Are You an Electrician? These are 5 Formulas You Should Know!
HTML-код
- Опубликовано: 14 май 2024
- Join this channel to get access to perks:
/ @electricianu
Being a great electrician requires a strong knowledge of math. We use it daily from bending conduit, to figuring out what wire to pull, even simply counting light fixtures or circuits. But which formulas should we be more familiar with? In todays episode of Electrician U, Dustin explains the top 5 formulas every electrician should know.
🤘⚡️MEMBERSHIP⚡️🤘
JOIN ELECTRICIAN U - become a member and get:
FREE Continuing Education every year
FREE Practice Exams
FREE Monthly Video Courses
FREE Weekly Live Instructor-Led Classes
FREE Monthly Educational Newsletter
Premium Members-Only Content
Private Discord Channel
Monthly Members-Only Discord Chats
Sign up here --- www.electricianu.com/electric...
🎧🎹MUSIC AND VIDEO:🎹🎧
/ descantmv
🎬✍️ART AND ILLUSTRATION:✍️🎬
www.daverussoart.com
First on the list is Ohms law. This formula is the relationship between Voltage, Amperage, and Resistance. In many cases, we are not given ALL of the information for a piece of equipment, but still need to determine either the voltage or amperage of it. Ohms law is simply E (voltage) over I (amperage) times R (resistance). So, draw a circle and put a large T in the center of it. Above the horizontal line of the T draw an E. On the left of the vertical line draw an I and on the right of the line draw an R. To assist you, cover up the letter you are attempting to solve. For instance, if I covered up E (voltage) it would leave me with I (amperage) multiplied by R (resistance). If I had a 20a piece of equipment with a 6 ohm resistance (20 x 6) it would be running at 120v!
The next formula is Joules law. This one is slightly different than Ohms law and is the relationship between Wattage, Amperage, and Voltage. The circle is the same as above but with a P on top, an I on the left, and an E on the right. The math is the same also. So, if I was attempting to see how many watts were on a given circuit, cover up the P and I am left with I (amperage) times E (voltage). For a 20a circuit operating at 120v, I would have 2400w. Both of these formulas are very useful because we don’t always get all of the information we need on the equipment nameplate.
Voltage Drop is something that every electrician should know how to figure out. For a single phase circuit the formula is 2 x K(conductor) x I (circuit amperage) x L (length) divided by the circular mils of the conductor you are attempting to use. For 3 phase replace the 2 with a 1.732 (the square root of 3). If you have a copper conductor use 12.9 and use 21.2 if you are using aluminum conductors. The circular mils for electrical conductors can be found in the NEC codebook in Chapter 9 Table 8. The resulting number after crunching the equation is the amount of volts that are lost. You may find that you may need to upsize your wire (and perhaps the conduit) to get your voltage drop down to a reasonable level.
Resistance formulas are needed for every electrical theory class! For a series circuit the total resistance is the sum of all the resistances. For a parallel circuit, it’s the reciprocal of the sum of all the reciprocals. So, 1 divided by 1/R1 + 1/R2 + 1/R3 + etc. An easier way for that last one would be product over sum formula. So, if you had a parallel circuit with resistances of 2, 3, & 4 the formula would be 2 x 3 x 4 divided by 2 + 3 + 4. Much Simpler!
Lastly is Horsepower. Something to just keep in mind is that 1 HP is equivalent to 746 watts. For single phase motors the formula is HP= E (voltage) x I (amperage) x EFF (efficiency) x PF (power factor) divided by 746. For a 3 phase motor, simply insert 1.732 (the square root of 3) in front of the E. Efficiency you can find on the nameplate of the motor. If you have a completely balanced load that isn’t running a ton of motors you may have a power factor of close to 1. The more motors you have on any given system, that number goes down (say to .7 or even more).
We hope this has been helpful in understanding the 5 most popular formulas an electrician uses frequently. Is there a topic you would like to see discussed on Electrician U? Leave us a comment in the comments section and let us know. Please continue to follow Dustin and Electrician U as we are constantly updating our content to assist our followers in becoming the best electricians that they can be.
#electrician #electrical #electricity #these #are #five #formulas #you #should #know
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!
Hello, from Dominican Republic
I'm an electrician. But I am watching this video to improve my english.
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
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.
Hi Dustin as an engineering student I appreciate what you are doing Keep it up its just another tool we need in our field.
I took electrical theory in high school. 30 years ago. Yes, it's time to brush up. Thanks. Good job.
Excellent formula review boss! I love the way you teach. Much appreciation
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
Starting my apprenticeship over after leavening during my 1-2 year. These videos are amazingly helpful for review.
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
Thank you for taking back to school. Learned these formulas and more in my electrical engineering degree
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👍
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
thanks for the help, you explain things so well that i didn’t get bored or leave the video the whole way through!
Great presentation I sent it to my son he’s heating and air tech. He owns his own business. Thought it would be helpful him too. Enjoyed it.
Really appreciate the videos!! Takes me back to apprenticeship school
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.
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.
Great video, easy to understand
Thank you
I just hope you put code references especially for VD where you find the numbers etc.
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...
@@epolpier the title of the video is “for electricians”
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
This video took me back to my EET years. Thanks for the review.
Awesome job. Thanks for explaining your process.
