DonC said:approx how many amps will 10 minutes of a small microwave consume? In summer heat or winter cold, my frig or heater takes my batteries down to 70 - 75% each night. Any lower than that from running a microwave and I worry about my 200W of solar being able to fully recharge the batteries by the end of the day, especially if there is a cloudy day.
A small 600 watt microwave isn't really a 600 watt load on your inverter. That 600 watts rating is cooking power, In reality a small microwave draws about 1000 watts from the power source. To figure out approximately how much current is being drawn by your inverter, let's presume it's 100% efficient (it isn't, but close). We'll also presume you have no voltage loss in your cable run to the inverter. To calculate current drawn from your batteries, divide the watts by your battery voltage. Let's say your batteries are at 13.2 volts and your inverter is drawing 1000 watts. 1000/13.2 = 75.75 amps. 75.75 amps a significant amount of current (why you need large wire for the inverter), but it's only being drawn for a couple minutes. Once the inverter is off, the batteries will recover most of their capacity in a few minutes. If you're going to run an inverter, I would suggest at least 2 100 AH batteries in parallel (YMMV).DonC said:approx how many amps will 10 minutes of a small microwave consume? In summer heat or winter cold, my frig or heater takes my batteries down to 70 - 75% each night. Any lower than that from running a microwave and I worry about my 200W of solar being able to fully recharge the batteries by the end of the day, especially if there is a cloudy day.
K6ON said:A small 600 watt microwave isn't really a 600 watt load on your inverter. That 600 watts rating is cooking power, In reality a small microwave draws about 1000 watts from the power source. To figure out approximately how much current is being drawn by your inverter, let's presume it's 100% efficient (it isn't, but close). We'll also presume you have no voltage loss in your cable run to the inverter. To calculate current drawn from your batteries, divide the watts by your battery voltage. Let's say your batteries are at 13.2 volts and your inverter is drawing 1000 watts. 1000/13.2 = 75.75 amps. 75.75 amps a significant amount of current (why you need large wire for the inverter), but it's only being drawn for a couple minutes. Once the inverter is off, the batteries will recover most of their capacity in a few minutes. If you're going to run an inverter, I would suggest at least 2 100 AH batteries in parallel (YMMV).
Camelracer: Dick, the only reason they don't want you to run an inverter through your controller is the wire size is too small and you'll let all the smoke out.
Bombsight: A 3000 watt inverter is overkill, a 1500 watt inverter will cover pretty much anything you'll run in your camper.
Thanks for that input. I really didn't want to drop a grand on something I will seldom use. I think instead, I'll get the 2000 watt Go Power.K6ON said:Bombsight: A 3000 watt inverter is overkill, a 1500 watt inverter will cover pretty much anything you'll run in your camper.
When filling a tire with an air tank and watch the air pressure gauge you will notice that the air pressure drops when you put the chuck on the tire and rebounds when you remove it. How far it rebounds depends on how much air you take out of the air tank. The same differential equations govern an electric system: pressure ≈ voltage, air volume ≈ amperage. Take a little air/amps out, pressure/voltage will rebound to almost where it was before starting. Take a lot of air/amps out and the pressure/voltage will not rebound as much.Wallowa said:. . . you stated that the batteries will "recovery most of their capacity in a few minutes"...you lost me there; what are the assumptions? . . .
JaSAn said:When filling a tire with an air tank and watch the air pressure gauge you will notice that the air pressure drops when you put the chuck on the tire and rebounds when you remove it. How far it rebounds depends on how much air you take out of the air tank. The same differential equations govern an electric system: pressure ≈ voltage, air volume ≈ amperage. Take a little air/amps out, pressure/voltage will rebound to almost where it was before starting. Take a lot of air/amps out and the pressure/voltage will not rebound as much.
Same equations govern electrical, hydraulic, & pnumatic circuits. As a mechanical engineer, when confronted by an electrical circuit that I didn't understand I would convert to the hydraulic analogs because I understand those circuits better.
jim