Couple of questions i can't find in the 05 nec

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Night65

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Hi guys i'm new on the site so if this is not how to ask a question please let me know.
1. What is the minimum size parallel conductor allowed in the code book
2. Why does code say parallel conductors must be in same pipe
3. how do i figure out the length it takes for a vd if the wire is #12 and resistance is 8 amps
4. How do you figure out total amps on a question that asking 120v/240v , 35000va
5. In a equipment room with a 1000v switchboard is the ceiling clearance 6'6"

Thank you
 
Welcome to the forum.

I will tell you we really expect the person asking the questions to make an attempt at it.

In other words no free rides.

I will get you started.

Night65 said:
Hi guys i'm new on the site so if this is not how to ask a question please let me know.
1. What is the minimum size parallel conductor allowed in the code book
2. Why does code say parallel conductors must be in same pipe
Click to expand...

Take a look at Article 310
 
Night65 said:
Hi guys i'm new on the site so if this is not how to ask a question please let me know.
1. What is the minimum size parallel conductor allowed in the code book
2. Why does code say parallel conductors must be in same pipe
Click to expand...


310.4 for both.

Night65 said:
3. how do i figure out the length it takes for a vd if the wire is #12 and resistance is 8 amps
Click to expand...

VoltageDrop.jpg


Night65 said:
4. How do you figure out total amps on a question that asking 120v/240v , 35000va
Thank you
Click to expand...

ohmlaw.gif
 
Last edited:
Thank you for answering so quickly

Thank you for answering so quickly

i did look in the 310.4 and saw it says 1/0awg but it goes on to say how other smaller wires are permitted in other cases. i also know found that ceiling height question but i found it where it did not say wheter it be more than 600v
 
Thank you you been a big help

Thank you you been a big help

i know the ohms law chart but i don't understand if you have 35000va = 291 if divided by 120 but 145 if by 240 how do i know which one to use to determine how to get total amps for a house
 
Night65 said:
i know the ohms law chart but i don't understand if you have 35000va = 291 if divided by 120 but 145 if by 240 how do i know which one to use to determine how to get total amps for a house
Click to expand...

If the voltage comes into the house at 240 you would use 240.
 
Thanks

Thanks

i know this probably is a dumb question but in the calc. problems do you always use the higher voltage for the answer
 
Night65 said:
i know this probably is a dumb question but in the calc. problems do you always use the higher voltage for the answer
Click to expand...

It depends what you are calculating. If you are calculating the entire premise for a service then you would use that voltage.

If I want to calculate the load for a 1200 watt toaster at 120 volts then I would use the voltage of the unit.

1200/120= 10 amps--- you would not use the 240 in this case.

Its not a dumb question.
 
Thank you that answers that perfectly

Thank you that answers that perfectly

maybe it's me but i don't understand the whole voltage drop thing , if i know

vd= 2 x 52.24 x 8 x 151 = divid by 6530 the answer is vd 19.328 so on a 120v line it would be 100.672 which is no good cause only 3% is allowed then how do i figure out the lenght so it is max lenght allowed
 
Night65 said:
maybe it's me but i don't understand the whole voltage drop thing , if i know

vd= 2 x 52.24 x 8 x 151 = divid by 6530 the answer is vd 19.328 so on a 120v line it would be 100.672 which is no good cause only 3% is allowed then how do i figure out the lenght so it is max lenght allowed
Click to expand...

If you are solving a VD for 120volts at 3% then insert 3.6 in the equation for VD and solve for the Length. Try it -- if you need help let me know

If you are solving a vd for 240volts at 3% then use 7.2 for VD in the equation
 
that wasn't clicking in my brain to me it sounded like

vd= 2 x 52.24 x 8 x 3.6 = / by 6530 = .4608 but that can't be it cause that would mean you can't even run a 12 wire 6 inches without drop
 
Night65 said:
maybe it's me but i don't understand the whole voltage drop thing , if i know

vd= 2 x 52.24 x 8 x 151 = divid by 6530 the answer is vd 19.328 so on a 120v line it would be 100.672 which is no good cause only 3% is allowed then how do i figure out the lenght so it is max lenght allowed
Click to expand...


Allowed by who?
 
Your formula is all wrong.

Try 3.6=2*12.9*8*L/6530.

Solve for L.

First, reduce to simple numbers

3.6=(2*12.9*8)*L/6530 becomes 3.6= 206.4*L/6530.

Now, get rid of the division. Multiply both sides of the equation by 6530.

(3.6*6530) = ((206.4*L/6530)*6530) The last two 6530s cancel each other out.

Now you have 23508=206.4*L

Now you can solve for L.

L=23508/206.4.

L=113.83 feet.


To double-check, enter 113.83 as L into the original VD formula (VD=2*K*I*L/6530) and see if you get 3.6.
 
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thanks for clearing that up, i'll keep that in mind. but still at hand in the problem the question i got was ( allowing a 3% drop with a #12 awg with a resistance of 8 amps what is lenght )

then answers they gave me was
0-150
151-250
251-350
351-above

how would i go about answering that
 
Night65 said:
that wasn't clicking in my brain to me it sounded like

vd= 2 x 52.24 x 8 x 3.6 = / by 6530 = .4608 but that can't be it cause that would mean you can't even run a 12 wire 6 inches without drop
Click to expand...

CM=2xKxIxD / VD

Thus D= cmxVD/ 2xKxI

D= 6530 x 3.6/ 2 x 12.9 (copper) x 8 amps

D= 23508/206.4= 114 feet.
 
Night65 said:
thanks for clearing that up, i'll keep that in mind. but still at hand in the problem the question i got was ( allowing a 3% drop with a #12 awg with a resistance of 8 amps what is lenght )

then answers they gave me was
0-150
151-250
251-350
351-above

how would i go about answering that
Click to expand...

I don't understand their answer-- given a 3% drop with #12 wire and 8 ohms you should be getting 64 feet as the max. The online calculator comes up with 60 feet. Close enough---
 
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