Thank you for making this video. I'm taking my Journeyman exam next week and this helps alot.
So, how did the exam go?
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.
Good teaching here as a field tech in the phone company I used a vom every day when troubleshooting t-carrier lines.
Interesting, informative and worthwhile video. A very worthwhile, simplified approach.
Dustin thank you very much for your great work.
Watching this was very useful. Thank you!
Hello Professor
Thank you so much for your help and advice,
i do appreciate your job.I wish you peace and happness under the sky of prosperity. Take care and have a good time. All the best.
Your Student from Algeria.
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
😉😉
Excellent vid. Wish I had this type of instructional resource when I was in school.
Yes yes yes thank you!!
Great to have them in one spot.
🎉
The more I watch the more I want to become an electrician. Love your style of teaching!!
Hahhahahshahahahhahahaha. As a decade long union electrician it’s a nice way to make good money, good insurance, can work when you want, quit when you want, travel where you want. But the work is PHYSICAL. I would use the job as a plan B. You can always come back to it and make good money. Also it’s nice to know it because everything is run off electricity so you’ll be hot shit to people who know nothing bout it. Get your Journeyman card then figure out a way to make money while you sleep cause it will wear your body out.
@@Jumpingjackflash123 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.
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 RUclips. I want to make sure I can pass the aptitude test.
Thank you so much Sir, that’s awesome! Very precise explanation and easy to remember formula.
More power to you and God bless !!!
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.
Can you do service calculations? I need a good step by step and you do the best explanations on every video
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
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 👍🤞
Thanks ! Very helpful.
We call the second one the Power watts formula in industrial use ;) Thanks for sharing this
this has been my week in class for my apprenticeship
Great teaching method....even for an old timer.
Thank you bro this video really helped 🙏
Please teach at the IEC in Austin. You would be a great instructor!
Passed my first two tests. One more to go. This video came at a great time. Needed the refresher!!
Awesome! Good luck!!!
I inspired to know more about electrical. Great! 😃👍
Interesting ! You make me more smarter in electrical work..❤😊
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?
Thank you for this information.
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 true!! Series resistance is the sum of each no matter how many total. However for a real world electrician you will NEVER have loads connected in series. Series circuits are typically used only in logic board type circuits, the only industrial or residential series circuits I have ever seen were the elements for resistive heaters, but the power connections were all contained in the mfg wiring.
Parallel uses reciprocal formula, Because the total resistance will always be less that the lowest resistor in the circuit. Think of it like parallels pvc drain pipes, each path greatly reduces the resistance ( or allows more GPM).
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.
First time introduced to Voltage Drop formulation.
Resistance formula with two (2) in parallel. Product divided by the sum.
Great intro to these formulas! One suggestions...add some units so it's easier to see what the numbers stand for.
Great video. Clear - Consistent - Common sense
great, congrats good job, I didn't know how to calculate Vd, thanks
Thank You 💪🏽❤️🙏🏽
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
Great work bro 🙌🙌🙌🙌
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
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!
Those pie charts. My apprenticeship teacher never mentioned just covering the letter. Now it makes sense why it's drawn like that.
Question: When you were talking about the voltage drop calculation, Would a person be able measure that voltage drop with a meter and if so would you see an 8 volt drop at the load or would you only see approximately half that drop at the load since the 100 ft conductor out to the load would only be half the resistance (allowing a little variance for the resistance of the load itself) and the other half of the resistance would be the conductor back to the source? You should to a video demonstrating this fact? :o). Thanks for the interesting content.
Make a video about Motor Drives, and how they work.
Variable Frequency Drives.
Thank you brother
Thank you, Dustin.
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.
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?
This is realy helpul sir
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.
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
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.
This is much better. Bravo. Watching the kick video I was worried what I’d find here.
I hope you keep finding things you like. More than you dislike.
Thanks for your videos Dustin good info
So have you designed your electrical tool belt yet?
Thanks!
What about Pythagorean theorem..? I use that on the field alot more than any of those that you mentioned... Helps alot with rolling offsets..
You are teaching the right thing
Well said, teach! Can you put the dimensions of the numbers in the equations next time? Thanks!
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.
I feel like some of these formulas are for second or third year apprentices, cause I have only being taught Ohm's law, Voltage Drop and finally Capacitance although not called it in Canada or at least not they call it while teaching it. Overall great video Dustin.
Thanks for the breakdown on the Series & Parralel - the Product / Sum method is waaayy easier. Quick Question though, the Voltage Drop formulae - i note you've used Feet for the Length or Distance - can im in Australia - can i just interchange it with Meters?
Thank you
Thanks for sharing your knowledge and spending your time making these videos they’re great
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 🙏🙏💐🌷🌺🌷🌺💐🌷💐🌺
Wonderful instructions! Question, shouldn't the formular for parallel be (product/sum of products)?
You’re great at this
We all really needs it🎉
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.
Outstanding may God continue to bless you.😅😊
Good to refresh my skills
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
@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.
GOOD CLASS
thank you
Thank u 4 joules law. Never would've heard of it otherwise
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.
Thanks talk about calculation of solar and batteries please
thank you masterelectricianU,☆;
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
Yeah I came up with the same thing. Product over sum doesn't work unless we are missing something.
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.