured in Mills or circular  Mills so a circular conductor is going to be   circular Mill so every conductor has a certain  size circular mils number 12 I know for 20 amps   is going to be 6530 just remember that remember  that number because most of the time you're going   to be dealing with number 12 conductors and  trying to see what the voltage drop is on that   sized conductor so the cross-sectional  area of number 12 is going to be 6530.   um now one thing to keep in mind if you're ever  dealing with like 250 KC mil or 250 MCM or like   500 600 you're talking big conductors the 600  actually means 600 000 circular Mills so those   are really easy once you get past 4i you start  250 300 350 all of that that's actually talking   about the circular Mills so uh otherwise chapter  nine table eight NEC um it's going to have all   of these values in there for each specific one  so beyond all of that like that's what we're   plugging into the formula so an example is say  we've got a single phase voltage drop that we're   trying to calculate we've got a hundred foot  distance there's two conductors they're copper   so it's 12.9 it's a 20 amp load divided by number  12 conductors which is 6530 we get eight volts so   in this situation over a hundred feet if we have  a 20 amp motor on copper conductors it's number 12   we're going to experience an 8 volt drop when we  hit that thing and actually start it up it's going   to drop the whole voltage of the circuit now for  three phase again we see this 1.732 thing square   root of three right because it's three phase so  everything three phase we've got to add this 1.732   thing so we have square root of 3 times 21.2 this  time instead of copper we're going to say we're   using aluminum just to see what that does still  a 20 amp load still 100 feet away and we're still   using number 12 it's just number 12 aluminum  this time do all of the math and we figure   it's an 11 volt drop so other than we changed and  put 1.732 instead of two this voltage drop if it   were both single phase just changing the aluminum  conductors would have still made this thing drop   a lot more volts it's actually less of a voltage  drop because we're in a three-phase circuit so it   would probably be higher maybe 12 volts if we were  just doing single phase but even when you go up in   aluminum for that higher coefficient you go down  because it's three phase just slightly next up   you're going to see this on every damn test you're  ever going to take electrical Theory this stuff   sucks especially once you start getting into RLC  circuits where you've got resistance inductance   and capacitance in series and in parallel and  sometimes when you're doing capacitance these   formulas for capacitance are actually completely  inversed so capacitance in a parallel circuit is   going to look like this capacitance in a series  circuit is going to look like this but not to   digress series and parallel resistance if you're  trying to figure out the total resistance of a   series circuit you just add all the resistances  together it's the sum of all of them it's super   easy to do you're in series you're just adding one  after the other so the total resistance or RT is   equal to the first resistance second resistance  third resistance and as many resistances as you   got you just keep adding all together where it  gets kind of wonky is when we get to parallel   resistance parallel resistance you have to take  the inverse of the sum of inverses there's a   different method product over sum we'll get to  that here in the next slide but it's a weird thing   right so I just usually do like one divide if it's  a 3 Ohm resistor or something I do one divided by   three plus one divided by two plus one to whatever  and you get that sum and then you take one divided   by whatever that sum is and that's what you get so  you'll see the dramatic inverse effect on putting   resistors in a series circuit than you will in a  parallel circuit all right series and parallel so   here we have a series circuit we've got R1 R2 and  R3 there are three different resistors our one is   two ohms R2 is 3 ohms R3 is 4 Ohms so we're doing  Series right they're all in series so we're going   to do the sum of all resistances method we're  going to take the total resistance equals R1   R2 and R3 all added together so we got two plus  three plus four two three four that's nine ohms   of resistance there's a lot of resistance  because we have each one of them in series   so it's adding more and more resistance to the  Circuit when we try to apply pressure or voltage   the next method is the reciprocal over the sum of  reciprocals so we were going to take still R1 and   R2 and R3 but in this circuit we're in parallel  rather than in series so we've got resistor one   two ohms still three ohms still and four ohms  still we're just putting them in parallel in   the circuit instead and what we get is one over  one half right because the two plus one third   because of the three plus one fourth because  of the four we add all that up and then we take   the reciprocal of it and it's 0.9 ohms rather  than nine ohms so we actually get