Fridge wiring requirements

Submitted: Wednesday, Apr 02, 2008 at 00:45
ThreadID: 56174 Views:2422 Replies:4 FollowUps:1
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I have been doing alot of research, and just a little confused and hoping for a simple response. I am wiring up for a fridge (Engel 40L) outlet, and have 6mm wire or 4.58mm2? Is this sufficient? All coming off a 2nd battery, just wire size is question.
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Reply By: Derek from Affordable Batteries & Radiators - Wednesday, Apr 02, 2008 at 08:28

Wednesday, Apr 02, 2008 at 08:28
For no more than 5m yes it is fine.

I would recommend 6mm2 but the 40L Engel will be fine.
AnswerID: 296088

Reply By: Keith_A (Qld) - Wednesday, Apr 02, 2008 at 13:33

Wednesday, Apr 02, 2008 at 13:33
Hi Gold - the main issue is voltage drop between the battery and the fridge, and it depends on 3 things - current draw; length of cable and size (cross section) of copper in the cable.

Here is the web site that will help : Collyn Rivers and look at the article on wiring woes and also fridges.

.................Regards..........Keith.
AnswerID: 296134

Reply By: Gold 4.2 Patrol - Wednesday, Apr 02, 2008 at 20:21

Wednesday, Apr 02, 2008 at 20:21
Awesome!! Thanks guys, good help, and great link. Thanks again
AnswerID: 296208

Follow Up By: Member - Phill E (NSW) - Wednesday, Apr 02, 2008 at 21:11

Wednesday, Apr 02, 2008 at 21:11
colly rivers is the go.. copy of it below here.

There is 'rounding up' discrepancy in most published AWG/ISO conversion tables (not in mine). Use one AWG gauge size larger (ie. numerically lower) than the conversion shows. For example, ISO 2.5 is AWG 13 (not 14 as usually shown) etc. Using ISO gauges is easy. The actual gauge number (eg. 1.5, 2.5, 4.0, 6.0, 10 etc) is the actual cross sectional area of the conductor, in sq.mm.
Another very easy way to work out the voltage drop for any given length of cable, current flow, and cable size is the simple formula shown below. I always use this in conjunction with my published tables. This formula is simply:
Voltage drop equals (cable length (in metres) X current (in amps) X 0.017) divided by cable cross-section in mm.sq.
For example: 10 metres X 5 amps X 0.017 = 0.85. Divided by (say) 2.5, the voltage drop is 0.34 volt. This is just acceptable.
The above, and wire-tables indicate voltage drop across a single conductor. For chassis (earth) returns, the resistance of the return path can be ignored. Where there's another conductor for the return path, the total cable length must be taken into account. In other words, if there's a separate conductor for the earth return that's ten metres of cable, so you do the sum as if it were ten.
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FollowupID: 562290

Reply By: wheeleybin - Thursday, Apr 03, 2008 at 18:03

Thursday, Apr 03, 2008 at 18:03
Collyn uses 0.36V drop as the basic acceptable for 12V.
5M of 4.5MM2 at 12V with a 3A load will give a voltage drop of 0.102 Give a terminal voltage at 11.898V so you have some up your sleeve.
10M of 4.5MM2 with a 3A load will give a voltage drop of 0.203 and give a terminal voltage of 0.203 so you have some up your sleeve.
The Amp load is the critical factor.
AnswerID: 296424

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