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Happy to help!

I'm glad you like it. Thanks for the feedback!

Watch the video I explain it.

javujo@gmail.com

Great question! Gas powered appliances still need an outlet to ignite the flame and help the electronics in the appliance work, this is known as the "power for supplemental equipment and lighting." Check out Article 210.52 (B) (2) Exception 2. This exception tells us that we can add an extra outlet to provide power for the "supplemental equipment and lighting" on gas ranges as well as gas ovens and counter-mounted cooking units. Notice that this is found in section (B) of Article 210.52. Section (B) is Small Appliances (Small Appliance Branch Circuit ... SABC) this means that this extra outlet can be added on one of the SABCs. In conclusion you do not need to add any additional load for a gas powered appliance. The circumstance where you might end up adding additional load for a gas powered appliance is if someone mentions they want it on a dedicated circuit. Since it is located in the kitchen the plugs will need to be on the SABC meaning you would count another SABC at 1500va. You might see this in real life with a customer.

That example was just to introduce the concept. There is a reason why I did not solve that question all of the way like I did with the other ones. But to answer your question: You could fit 2,165 #12 THHN conductors in a 12" x 12" gutter, again 40ft is irrelevant in this calculation since we are using square inches. (All explained in the video)

@@jarodvelectric7431 yes but the 40 feet shouldn't be irrelevant. A lot of these codes are dumb and don't really make sense as to why they were the way they are. You should definitely be able to fit more wires in a 40 foot long gutter compared to a 2 foot long gutter.

@@jimbush1581 Yeah the code book does not always make sense logically. They way I look at it is like a law, it may not make sense but it is the rule.

Neither is music! But as you continue it becomes less of a "foreign language" and you begin to connect the dots. You got this!!

30:57

Yes I have been studying this for over 6 years and have years of experience on the job as well. That is a great question and I am glad you enjoy doing these because that should benefit you in life. Are you an electrician? As an electrical contractor you will need to be able to solve these to get licensed and also to know how to do your job. There could be positions where you solely do load calc but I have never heard of them ... I'm guessing the best chance at finding a position like that would be for the union. However it is good to know positions like that would usually not be for the new employees but instead reserved for those with lots of experience and years with the company.

@Brandom84 Good job! You also passed because you studied and took it seriously. I'm really glad my videos helped!! And thank you for the feedback.

Yes, it would be 118A as amps get rounded up if 0.5 or higher, according to Article 220.5 (B). Yes correct, 125A Service OCPD with #2 cu. or 1/0 al.

@@yasserabisaipineda9700 125% is for commercial load calculations, not residential.

@@JesusCruz-wt2zp Thanks!!

Never mind. You explained it in the next video! Lol.

Yeah that whole thing is ridiculous, I think it is a cheap question. I'm glad you saw the explanation in my other video, I hope you pass the next time!!

@@jarodvelectric7431 thank you! After watching your playlist on residential load calcs, it helped me understand much better. I’m feeling pretty confident now. Also you’re spot on about some of these “practice tests” and “study guides”. I appreciate you man.

No you do not use the AC in the 25% load calculation. You choose the next largest motor below that. Even Mike Holt teaches this. Read this: forums.mikeholt.com/threads/selecting-largest-motor.2440/ read the comment by "txinspect"

@@jarodvelectric7431 thank you! Second question (and I’ll probably ask you a lot more now that I have found you as a resource) when doing dwelling calculations (single or multi family) am I ever going to calculate heating or cooling as a CONTINUOUS LOAD and multiply them at 125%?

@markjohnsonkeys Yeah no prob! Ask away! No, continuous load demand factors are only when you get into commercial load calculations.

Do you apply 83 percent rule after you have done standard or optional method? Or is that already applied in table 310.12, 100-400 amps.

@@wakeandbakw it is already applied in T310.12

This is just an example. These are separate examples to break down each section of the calculation into separate sections. The point is to show you if you had 32,800 VA, what you would do with it in the optional method. At 15:17 there is an actual example which includes everything all at once.

