Wire Diameter

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Apr 17, 2016
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Peterborough, Cambridgeshire
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Carthago Chic E Line
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Since March 2004
Is there a formula for working out the diameter that a wire should be? eg Battery negative to shunt. Battery positive to bus bar and then to inverter etc.
 
Not having a great deal of knowledge about the mysteries of electrickery, if the battery is 200 Ah does that mean that the positive terminal has 200 amps and then match the negative the same?
 
.if the battery is 200 Ah does that mean that the positive terminal has 200 amps and then match the negative the same?
Not exactly. The 200Ah is a measure of current and time. 200Ah (200 Amp hours) is the battery's (theoretical) capacity. To use your 200Ah figure it means that it could (in theory) deliver 1 amp for 200 hours or 50 amps for 4 hours, 10 amps for 20 hours.........etc. Not all of a battery's capacity will be usable because as it discharges the voltage will drop.
To calculate the size of a cable you need to know the max current in amps (amps = watts divided by volts) that the connected equipment will be drawing and the length of the cable run.

See points 1 to 5 in XxAnthxX 's post above.

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Starting to understand a little. I have a 216 Ah lithium going to an 1800W inverter. So accepting a loss of 3% on a 12V system, the amps (Watts divided by voltage i.e. 1800W (inverter) divided by 12v (being system voltage) gives 150 amps. Half metre cable run (maximum) it calculates 35mm. Does that seem too heavy? or have I got it wrong. I much appreciate the guidance given by you guys that have replied so far. Thanks for taking the time.
 
Edit: I originally posted the total length ie battery to the appliance and back to the battery needed to be entered, but then looking at the calculator in #3, they state battery to appliance only, so they must double that in their calculations.
 
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Starting to understand a little. I have a 216 Ah lithium going to an 1800W inverter. So accepting a loss of 3% on a 12V system, the amps (Watts divided by voltage i.e. 1800W (inverter) divided by 12v (being system voltage) gives 150 amps. Half metre cable run (maximum) it calculates 35mm. Does that seem too heavy? or have I got it wrong. I much appreciate the guidance given by you guys that have replied so far. Thanks for taking the time.
I'm not familiar with inverters and your calculation is probably right but my understanding is that the 1800W is the max output power that it can deliver to connected equipment (hair dryer, coffee maker etc.) at its output voltage of 240 volts, and is not the current that it draws from the battery.
I'll duck out now and leave to the inverter experts for further info. ;)
 
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Starting to understand a little. I have a 216 Ah lithium going to an 1800W inverter. So accepting a loss of 3% on a 12V system, the amps (Watts divided by voltage i.e. 1800W (inverter) divided by 12v (being system voltage) gives 150 amps. Half metre cable run (maximum) it calculates 35mm. Does that seem too heavy? or have I got it wrong. I much appreciate the guidance given by you guys that have replied so far. Thanks for taking the time.
That sounds about right although the cable could probably be reduced to 25mm2 but for such a short length I would go for the heavier.

I use this calculator but they are all probably similar but check the cable length as it is often the total length there and back.

 

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