5 Formulas Electricians Should Have Memorized!

Summary

This video covers five essential formulas every electrician should know, starting with Ohm's Law, which explains the relationship between voltage, current, and resistance. It then explores Joule's Law, focusing on power transfer, and provides examples for calculating wattage and current. Voltage drop formulas for both single-phase and three-phase circuits are discussed, followed by methods for calculating resistance in series and parallel circuits. Lastly, the video covers horsepower calculations for motors, highlighting how efficiency and power factor impact motor output in both single-phase and three-phase systems.

Takeaways

• ⚡ Ohm's Law explains the relationship between voltage, amperage, and resistance. It's often written as E = I * R.
• 🔌 To solve for a specific variable in Ohm's Law, cover the one you're solving for and use the appropriate equation, like E = I * R for voltage.
• 💡 Joule's Law relates to the amount of energy transferred between systems, commonly expressed as P = I * E.
• 🔋 Voltage drop can be calculated using formulas based on Ohm's Law, with additional factors like conductor length and material affecting accuracy.
• 🔗 In three-phase circuits, voltage drop calculations introduce the square root of 3 (1.732) for more accuracy.
• 🧮 For resistances in series, simply add them together. In parallel circuits, use the sum of reciprocals for total resistance.
• 📏 The product-over-sum method is an easier alternative to calculate parallel resistance compared to the sum of reciprocals method.
• ⚙️ One horsepower equals 746 Watts. However, motor efficiency and power factor influence the actual horsepower output.
• 🏎️ Three-phase motors generally produce more horsepower than single-phase motors due to more efficient power transfer.
• 📚 Advanced topics like capacitance, inductive reactance, and RLC circuits require additional study but are important for more complex electrical systems.

Q & A

What is Ohm's Law, and how does it relate voltage, amperage, and resistance?

Ohm's Law describes the relationship between voltage (E), amperage (I), and resistance (R). It states that E = I × R. If you want to solve for one variable, you can rearrange the equation accordingly, such as I = E / R or R = E / I.

How can Ohm’s Law be used to calculate the amperage in a circuit?

To calculate amperage using Ohm’s Law, divide the voltage by the resistance. For example, if you have 120 volts and a 6-ohm resistor, the amperage would be 120 / 6 = 20 amps.

What is Joule's Law, and how does it differ from Ohm’s Law?

Joule's Law relates to the amount of energy transferred, particularly from electrical to another form, like light or heat. It is represented by P = I × E, where P is power (watts), I is current, and E is voltage. Unlike Ohm’s Law, which focuses on voltage, amperage, and resistance, Joule’s Law introduces power.

How can you calculate the total wattage of a circuit using Joule's Law?

Using Joule's Law (P = I × E), if you know the voltage and the amperage, you can calculate total power. For instance, in a 120-volt circuit drawing 20 amps, the total power would be 120 × 20 = 2400 watts.

What is voltage drop, and how is it calculated in single-phase circuits?

Voltage drop is the reduction in voltage in a circuit between the power source and the load. In single-phase circuits, it can be calculated using the formula VD = 2 × K × I × L / CM, where K is a material constant (12.9 for copper and 21.2 for aluminum), I is the current, L is the length of the conductors, and CM is the circular mils of the conductor.

Why is 1.732 used in three-phase voltage drop calculations?

The value 1.732 is the square root of 3 and is used in three-phase voltage drop calculations to account for the three-phase system’s different power transfer characteristics compared to single-phase systems.

What is the difference in voltage drop between copper and aluminum conductors?

Copper has a lower resistance than aluminum, leading to less voltage drop. The constant for copper (K = 12.9) is smaller than for aluminum (K = 21.2), meaning aluminum conductors experience more voltage drop for the same current and distance.

How do you calculate resistance in a series circuit?

In a series circuit, the total resistance is the sum of all individual resistances. For example, if you have three resistors with values of 2 ohms, 3 ohms, and 4 ohms, the total resistance would be 2 + 3 + 4 = 9 ohms.

How do you calculate resistance in a parallel circuit?

In a parallel circuit, resistance is calculated using the reciprocal method: 1 / RT = 1 / R1 + 1 / R2 + 1 / R3. For example, if you have resistors of 2 ohms, 3 ohms, and 4 ohms, you would calculate 1 / RT = 1 / 2 + 1 / 3 + 1 / 4. The total resistance would be the reciprocal of that sum.

What is the relationship between watts and horsepower?

One horsepower is approximately equal to 746 watts. This means that to convert watts to horsepower, you divide the total watts by 746. For example, a motor producing 2400 watts would have a horsepower of 2400 / 746 ≈ 3.2 horsepower.

Outlines

00:00

⚡ Understanding Ohm's Law

The first paragraph introduces Ohm's Law, explaining the relationship between voltage, amperage, and resistance. It provides formulas for calculating each based on known values: E = I * R (voltage), I = E / R (amperage), and R = E / I (resistance). These formulas are essential for electricians to solve for missing electrical variables efficiently.

05:04

💡 Joule's Law and Power Calculation

This section dives into Joule's Law, which deals with the transfer of energy, such as when electrical energy converts to light or heat. The formula P = I * E (power) is introduced, which helps electricians calculate the total wattage in a circuit. The example given demonstrates using a 120-volt, 20-amp circuit to determine total power output. The explanation concludes with methods for calculating voltage, amperage, and wattage in practical settings.

