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#78432 - 09/23/01 09:28 PM Conductors in Parallel
Joe Tedesco Offline
Member
Registered: 10/07/00
Posts: 2749
Loc: Boston, Massachusetts USA
Subject:
Question
Date:
Sun, 23 Sep 2001 16:47:20 -1000
From:
"Joe Wilhelm" <jrwil@attglobal.net>
To:
<joetede@ix.netcom.com>


Hello,

I would appreciate your help with the following:

If I understand correctly, a 14 gauge wire is rated for 15 amps and 8 gauge
is rated for 40 amps. If I combine the 14 and 8 wires in parallel does that
then provide 55 amp load capability?

Thank you.

This installation is contrary to common practice. Please see Sections 110-8 and 310-4 in the 1999 NEC.
_________________________
Joe Tedesco, NEC Consultant
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#78433 - 09/24/01 03:41 AM Re: Conductors in Parallel
sparky Offline
Member
Registered: 10/18/00
Posts: 5303
me thinks the #14 would be in trouble
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#78434 - 09/24/01 04:29 AM Re: Conductors in Parallel
Redsy Offline
Member
Registered: 03/28/01
Posts: 2056
Loc: Bucks County PA
First off 310-4 only permits conductors 1/0 and larger to be paralleled. It also requires they be the same diameter.
Secondly, the values suggested for ampacity are ambiguous, due to lack of conductor & application detail.
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#78435 - 09/24/01 07:01 AM Re: Conductors in Parallel
resqcapt19 Offline
Member
Registered: 11/10/00
Posts: 2148
Loc: IL
While this installation is not code compliant, neither conductor would be overloaded. If you use the AC resistance from Chapter 9 Table 9 you can solve this as parallel resisitor problem. If we make the circuit 100' long and assume a 55 amp load the resistnace of #14 is 0.31 ohms and the the #8 is 0.078 ohms. The total resistance of these two conductors in parallel is:

1/0.31 + 1/0.078 = 1/Rt = 16.046

Rt= total resistance

1/16.046 = 0.0623 ohms = resistance of 100' of #14 in parallel with 100' of #8

Now if we use E=I(R) to solve for the voltage drop we get E= 55(0.0623) = 3.426 volts

We can now use this voltage drop to solve for the current in each of our conductors.
Again E = I(R) 3.426 = I(0.31) = 11.05 amps for the #14.

3.426 = I(0.078) = 43.92 amps for the #8

These two values do not exactaly equal 55 amps due to rounding. They do total 54.97 amps.

Don(resqcapt19)

[This message has been edited by resqcapt19 (edited 09-24-2001).]
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Don(resqcapt19)
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#78436 - 09/24/01 01:04 PM Re: Conductors in Parallel
Joe Tedesco Offline
Member
Registered: 10/07/00
Posts: 2749
Loc: Boston, Massachusetts USA
Subject:
RE: Follow-up to Answers to your question
Date:
Mon, 24 Sep 2001 10:11:25 -1000
From:
"Joe Wilhelm" <jrwil@attglobal.net>
To:
"Joe Tedesco" <joetede@ix.netcom.com>


Joe, Thank you very much for your help! While I get around on the internet,
I don't know how to "post" or put up a reply or follow-up question... so if
I could ask you to please post this follow-up:

Thank you all for your helpful comments, especially "resqcapt19" for your
calculations! In light the reply's, could you please tell me how much
current 8 gauge will carry at 220V and 110V over a 50' long run? I'm
thinking of using a 50 amp double pole breaker (220V) by the main panel to
feed the 220V to another box 50' away with 4 number 8 wires (2 hot, 1
neutral, 1 green). The box at the end of the 50' will have at least 1 double
pole 20 amp breaker for a 220V line and some 20/15 amp breakers for 110V
lines. Will this work ok? Would it meet "code"? Someone told me that a
number 8 wire is only good for 40 amps so that's why I was thinking of
combining the 8 and 14 wires (a 3/4" conduit is already in place and I don't
think I can get 4 number 6 wires through it). Thank you all for your help!
_________________________
Joe Tedesco, NEC Consultant
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#78437 - 09/24/01 01:43 PM Re: Conductors in Parallel
Bill Addiss Offline
Member
Registered: 10/07/00
Posts: 3875
Loc: NY, USA
Joe,

It's quite easy to post a reply here. You just have to register first. You just fill out a simple form where you choose a "Name" and a password. Then you only have to click on "Post Reply" to respond to something already here or "Post New Topic" will start a totally new Topic.

BTW, If I'm not mistaken, according to table 250-122 you could run a #10 for your ground. That would be 3 #6 and a #10 in your 3/4" conduit for the 50A circuit.

Bill
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#78438 - 09/24/01 02:05 PM Re: Conductors in Parallel
resqcapt19 Offline
Member
Registered: 11/10/00
Posts: 2148
Loc: IL
Joe,
The #8 is limited to 40 amps unless the terminations at both ends of the wire and the wire itself is listed for 75°C. Four #6 THWN will fit in a 3/4" conduit. Also Table 250-122 only requires a #10 ground wire for a 40 or 50 amp circuit. You could install three #6 and one #10 to make the pull a little easier. The voltage drop on a 120 volt circuit of #8 50' long with 40 amps of load would be 3 volts of 2.5%.
Don(resqcapt19)
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Don(resqcapt19)
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#78439 - 09/24/01 04:39 PM Re: Conductors in Parallel
jw Offline
Junior Member
Registered: 09/24/01
Posts: 1
So the voltage drop of #8 for 50' is 3 volts... what is the resistance of 50' of #8?
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#78440 - 09/24/01 06:21 PM Re: Conductors in Parallel
resqcapt19 Offline
Member
Registered: 11/10/00
Posts: 2148
Loc: IL
jw,
I used a voltage drop calculator on another site to figure the voltage drop. I didn't use the resistance, but it would 0.039 ohms.
Don(resqcapt19)
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Don(resqcapt19)
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