Thanks

Four 20A kitchen circuits would be 6,000 va. Kitchen circuits = small appliance branch circuits. They are 1500va each, with 2 minimum. So if you are not told how many you have it is a minimum of two at 1500va each. Since you have four then there are four at 1500va each. No, small appliances are only related to the kitchen. The way to do load calculations is taught in my videos and I never mention bathrooms because that is not a part of the load calculation. To understand this more read this: forums.mikeholt.com/threads/bathroom-circuit-added-to-load-calculation.24798/ And make sure to check out my newer load calc vids!

Q) If we know the total VA of general light and receptacle, can we know exactly how many lights and outlets are there in that dwelling unit? A) The exact amount would not be known but there are minimums. Such as a minimum number of branch circuits on General Lighting using #14 AWG or #12 AWG. You would figure this out by getting your sqft multiplied by 3va. Then From this number dividing it by 120v then divide that by the amps you are working with, 15a or 20a. This will tell you how many branch circuits you have with this portion plus an additional FOUR 20-amp Branch Circuits for: 2 SABC, 1 Laundry Circuit, and 1 Bathroom circuit. For example: if you had a 2,500 sqft dwelling then you would first do 2,500 x 3 = 7,500 from here you divide by 120. 7,500 / 120 = 62.5 from here you divide by the amps you are working with, first I will do 15A for #14 AWG. 62.5 / 15 = 4.166 You always round up to the NEXT whole number. So even though you normally would not round up 4.166 in this case you always round up to the next whole number. 4.166 becomes 5. This means you need 5 Branch Circuits for this 2,500 sqft dwelling plus an additional FOUR 20-amp Branch Circuits as explained above. This is a complicated area because they might ask how many general purpose branch circuits you would need... this means instead of the additional 4 you would only count the 5 for the lighting portion. If you had 20A for #12 AWG then you would divide accordingly: 62.5 / 20 = 3.125 again round up to the NEXT whole number Now you can have 4 General Purpose Branch Circuits for this 2,500 sqft dwelling and if you are doing the entire General Lighting Load you should know the additional FOUR 20-amp branch circuits explained above.

My last comment was incorrect I thought you just typed "appliance circuit" the small appliance calc would be 6,000va with four 20A kitchen circuits.

Thanks!

That is true but it is not common that you will have to figure out that number on your own without using this formula to solve for CM. To find CM (Circular mils) in the codebook go to Chapter 9 Table 8. In the first column you will see Area, and below it is Circular mils that correlates to each wire gauge to the left.

Thank you!

Great comment, thanks!

Thank you, I appreciate that!!

Gracias por el comentario. Muy apreciado. :)

ruclips.net/video/raJksPKDeJc/видео.htmlsi=j72WXC396lnOUGo5 here is a video I made on my other channel before I created this channel.

But I see you already commented on that video. I could make a new video as I do have some good info on it.

@@jarodvelectric7431 did i sir? i can't remember. i'll just look into your channel. thanks anyway

Thank you!! I am really glad you passed, and first try is pretty rare so good job on that! Thanks for the feedback it means a lot. 😃

Thanks for the comment. I am using the 2020 Edition. I forgot to mention that in this video, I will remember to next time!

Then instead of 2 K I L / CM ... it is the square root of three in place of 2. √3 K I L / CM

Square root of three is something you will often see used in 3 phased systems. The number used is 1.73 or a step further 1.732.

Do you have an UGLYs book? It has all the formulas in it

copper is 12.9, aluminum is 21.2

I should have used 12.9 for copper in this video because that is what you do in real life.

@@jarodvelectric7431 that's more correct because more people are watching sir so you must pay attention on the correct information so as not to mislead the beginner. thanks for the correction

@@manuellastrollo2168 Yeah I agree but it was non intentional. The reason I used 12 is because I went to school for this and they taught me the number was 12. This is why it is always good to double check what you are being taught to make sure it is correct.

