if i have a draw of 3000 watts on the inverter and my pump kicks on will have the 4500 watts needed to start the pump?
1596844
It's kinda hard to say for sure, but I don't believe you should have any problems once your system is set up properly by that I mean proper wire sizes and all that and good batteries and providing the does not draw that for very long and also does not draw too close to start up power to run, make sure it's in proper condition before you put it to work.
1596845
You need to determine your amp draw on the equipment. Remember that start-up amp on most compressors will be almost time and a half the normal running usage amps.
1596846
Gary,If your pump requires the 4500 watts for more than a split second it will not start.Joe
1596847
Although it is not easy to know this without a test, I was running a 1000 watt compressor and welded with a 90 amp welder without crashing the inverter. It was quite a drain but it didn't quit. I'm sure your pump would start especially if it only draws the startup wattage for a second before dropping down to running wattage. I do have #1 wire on all 4 posts and a bank of capacitors on the battery side. It is vital you use the largest and shortest wires possible for the high amp draw.
136850
Sorta. AC watts output and DC watts input are not quite the same thing. I would be thinking "amps" instead of "watts". First, this is a modified sine wave inverter. It will work on most appliances, but not all. 3,000 AC watts in appliances is within specs for this inverter. However, if you think one or two batteries connected to it will work...I wouldn't even try. To run that much power, you are going to need at a minimum of 4-6 batteries wired in parallel (12v) with very large battery cables (0-gauge or thicker). At 3,000 AC watts this inverter will likely draw over 250 amps from the batteries. A typical deep cycle marine battery will have 100-125 amp/hours. When you parallel, you add amp hours. Two batteries (200-250) or four batteries (400-500 amp hours). Running 3,000 watts AC out of this inverter is possible, if you have enough batteries, thick battery cables, and something to recharge them...cause you will drain them QUICK. Your battery cables must be rated to handle this much amperage. Download the Power Bright Owners Manual, call them, or do a Google search...be safe. Good luck.
937543
Sounds like the 3000 should work fine! I've found this inverter to be very capable!!
686896
Full power would be speed setting 3, which my kill a watt meter measured to consume about 35 watts. Assuming you have consistent 120 volt outlets (sometimes they are a bit off, but that should not effect calculations too much), we get p=i*v. Plugging in: 35=i*120--&gt; i=.3 amps. Again, using Tony's answer, worst case scenario the initial amperage pull when starting would be 1 amp, then it'd settle down to .3 amps. But when starting the motor I've noticed that it takes a few seconds to get up to speed so the initial pull shouldn't be anywhere near worst case.
222843
Amps are not Amps. WATTS are what need to be conserved. If you have 10000 Watts of AC available on the output, it will require OVER 833 Amps on the input. If fact, it will require even more than 833 Amps as a result of the efficiency (lack of perfect conversion) of the inverter. I'd bet the overall conversion eff. to be less than 80%, which means it would be 10000 Watts/0.8 = 12,500 Watts, and for the sake of easy math 1000 Amps @ 12.5 Vdc input. Thats some serious current and is going to be a real problem when attempting to minimize the voltage drop on the input cables! (Which when it occurs will result in even more current being required for the same number of Watts! Amps are not Amps. WATTS are what need to be conserved. If you have 10000 Watts of AC available on the output, it will require OVER 833 Amps on the input. If fact, it will require even more than 833 Amps as a result of the efficiency (lack of perfect conversion) of the inverter. I'd bet the overall conversion eff. to be less than 80%, which means it would be 10000 Watts/0.8 = 12,500 Watts, and for the sake of easy math 1000 Amps @ 12.5 Vdc input. Thats some serious current and is going to be a real problem when attempting to minimize the voltage drop on the input cables! (Which when it... » Read More Amps are not Amps. WATTS are what need to be conserved. If you have 10000 Watts of AC available on the output, it will require OVER 833 Amps on the input. If fact, it will require even more than 833 Amps as a result of the efficiency (lack of perfect conversion) of the inverter. I'd bet the overall conversion eff. to be less than 80%, which means it would be 10000 Watts/0.8 = 12,500 Watts, and for the sake of easy math 1000 Amps @ 12.5 Vdc input. Thats some serious current and is going to be a real problem when attempting to minimize the voltage drop on the input cables! (Which when it occurs will result in even more current being required for the same number of Watts! « Show Less
1319299
No Way. the peak load for an AC unit is probably (3) times the running load (which already exceeds the 500 watt rating of the inverter).
2154616
I don't know. I don't have such an inverter. I seriously doubt that the battery would produce many watts for very long anyway.
65462
It depends on how long you want to use the power inverter and how many watts per hour you are going to use and will you be charging the battery at the same time or just use a battery without charging .The formula is (amps x 120 v = watts X hour) amp draw should be listed on tag by the model number
1510645
No. That air pump draws 1.45 amps, or 120v x 1.45a = 175 watts. This inverter would go into overload protection mode and shut down.
686897
Full power would be speed setting 3, which my kill a watt meter measured to consume about 35 watts. Assuming you have consistent 120 volt outlets (sometimes they are a bit off, but that should not effect calculations too much), we get p=i*v. Plugging in: 35=i*120--> i=.3 amps. Again, using Tony's answer, worst case scenario the initial amperage pull when starting would be 1 amp, then it'd settle down to .3 amps. But when starting the motor I've noticed that it takes a few seconds to get up to speed so the initial pull shouldn't be anywhere near worst case.
2165852
Thanks for your inquiry. No, this unit offers 720 watts from it's battery before it's depleted. If you have a pump that can use 720 watts in 3 hours then this would be for you.
1714715
I would, 2000 watts @ 12 volts = 166.6 amps. The hot leg of any circuit (AC or DC) should always be protected with a fuse or circuit breaker. If you don't have a manual on your inverter search the internet to try to find its spec sheet to find out its max load in amps then size your fuse (or breaker) just slightly larger then it's max amp load
303305
Assuming there is a constant 12 volt direct source, the pump is going to draw 3 to 5 amps. amps x volts = watts. At 5 amps that would be 60 watts. With a battery in line it would work. Straight off a 50 watt solar panel probably not so well.
1990303
Normally you will have to figure out how much power your machine or device will take and then go from there . in order for the size of inverter you need
1979311
probably will. Pumps are high inductive loads that pull a lot of power to start up. You might also consider a 12 volt bilge pump connected directly to the marine battery. Will require a bit of plumbing, but more energy efficient since you don't have to convert voltage. Also, you could set up the bilge pump with it's own float switch so it comes on automatically. What if you are not home when the power goes out? Who will move the plug for the 120 Volt bilge pump from the wall to the inverter?
172599
Check to see what the peak start up surg of your mower is. Sounds like it may be pretty high if your normal running watts are 1400. It should run as long as you don't exceed the surg capabilities of inverter. Also be sure you have the heavy enough cables on inverter side to carry the amps needed to support that type of load.
2022153
I promise you, this inverter cannot handle 200 Amps. It's output rating is around 4 Amps (400w/110v=3.6A) at 75% efficiency, the input current would be around 45 Amps (400w/12v=33.3A /0.75=45A).