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[Clansgian AT wmconnect.com]

> I get confused when words like
> resistance and amperage get thrown around.

Bev, here's the crash course:

voltage is the measure of electrical force (potential)
amperage is the measure of electrical flow (quantity)
wattage is the measure of electrical power

wattage = voltage x amperage

A 100 watt lightbulb connected to 100v (rounded down to simplify) AC house 
current is using 1 amp of current.

100 watts = 100 volts x 1 amp

Resistance is the 'load' on the circuit, it takes the "pressure", the 
potential, of one volt to make one amp flow through one ohm of resistance.

So our above lightbulb has a resistance of 100 ohms.

100 ohms = 100  volts / 1 amp

This is germane to the alternate energy discussion in that we are most often 
generating our laternate energy at a low voltage, 6, 12, or 24 volts DC and 
most of the devices we want to power require 120 volts AC.  A lot of things 
happen when we try to store, convert, and transport electrical energy.  For 
example, if you have a device that requires 120 v AC and it requires a 
continuous 
draw of 3 amps and if your alternate source generates 12 v DC, let's ignore 
losses for the moment and thus the power is constant.

Power (of the source ) = 12 v X some amps.
Power (of the device ) = 120 v X 3 amps.

So:

12 v X some amps    =    120 v X 3 amps

some amps = (120 x 3)/12
 
some amps = 30 amps

You'd need a source capable of delivering 30 amps ( a lot!).

When you convert from DC to AC (or vs versa) you lose some of your power.   
Transmission lines and cords have their own resistance (load) and so they use 
up some power.  Line will lose power as the resistance goes up and the 
voltage 
goes down, this means that a given wire will lose a greater percent of its 
power delivering 30 amps @ 12 v than it would delivering 3 amps @ 120 v.   
The 
line losses for DC are greater than for AC.

This means, for example, that in the field running a line from your truck 
battery to your astronomy site to an inverter will lose a much greater  
percent 
of the available power from the battery than inverting the power at the 
battery 
and running the resulting AC at a higher voltage to the site.

----------------------------------------

Like all other analyses, such as we've suggested for canning, just a little 
math helps put things in persepctive.   Can you save money (or power) 
recharging, say AA NiMH batteries with a solar charger?   A full charge is 
typically 
2200 milli-Amp-hours at 1.25 volts.  But if we are charging from house 
current, 
it is 120 v (roughly 100 times the voltage).  So our above formula tells us 
that it will take 1/100th of the current for a full charge (ignoring losses 
for 
the moment), or 2.2 thousandth's of an amp for an hour, that is, about a 
quarter of an watt-hour.   (.0022 amp x 120 v).  If electricity costs 10 
cents kWh, 
you charge that battery for about 1/400th of a cent.

If you are living completely off the grid or far afield from your power 
source, charging your batteries vs not charging them is a big thing.  But if 
you 
are already on the grid, the use of solar to charge small batteries is 
meaningless except for an amusement.

On the other hand, doing the same analysis for, say, an electric dish washer 
shows that doing dishes by hand saves a huge amount of resources (running the 
washer, heating the water, using more water) etc.  

Refraining from using is an absoulute trump in the alternate energy game.