That is very kind of you, thank you!!

Thank you, I really appreciate that! I am editing some videos on the Optional Method right now :)

Thanks for the feedback! Yeah it is normally difficult finding specific videos for electrical theory so that is what I am trying to offer here. That is a great suggestion/idea! I definitely am proficient on the whole grounding/bonding topic and will put that on my bucket list for videos to make in the future. I will furthermore make an entire video on the equipment grounding conductor and another video for the grounding electrode conductor. Unless it is a misconception, it is common knowledge that the ground rod can be helpful when lighting strikes ones house to help the current go into the ground. But in terms of safety in the house I would think of GFCI protection.

@@jarodvelectric7431 By safety, I meant, the electrode doesn’t help with clearing a fault current. Most non-union guys I’ve worked with simply believe that “ground” means the same thing in all contexts; ground rod, grounding, grounded, EGC, ground fault. To be fair, it’s the fault of the pedagogy of the NEC and the vernacular that the industry uses which causes people to be confused. Other things guys are getting wrong are conduit fill, derating, load calc & voltage drop. I’d suggest creating a downloadable pdf worksheet for each of these scenarios you cover (create a website that you can later monetize) specifically the calculations. Mike Holt’s calculation course costs $500! That’s a shit ton of money. Plenty of room for you to cover ground and sell eHandbooks and test questions. I’ve been thinking of doing the same for some time now, complete with “how to” videos, but I’m nowhere near qualified yet. Good luck. I subbed.

And AFCI! ^^

Thank you so much for all of the support as well as positive comments!! I am planning some videos about Load Calc for Duplex and Multifamily for both the Standard Method and also the Optional Method. They should all be out by the end of the month

Yes I’m looking forward to your new videos. Let’s Do It

Good luck! Make sure to have your 2020 UGLYs book because that will have formulas in it you can use on the test! Also, pick up a key term index for your RCW/WAC if you do not already have that it will help you a ton!

Hey man just wanted to let you know that I passed my administrators exam all due to your videos. You explain things so much easier than any other of the big names here on RUclips. God bless man .

​ @user-zx8zt4pw3e That makes me really happy to hear that. I would have not made this channel if I felt like the big names taught well enough. So for you to tell me this means a lot! Thanks for sharing this with me. :)

It is hard to work with someone with remarks like that... I have a newer video if you are trying to learn. ruclips.net/video/er07gUB1l7g/видео.html&t

@@jarodvelectric7431 you’ll never get to a 200 amp panel at 1200’ home.

@@MarioLopez-iv2nw The point is not the answer it is how you got there... You must be very difficult in real life.

@@MarioLopez-iv2nw Go make your own video then buddy.

I am glad that you are getting use out of these! That makes me really happy to hear. If you do not already have it, make sure to pick up a copy of your version of UGLYs! It has mostly all of formulas I teach in my videos.

Thanks! Remember to include 25% of your largest motor when you are done with all of this. Check the pinned comment for a more in depth explanation plus I will be uploading a video on it soon on my channel so look out for that.

Glad to help! Remember to include 25% of your largest motor when you are done with all of this. Check the pinned comment for a more in depth explanation plus I will be uploading a video on it soon on my channel so look out for that.

I cannot provide a copy of the worksheet. But it is more of a burden than a help, here is my newer video on Residential Load Calc that explains a method that relies on the book which is more accurate to a testing scenario. (Vid Below) Residential Load Calc:ruclips.net/video/er07gUB1l7g/видео.html&t Optional Method Video: ruclips.net/video/qjgStSt4xJo/видео.html&t

Thanks! Yes it is one of my most recent videos, here is the link:ruclips.net/video/qjgStSt4xJo/видео.html&t Also, I have a newer video on the standard method where I show the process without a worksheet, which is more realistic to the actual test so I highly recommend that video too. It is called Residential Load Calculations, here is the link:ruclips.net/video/er07gUB1l7g/видео.html

Yes! And good luck!