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#167128 - 08/06/07 04:26 PM Parellel Feeds
luckyshadow Offline
Member

Registered: 01/04/05
Posts: 305
Loc: Maryland USA
I am looking for the calculation that show what the amp draw will be on Each leg of a parellel feeder. Specifically if 1 leg is shorter then the other leg.
I'm sure someone here has that calculation.

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#167133 - 08/06/07 05:41 PM Re: Parellel Feeds [Re: luckyshadow]
renosteinke Offline
Cat Servant
Member

Registered: 01/22/05
Posts: 5305
Loc: Blue Collar Country
I know there's a reason the NEC requires the feeders to be the same size, length, and type of wire ....

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#167145 - 08/06/07 09:51 PM Re: Parellel Feeds [Re: renosteinke]
Samurai Offline
Member

Registered: 05/04/07
Posts: 45
Loc: Fl.
I=E/R (R including the conductor resistance)?

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#167157 - 08/07/07 07:52 AM Re: Parellel Feeds [Re: Samurai]
resqcapt19 Offline
Member

Registered: 11/10/00
Posts: 2209
Loc: IL
In general with everything else being equal the current will divide in inverse proportion to the percentage of length. Add the lengths of both (or all) runs, and divide the length of each run by the total. The most current will be on the shortest run. For example, with runs of 48' and 52', 52% of the total current will be on the 48' run and 48% on the 52' run. If you have 3 sets of 31, 33, and 36', you will have 36% of the current on the 31' run, 33% on the 33' one and 31% on the 36' run. However as reno said, the code requires that the runs of parallel conductors be of equal length.
Don
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#167170 - 08/07/07 12:43 PM Re: Parellel Feeds [Re: resqcapt19]
earlydean Offline
Member

Registered: 12/22/03
Posts: 749
Loc: Griswold, CT, USA
Use ohm's law, I = E / R, first determine the resistance using Table 8 (or 9) in Chapter 9 of the NEC. In the longer runs, we determine that the conductors do not need to be exactly the same length, in the short runs you had better be very close.
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Earl

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#167174 - 08/07/07 01:39 PM Re: Parellel Feeds [Re: earlydean]
resqcapt19 Offline
Member

Registered: 11/10/00
Posts: 2209
Loc: IL
Earl,
There is no need to use ohms law and look up the resistances...the footage method provides the same answer. Much quicker and easier.
Don
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Don(resqcapt19)

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#167176 - 08/07/07 01:52 PM Re: Parellel Feeds [Re: resqcapt19]
RODALCO Offline
Member

Registered: 12/08/05
Posts: 863
Loc: Titirangi, Akld, New Zealand
If the cable has a reasonable length, more than 20 metres, 30 cm is not that much of an issue (1.5% of length).
The shorter cable will take a bit more current, heats up a little more, resistance increases I²R and will equalise currents with the longer cable.

Ideally they should be the same length (as per code).
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#167188 - 08/07/07 06:15 PM Re: Parellel Feeds [Re: RODALCO]
luckyshadow Offline
Member

Registered: 01/04/05
Posts: 305
Loc: Maryland USA
Thanks resqcapt19
Thats the formula I was looking for!
I know ohms law but knew there was another one.

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#167235 - 08/08/07 10:32 AM Re: Parellel Feeds [Re: luckyshadow]
earlydean Offline
Member

Registered: 12/22/03
Posts: 749
Loc: Griswold, CT, USA
Don,

Your formula is easy only if (like in your examples) the total length of the parallel paths add to 100 feet.

The question asked was how do you calculate the current flow, not how is the current proportioned.

But, it turns out your answer was what luckyshadow meant to ask.

You two must be on the same wavelength.
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Earl

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#167245 - 08/08/07 01:03 PM Re: Parellel Feeds [Re: earlydean]
resqcapt19 Offline
Member

Registered: 11/10/00
Posts: 2209
Loc: IL
Earl,
It works with any combination of lengths, but yes I chose easy numbers so I wouldn't need a calculator. It is always easier than ohms law because you never need to look up the resistance of the wire. Even to use ohms law, you need to know the total current to come up with the current on each of the parallel runs.
Don
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Don(resqcapt19)

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