10:08

🔋 Voltage Drop Calculations

This paragraph focuses on voltage drop and introduces the formulas to calculate it. It explains the use of constants for different conductor materials, such as copper (K = 12.9) and aluminum (K = 21.2). An example of a voltage drop calculation over a 100-foot distance is provided, illustrating how conductor type and amperage affect voltage drop in both single-phase and three-phase circuits.

15:13

🔀 Series and Parallel Circuits

This section explains the difference between series and parallel circuits, particularly focusing on how resistance is calculated in each. In series circuits, resistances are simply added together, while in parallel circuits, the inverse of the sum of inverses method is used. The paragraph also mentions an alternative method, the 'product over sum' approach, which simplifies the calculation of total resistance in parallel circuits.

🏇 Horsepower and Motor Efficiency

The final paragraph covers the calculation of motor horsepower, specifically comparing single-phase and three-phase motors. It explains the importance of factors like efficiency and power factor in determining motor output. The key formula used is horsepower = (voltage * amperage * efficiency * power factor) / 746. Examples are provided to show the difference between calculating for single-phase and three-phase motors.

Keywords

💡Ohm's Law

Ohm's Law defines the relationship between voltage (E), current (I), and resistance (R) in an electrical circuit. It is fundamental to electricians as it helps determine one of these quantities when the other two are known. In the video, it's explained as a tool to solve for voltage, amperage, or resistance by covering the one you're solving for in the formula E = I * R.

💡Voltage

Voltage, often denoted as E, refers to the electrical potential difference between two points in a circuit. It is essential for determining how much potential energy exists to move electrons through a conductor. The video discusses how to calculate voltage using Ohm's Law and its relationship to power in circuits.

💡Amperage

Amperage, denoted as I, refers to the flow of electric current in a circuit, measured in amperes (amps). The video uses examples of circuits with 20 amps to explain how to apply Ohm's Law and calculate power and resistance based on known amperage.

💡Resistance

Resistance, denoted as R, measures how much a material resists the flow of electrical current. It is a critical factor in determining how voltage and current interact in a circuit. The video frequently references resistance when discussing Ohm's Law and examples with 6 Ohm resistors to show how it affects voltage and current.

💡Joule's Law

Joule's Law relates to the amount of energy produced or transferred by electrical current. It is often used to calculate the power output in electrical devices, such as light bulbs, where electrical energy is converted to light and heat. The video connects this concept to power calculations in circuits with real-world examples, such as light bulbs.

💡Power

Power, denoted as P, refers to the rate at which electrical energy is transferred or used in a circuit, usually measured in watts. The video explains how power is calculated using the formula P = I * E (current times voltage) and provides examples like calculating the total wattage of a circuit.

💡Voltage Drop

Voltage drop refers to the reduction in voltage as electrical current travels through a conductor. This is important in maintaining efficiency in electrical systems. The video explains how to calculate voltage drop using Ohm's Law and provides examples for single-phase and three-phase circuits.

💡Three-phase Circuit

A three-phase circuit is a type of electrical distribution that uses three alternating currents, often used in industrial and high-power applications. The video describes how voltage drop calculations differ in three-phase circuits and explains the importance of the square root of three (1.732) in these calculations.

💡Series Circuit

A series circuit is a circuit in which components are connected end-to-end, so current flows through each component sequentially. The video covers how resistance is additive in a series circuit, meaning the total resistance is the sum of the resistances of individual components.

💡Parallel Circuit

A parallel circuit is one in which components are connected across common points, allowing current to flow through multiple paths. The video highlights how resistance is calculated differently in parallel circuits, using the reciprocal of the sum of inverses for resistances.

Highlights

Ohm's Law defines the relationship between voltage, amperage, and resistance, without considering power.

To solve for any variable in Ohm's Law, cover the variable you're solving for: E = I * R for voltage, I = E / R for amperage, and R = E / I for resistance.

Joule's Law explains the transfer of energy between systems, such as electrical energy converting to light and heat in a light bulb.

Pi formula (P = I * E) is used to calculate power in electrical circuits, commonly used to find wattage in a circuit.

In field calculations, understanding voltage drop is important, and both Ohm's Law and specific voltage drop formulas can be applied.

In three-phase circuits, the square root of three (1.732) is introduced into the voltage drop formula, which generally results in a smaller voltage drop than in single-phase circuits.

The constant 'K' in voltage drop calculations varies depending on the conductor material: 12.9 for copper and 21.2 for aluminum.

For common calculations in the field, a #12 copper wire has a circular mils value of 6530, which is used in voltage drop formulas.

Series circuits' total resistance is the sum of all resistances, while parallel circuits require the reciprocal of the sum of the reciprocals for total resistance.

The product-over-sum method simplifies the calculation of resistance in parallel circuits, producing the same result as the sum of reciprocals.

One horsepower equals approximately 746 watts, and efficiency and power factor are key factors in calculating the true horsepower output of a motor.

In a three-phase motor, efficiency is improved over single-phase due to the 1.732 multiplier in the horsepower equation.

Capacitance and inductance formulas in series and parallel circuits differ, with capacitance formulas being inverted in parallel and series configurations.

Capacitance and inductance are more advanced electrical concepts often tested in electrical theory exams, requiring an understanding of RLC circuits.

Horsepower output is greater in three-phase motors due to more efficient power transfer, compared to single-phase motors under the same conditions.

Transcripts

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