In example 1 series you have 2 12 inch leads out of the pack.
In example 2 parallel you use all the leads to connect them together and the
12 inch leads out of the pack are missing,  They would add .001 ohm each if
the same size wire which is a second 40 watts.
david

-----Original Message-----
From: EV [mailto:[email protected]] On Behalf Of Lee Hart via EV
Sent: Tuesday, June 07, 2016 11:24 PM
To: Larry Gales; Electric Vehicle Discussion List
Subject: Re: [EVDL] Off-grid solar house and electric car charging

Larry Gales via EV wrote:
> Thanks, I was somewhat aware of the increased use of copper, but not 
> to the extent that you specify, so it looks like AC is the way to go, 
> even for off-grid solar.

Lower voltage means higher current and bigger wires; but it's not as bad as
you think.

First, consider a motor or transformer. You would think that winding it for
a lower voltage / higher current would require more copper... but it
doesn't. Motors and transformers are exactly the same size, have the same
efficiency, same power rating, and use the same amount of copper no matter
what voltage they are built for.

Here's why: If you halve the voltage, you double the current (to get the
same power). But half the voltage requires half the turns. So the wire is
twice as think, but half as long. The total amount of copper thus stays the
same. This only breaks down if the voltage is so low that you need less than
1 turn, or if the voltage is so high that excessive amounts of space are
taken up by insulation instead of copper.

Now consider a pair of identical 12v batteries. You can wire them in series
(24v), or parallel (12v). For the same power, you'll have the same current
in each battery (since their voltages are all the same). 
So, the same wire size to every battery. For the sake of argument, let's
assume you connect a 12" piece of wire to every battery post, and it has
1 milliohm of resistance.

If they're in series, you have a total of 4 feet of wire total, all in
series, and so 4 milliohms of resistance. if the load is 24v at 100 amps,
then this 4 milliohms is burning up I^2R = 100^2 x 0.004 = 40 watts as heat.

If they're in parallel, the free ends of the + wires connect together, and
the free ends of the - wires connect together. Now you have two parallel
strings, each with 2 feet of wire in it; so each string has half the
resistance or 2 milliohms. But there are two of these strings in parallel,
so the total resistance is 1 milliohm. The same load power is 12v at 200a.
I^2R losses are 200^2 x 0.001 = 40 watts.

Exactly the same size and length of wire, and exactly the same losses!

The same thing happens with PV panels, power semiconductors, and just about
any power devices. Arranging them for low voltage/high current results in
the same losses as arranging them fro high voltage/low current.

The only time high voltage helps is when you need to have long wire runs. If
your PV panels are far from your inverter, then high voltage for the wires
between them will the reduce the amount of copper needed and/or lower your
losses. However, if you're using small low-voltage individual inverters
mounted right on each panel to one big central inverter located far away,
then the small inverters can "win" and use less copper overall.

You have to carefully consider the specifics of the situation, and not make
snap judgements about low voltages being automatically worse.
--
"IC chip performance doubles every 18 months." -- Moore's law "The speed of
software halves every 18 months." -- Gates' law
--
Lee Hart, 814 8th Ave N, Sartell MN 56377, www.sunrise-ev.com
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