Agree with:

* keep inverter as close as possible to the batteries to reduce voltage drop and to minimise costs of expensive large diameter wiring

* do not keep inverter in the same compartment as the batteries, although sealed AGM have very little acid air vented, but yours' are open (vented) lead acid by Trojan I suppose? The other side of a bulkhead is a good idea.

Do not agree with... or let me check:

* that amperage goes up when voltage goes down. Yes it looks that way when using the formula Amp (I)=watts/volts, but the wattage is variable and depends on the volts!!. In fact there is an resistance that needs to be taken into account and the formula. For instance, if the wattage is 700 at a voltage of say 12.5 Volt (as per specs of the unit). Then the formula is I=W/V, or 700/12.5=56 Amps. At that Voltage!! Then resistance is R=V/I=12.5/56=0.22 Ohm

Then when the voltage drops ie to 12 Volt, resistance remains the same, and therefore the amps go down to I=V/R=12.0/0.22=53.8 Amps..... yes, even if the resting voltage is 12.7 Volt) it will go down under such a load to 12 Volt or less.

In fact inverters are not 100% efficient and 'use' power. A good inverter and the best one I have seen is around 94% efficient (meaning a loss of 6%). Most have an efficiency of around 80-85%, so a loss of 15 to 20 %. So one could add these percentages (ie 6 to 20%) to the amps used.

If one is really pedantic, one can then calculate the real Wattage of the microwave at a lower Voltage. In this case at 12 Volt and 85% efficiency: 549 Watts. All figures rounded.

If you do not believe all the above, think of a lamp. When the battery goes flat (lower voltage) the light goers dim (lower amps).

Of course, if the voltage of the inverted AC voltage (110 Volts) remains the same, then Wattage remains the same, and the amperage has to increase dramatically to make up for the voltage losses. So 'Norah' was right, although intuitively it did not look that way.

Sorry, this was not really the question that was asked.

Clarification:

* previous posters used the term "large" cables. I am sure they mean "large diameter" cable.

Additional info:

* Generally cheap inverters have low efficiency rates, meaning have big losses, or/and use higher amps when in standby mode.

* additional switches or connectors in the 'large current' circuit introduce resistance, and power is lost in form of heat, resulting in reduced voltage or voltage drop.

* Similarly voltage drop goes up substantially by using undersize cabling

* generally when using 100Amps from a battery one has to put approx up to 120 Amps back in again due to battery losses

* Before installing equipment that draws large currents, ensure you have the charging methods all worked out, and can replenish the amp/hrs used.

my 2 cts