the inverse   number um because it's going through current is  traveling through the circuit differently it's not   just one resistance being added to each current's  going to go everywhere in this circuit but there's   nothing impeding it um in a row in order so you  can sit and mess around with that and have fun   if you want to but I find an easier method of  going about it is just to use the product over   some method so you can do this craziness if you  want to but it's also possible to just do each   one of the resistances times each other and then  put over each one of the resistances added to each   other so the product 2 times 3 times 4 the product  over the sum to plus 3 plus 4 and you get the same   exact number so super helpful you're going to  run into lots of that once you start getting into   taking your electrical Theory exams so have fun  last one we're going to talk about is horsepower   and we're specifically talking about the output of  a motor what is it going to put out based off the   conditions of the circuit and the conditions of  the motor talking about just converting a watt to   a horsepower because one horsepower equals roughly  746 Watts that's another number just remember that   if you're ever trying to think of like hey one  horsepower motor what does that produce 746 Watts   roughly but there's a lot more things to factor  with a motor some Motors are a lot more efficient   than other Motors a lot of them have a rating  on them and it'll say eff and it'll give you   efficiency of like 90 or 0.9 so it's not a hundred  percent efficient it's only 90 efficient so that's   going to change the Dynamics of the output of  that motor another thing is you could have a   power factor problem so you might be in a building  that's got a lot of inductive loads like maybe   tons of motor loads well the overall power factor  of that circuit is going to change the nature of   the output of that motor so um we're going to be  putting in voltage amperage the efficiency of the   motor the actual power factor for the circuit  and we're going to divide by 746 right because   one horsepower equals 746 Watts so to figure out  what the true horsepower of this motor is let's   look at an example so we're going to be single  phase three phase again we got the square root   of three thing right three phase so let's look at  a single horsepower motor we're going to take the   voltage times say it's a 240 volt circuit we've  got a 20 amp motor the efficiency on the motor   actually says it's only an 80 percent efficient  motor so not very great and the power factor is   kind of crazy at 0.7 so Unity power factor would  be one if it's a hundred percent power factor that   means your voltage and your your amperage when  you apply voltage is immediate amperage there's   no like delay or lag reactance that's happening  um then you're gonna have one but anything that's   worse than that is going to be like below  one so 0.7 is a pretty terrible power factor   then we're going to divide by 746 so we do all  the math single phase horsepower we get 3.6   horsepower now that's not the actual horsepower  rating sizing your conductors and doing amperage   and all of that stuff off of that you're going  to use 430 in the National electrical code uh   247 248 249 and 250 and that's going to be you  know single phase three-phase DC so now when we   look at a three-phase situation we have to add  the 1.732 so that's that's going to help our   number a little bit because we're actually adding  more it's not like two times everything but it is   1.732 times everything so square root of 3 is  1.732 times 240 volts same circuit uh 20 amps   0.9 on the efficiency this time so  this is a much more efficient motor   but power factor is still 0.85 which is still  better than it was and then we're dividing by   7.46 and we get 8.5 horsepower out of that  circuit now this is a three-phase circuit   right we've got three conductors going into it  it's on a three-phase breaker and we're hooking   it up so there's going to be just naturally more  power transfer between three-phase circuit and a   three-phase motor whereas a single phase we've  got two conductors that are probably pulsing a   lot differently but typically horsepower output  for a three-phase motor is going to be slightly   greater than it is for a single phase motor  if we were to keep all the numbers the same   the equations would be the same but you would  have 1.732 times the output of that motor   so it's actually going to be more output on a  three-phase motor than it is on a single phase   motor so those are all of your formulas that's  actually not all of them if you really wanted to   get into capacitance or you wanted to get into  like inductive reactants capacitive reactants   you want to get into RLC circuits there's more  stuff to learn there's quite actually there's   quite a few more like formulas that you might  want to know if you want to get into like trig   or calculus weird stuff like that but we don't  because this is YouTube and we're all dummies   kidding I love you YouTube all right I'll see you  crazy people in the next one thanks for watching