There is so much garbage out there, I'm really happy to find someone who is clear, easy to understand and stays focused on the question at hand. Thanks and keep on doing what you're doing!
As a designer for a contractor and doing thousands of calculations over the years, I always wanted to know the wire size needed in one calculation. I did not want to keep guessing until I got less than 3%. Less complicated method. 2xKxampsxLdivided by 3% of voltage= cm. Conductor properties table will give you needed wire size. Also, for balanced 3 phase 3 wire multiply single phase calculation at end by .86. 1.73/2=.865. And for balanced three phase 4 wire, multiply single phase calc at end times .5. If balanced load, then no neutral current. 1/2=.5. Respectfully submitted, Kevin
...Nice learning video. I also like using Z (the impedance) because it provides a glimpse for the human that impedance plays an important role in voltage delivery to the end point.
Thank you very much for the very clear, detailed and straight forward video lesson. This helps me a lot on my studies. NEC Table 9 have a title of: Alternating-Current Resistance and Reactance for 600-Volt Cables, "3-Phase", 60 Hz, 75C (167F) - Three Single Conductors in Conduit May I kindly consult if we can use NEC Table 9 for 1-Phase circuits? even though the title indicates "3-Phase" Thanks again and more power to your channel.
Thank you for this video. Question for you about the impedance for the single phase case. The Chapter 9 Table 9 of the NEC says in the title that it is for three phase power, but you used it in the single phase case. Is there something in the NEC that indicated that table 9 can be used on the single phase case or is there another table for the single phase case?
GOOD AFTERNOON, I HAVE THE FOLLOWING QUESTION: IF A THREE-PHASE FEEDER IS INSTALLED IN A METAL WIREWAYS, TABLE 9 IS APPLIED TO CALCULATE THE VOLTAGE DROP?
Both methods should give equivalent results. The reason why using table 9 gives you a slightly different result is because eddys current (Q) losses is not taken into account. But if (Q) is used both methods should be 99.9 % equivalent.
Jose ... you need to figure the drop for each run separate then add up the results for the whole run. If the numbers are too high (percentage of drop) then you need larger wire. You will almost never have a problem if you stick to 80% loading of a branch circuit and stay under 100 ft in a run. To ease the brain work use one of the voltage drop calculators on the web. You plug in wire size, load, and run length. They will tell you voltage drop and percentage. After you use it a few times you will get a feel for where you need to be.
good video! a couple questions tho. How did you get 2.75% from 13.20 in the single phase method (15:27) and your applied voltage cell reads 480. thanks
Why does no one bring up how to find voltage drop of wire larger than 4/0? I need voltage drop of 850ft run of single conductor aluminum 750kcmil, 3000A.
Please HELP looking for someone that's very good with voltage drop on extremely long runs on Main Service cable I'm running 4/0 4/0 4/0 2/0 aluminum direct burial to my shop 670 FEET away. I have 250 volts at my main panel I'm trying to understand how many effective amps I will really have when it gets to the shop I understand there's a big voltage drop and a loss of amperage but can anyone tell me what that would actually be or close to it I have tried voltage drop calculators online but I'm getting differentials that don't make sense. Trying to know if I would at least have enough amps to run a compressor welder and possible mini split with lights I understand I will be running close to capacity but need to know if it's really worth the extra $3,000 to do an extra run alongside my current run so that way I don't need to rip it out. How many usable amps should I expect for the shop.
Great video only issue with using the table 9 formula with impedance (z value) is that the impedance of Table 9 is only good for 3phase - Three single conductors in conduit as per the title of that Table so I’d assume it’s not totally accurate for use with Single Phase Voltage drop calculations. Sorry to be that guy.
Studi314 , I see what you’re saying. It would’ve been helpful to have the formulas with parentheses. But also under the rules of orders of operation in math, multiplication always comes before division. So the formula is correct with the parentheses or not but it is more helpful with the parentheses
circular mills is the circular cross section of the wire we are examining. You can find the CM for each size of wire in Chapt 9, Table 8 of the NEC. Check minute mark 12:38 and you will see that table.
![Voltage-Drop - Manufacturer and NEC Recommendations [210.19(A) Note 3, 2020 NEC]](http://i.ytimg.com/vi/7gKW7j7jLxk/mqdefault.jpg)
![Voltage-Drop - Manufacturer and NEC Recommendations [210.19(A) Note 3, 2020 NEC]](/img/tr.png)
There is so much garbage out there, I'm really happy to find someone who is clear, easy to understand and stays focused on the question at hand. Thanks and keep on doing what you're doing!
As a designer for a contractor and doing thousands of calculations over the years, I always wanted to know the wire size needed in one calculation. I did not want to keep guessing until I got less than 3%. Less complicated method. 2xKxampsxLdivided by 3% of voltage= cm. Conductor properties table will give you needed wire size. Also, for balanced 3 phase 3 wire multiply single phase calculation at end by .86. 1.73/2=.865. And for balanced three phase 4 wire, multiply single phase calc at end times .5. If balanced load, then no neutral current. 1/2=.5.
Respectfully submitted, Kevin
Thank you for putting these videos together! For someone getting underway in the trade I find these videos very helpful. Shout out to the 209 too!
Thank you for your time in making this video. It is easy to follow along and understand. Especially on impedance. Once again thank you.
Thanks for commenting. I appreciate the feedback.
I think you're Portuguese, but I love you're videos. Thank You.
...Nice learning video. I also like using Z (the impedance) because it provides a glimpse for the human that impedance plays an important role in voltage delivery to the end point.
Thank you for this video! It helped me a lot in understanding and feeling confident in my calculations.
Another good informational video👍
Your doing such a good job for those who want to learn more and improve.
