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#170854 - 11/13/07 05:15 PM Low voltage line loss
bwise121 Offline
Member

Registered: 09/11/03
Posts: 113
Loc: Sacramento, CA USA
I'm running some long 12 Volt circuits to rope lighting and I'm needing to figure out the voltage drop. I have already put in 12-2 12 volt landscape wire. I need to find out what the resistance for this wire is. It seems to have more strands than stranded #12 so I don't think that Ohm value is what I want.

Any help?

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#170863 - 11/13/07 06:04 PM Re: Low voltage line loss [Re: bwise121]
Trumpy Offline

Member

Registered: 07/05/02
Posts: 8540
Loc: SI,New Zealand
Do you have any loss figures for the wire itself?
In that, usually there are stated losses for a given size of wire given as mV per Amp per metre (foot?) this is then used to calculate the total voltage drop of the circuit.
BTW, if you are trying to find the resistance of that wire, you'll be needing a low-ohm meter or a really accurate multi-meter.
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#170864 - 11/13/07 06:12 PM Re: Low voltage line loss [Re: Trumpy]
bwise121 Offline
Member

Registered: 09/11/03
Posts: 113
Loc: Sacramento, CA USA
I have no intention of using a meter to calculate it myself. I know there are tables for cross-sectional are of conductors but like I said this wire is stranded with probably twice as many strands as #12 THHN stranded.

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#170865 - 11/13/07 06:16 PM Re: Low voltage line loss [Re: bwise121]
bwise121 Offline
Member

Registered: 09/11/03
Posts: 113
Loc: Sacramento, CA USA
I guess the question I'm really needing answered is does the voltage drop formula differentiate between stranded and solid wire.

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#170867 - 11/13/07 06:40 PM Re: Low voltage line loss [Re: bwise121]
bwise121 Offline
Member

Registered: 09/11/03
Posts: 113
Loc: Sacramento, CA USA
http://www.generalcable.com/NR/rdonlyres...dscpLghtgWr.pdf

That is about as close to specs as I can get. It is one of the #12 wires toward the bottom. I still don't think any of that information will help.

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#170877 - 11/13/07 07:32 PM Re: Low voltage line loss [Re: bwise121]
bwise121 Offline
Member

Registered: 09/11/03
Posts: 113
Loc: Sacramento, CA USA
Something isn't right. Can someone advise me on this.
Ed = (12*I*L*2)/CMA

Ed is allowable voltage drop
I is amps
L is length in feet
and CMA is Circular Mils

In this case:
I = 13.3 (8 20 watt bulbs 12Volt)
L = 40'
CMA is 6530 for #12

This means that Ed is about 16%(1.955 volts).

I know from experience we have put many out door lights on a transformer and have had no problems (watts ? can't remember). What gives?

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