Thank you! Im getting ready to take my state test here in CA this Saturday. I'm ready for a voltage drop question.
Ernesto Cortez, thanks for the comment. Good luck on your test.
Thank you for the clear discussion abou voltage drop.
Great explanation. Thank you very much.
Thank you very much for the very clear, detailed and straight forward video lesson. This helps me a lot on my studies.
NEC Table 9 have a title of:
Alternating-Current Resistance and Reactance for 600-Volt Cables, "3-Phase", 60 Hz, 75C (167F) - Three Single Conductors in Conduit
May I kindly consult if we can use NEC Table 9 for 1-Phase circuits?
even though the title indicates "3-Phase"
Thanks again and more power to your channel.
That was such a WELL done video. So clear and precise! great job. Others should follow as some ramble through their videos. Again ~ thank you!!!!
Rich Scibetta thanks for the great comment!
Great video. But can we use Vd = 2*L*R*I / 1000 formula?
Thank you for this video. Question for you about the impedance for the single phase case. The Chapter 9 Table 9 of the NEC says in the title that it is for three phase power, but you used it in the single phase case. Is there something in the NEC that indicated that table 9 can be used on the single phase case or is there another table for the single phase case?
When would one use the “coated” section?
GOOD AFTERNOON, I HAVE THE FOLLOWING QUESTION: IF A THREE-PHASE FEEDER IS INSTALLED IN A METAL WIREWAYS, TABLE 9 IS APPLIED TO CALCULATE THE VOLTAGE DROP?
And what about you having 2 sets of conductors with Z method? Please Help!!
Thanks for your help in solving voltage drop calculation.
Is table 8 is only applicable in DC system?
Awesome! Thank you once again.
where can I get the formula book? internet?
Super Helpful! Good Informational video!!
Both methods should give equivalent results. The reason why using table 9 gives you a slightly different result is because eddys current (Q) losses is not taken into account. But if (Q) is used both methods should be 99.9 % equivalent.
Thanks!! Keep them coming.
Thanks a lot. I find this very useful.
If using meters, for example with a 10AWG cable, would it be the same? 4.070*10380/1000m?
Thanks A lot , your method of teach is very unique I completely understand of VD calculations. 🌹🌹🌹
Thank you so much for this video! Subscribing now
Thanks! Glad the video was helpful!
Good job thank you for your help
How do you factor in multiple runs of conductors into that Voltage Drop equation?
Jose ... you need to figure the drop for each run separate then add up the results for the whole run. If the numbers are too high (percentage of drop) then you need larger wire. You will almost never have a problem if you stick to 80% loading of a branch circuit and stay under 100 ft in a run. To ease the brain work use one of the voltage drop calculators on the web. You plug in wire size, load, and run length. They will tell you voltage drop and percentage. After you use it a few times you will get a feel for where you need to be.
Why can't we use Chapter 9, Table 9 of the NEC?
good video! a couple questions tho. How did you get 2.75% from 13.20 in the single phase method (15:27) and your applied voltage cell reads 480. thanks
Matthew Gaskey also, this calculator may be helpful. www.calculatorsoup.com/calculators/algebra/percent-difference-calculator.php
Only thing is i couldnt find where to find chapter 9 table 8 in my nec book
Why does no one bring up how to find voltage drop of wire larger than 4/0? I need voltage drop of 850ft run of single conductor aluminum 750kcmil, 3000A.
Which cable is affected by the voltage drop?
All cable
Please HELP looking for someone that's very good with voltage drop on extremely long runs on Main Service cable I'm running 4/0 4/0 4/0 2/0 aluminum direct burial to my shop 670 FEET away. I have 250 volts at my main panel I'm trying to understand how many effective amps I will really have when it gets to the shop I understand there's a big voltage drop and a loss of amperage but can anyone tell me what that would actually be or close to it I have tried voltage drop calculators online but I'm getting differentials that don't make sense. Trying to know if I would at least have enough amps to run a compressor welder and possible mini split with lights I understand I will be running close to capacity but need to know if it's really worth the extra $3,000 to do an extra run alongside my current run so that way I don't need to rip it out. How many usable amps should I expect for the shop.
Great video only issue with using the table 9 formula with impedance (z value) is that the impedance of Table 9 is only good for 3phase - Three single conductors in conduit as per the title of that Table so I’d assume it’s not totally accurate for use with Single Phase Voltage drop calculations. Sorry to be that guy.
Is the second formula 2xZxDx(I /1000) the way you have it written or is it really ( 2xZxDxI)/1000?
Studi314, I’m not sure what you’re referring to because I have a written (2xZxDxI)/1000.
At the time 13:00, the formulas do not have any parenthesis in them.
Studi314 , I see what you’re saying. It would’ve been helpful to have the formulas with parentheses. But also under the rules of orders of operation in math, multiplication always comes before division. So the formula is correct with the parentheses or not but it is more helpful with the parentheses
CM in the equation. What are Circular Mills?
It's ok. Got it. Just wondering what's wrong with square mils. Or even square inches.
circular mills is the circular cross section of the wire we are examining. You can find the CM for each size of wire in Chapt 9, Table 8 of the NEC. Check minute mark 12:38 and you will see that table.
Great
Hola boss this Is José Marín can you please maje a vídeo for load calculacions for motors.
Thanks
hey Jose. hope all is going well with you... I will try and do a video on that subject.
@@AccessToPower sir what is the formula to get the percent?
nice vedios
You said single phase formula but the voltage drop was 480??? Wasn’t suppose to be 120v
Good point !
Your great but we are gonna have to ask you to stop teaching us using spreadsheets. We cannot use them while taking the exam.