I am installinga 4548 inverter. At 4500 watts, it puts out approximately 20 amp per phase. My generater that will feed the charging current for the batteries will put outapproximately 28 amps. Is the output of the inverter now 48 amps?

Here's a site that gives formulas for calculation. Looks like you need some other info.
I'll don't do math -


https://www.rapidtables.com/calc/electric/Amp_to_kVA_Calculator.html

Please see below courtesy of engineeringtoolbox.com

Electrical power is in general rated in watt (W) or horsepower (HP). A horsepower is a unit of power equal to 746 watts or 33000 lb ft per minute (or 550 lb ft per second).
A watt is equal to the power produced by a current of 1 amp across the potential difference of 1 volt. A watt is 1/746 of 1 horsepower.
Even if the watt is the base unit of electrical power, it is common to rate motor power in either horsepower or kilowatts.

Power in Watts

Direct Current Electric power supply to a direct current (DC) motor:
PkW = ?m U I / 1000 (1)
where
PkW = power (kW)
?m = motor efficiency
U = voltage (V)
I = current (A, amps)
Electric power supply to a alternating current (AC) motor:
Single Phase PkW = ?m U I PF / 1000 (1b)
where
PF = Power Factor
Two Phase Four Wire PkW = ?m 2 U I PF / 1000 (1c)
Three Phase PkW = ?m 1.73 U I PF / 1000 (1d)

Power in Horsepower

Horse power can be expressed as:

Php = PkW / 0.746 (2)
or
Php = (?m U I / 1000) / 0.746
= ?m U I / 746 (2b)
where
Php = horsepower (hp)

Yes
Voltage and wattage printed is on end of each element
If element shows 240V 4500 watt, then element Wattage at 208V is reduced to 3375 watts and will heat 16+ gallons per hour
If element is rated 208 and 4500w, then element will operate at 4500 watt and heat 20 gallons per hour
http://waterheatertimer.org/Figure-Volts-Amps-Watts-for-water-heater.html

http://waterheatertimer.org/images/Whirlpool-energy-smart-watt-rating-IMG_2600.jpg

http://waterheatertimer.org/How-to-replace-water-heater-element.html

Gene

If you need further help, I’m available over the phone at https://www.6ya.com/expert/gene_9f0ef4df2f9897e7

If you have (90) 50w lamps = 4500 watts total. Assuming a 120/240 panel, if you put 1/2 on one "side" of the panel and the other 1/2 on the other "side" of the panel, that would be 2250 watts on each half. The generator should be rated *at least* 125% of the load; 4500w x 1.25 = 5625W. Using a 4500W generator on this load will cause it to overheat and shorten its life as it is running at 100% of capacity all the time..

One half of the panel is 120V to neutral, and the other is 120V to neutral - or 240V between both circuit breaker terminals. Ohms law for DC circuits and purely resistive AC circuits says Volts x Amps = Watts; or Watts / Volts = Amps. So, 2250W / 120V = 18.75A on each pole of a 2 pole circuit breaker that feeds the sub panel. A #12 copper wire is rated for 20 amps; but as per National Electrical Code - must be de-rated to 80% of rating which means it is good up to 16 amps maximum. A #10 copper wire is rated for 30 amps, but it too must be derated to 80%, making it good for 24 amps maximum. So, if you are going to feed a sub panel supplying (90) 50watt lamps, you will need to run a #10/3 copper cable from a two pole 30 amp circuit breaker at the generator to a 120/240 volt "main lug only" sub panel rated for at least 30 amps.

Divide your load evenly across the sub panel - (4) 15 amps circuits via (2) two pole 15 amp circuit breakers on each "side" of the panel if you run (2) 14/3 cables out to the lights - or (4) single pole 15 amp circuit breakers if you run (4) 14/2 cables out to the lights. No circuit breaker terminal should have more than 23 lamps that means you have (2) w/ 22 lamps and (2) with 23 lamps. The circuit w/ 23 lamps will draw 23 lamps x 50w = 1150W. 1150W / 120V = 9.6A. The 22 lamp load will be 22 x 50w = 1100W. 110W / 120V = 9.2A. Which is well within the 12A maximum allowed (after derating as required by code) by a #14 copper wire rated for 15A.

Good luck!

I'm sorry that I can't help you, but if you're in a hurry, you can visit the Magnum website at: http://www.magnumenergy.com/Service/Servicefront.htm and try to find your model's online manual and search troubleshooting advice or you can email customer service at
[email protected] and they should at least be able to lead you in the right direction.

A few questions: Does this problem remain even when there is at least a 10+ watt load on your system? is your Magnum in a hot environment? Does your inverter feel extraordinaryily hot? Have your batteries outlived their lives? Have you cleaned all the connections? Even if they 'look' like they don't need it?

All the best to you.

Hello, First figure out what the load per leg is on your generator. Are you using the 240 Volts because this is two 120 AC volts legs and they are 120 degrees out of phase with each other.

You will need to figure what the power factor is for each phase on this generator. To figure the power/wattage by using ohm's law. Voltage times Current will give you Watts. Example: If the generator is generator 240 volts the maximum current available is about 23 amps. The maximum current for 120 volts will 46 amps. Therefore, the maximum current per 120 volt circuit is 23 amps for one circuit and 23 amps for the second 120 circuit..

I would try replacing the circuit breaker with the same amperage breaker. Also, watch what the generator load is per circuit. You can install current meter for each circuit. This will give a good indication of which circuit is pulling more load than the other circuit. GB...stewbison

if the generator has been sitting for a while you might have to excite the magnets to start producing power. plug in a reversible drill and with the generator running turn the drill in the reverse direction by hand while holding the trigger in the forward position.

Does your generator is single phase generator set or three phase four wire generator? Why your generator set is 110V at full load, it should be 220V or 240V to generator electricity. If it is used for industrial equipment, it should be 380V, 400V, 440V, etc.
Generator technical support: https://www.dieselgeneratortech.com/support/

If you want to get more precise, figure out everything in terms of power (watts).

Basic electrical rule 1, 2 and 3:

voltage x current = power

or re-arranged:

current = power divided by voltage

or re-arranged:

voltage = power divided by current


For example, 12V X 2 amps = 24 watts.

or another example, 400 watts divided by 120 Volts = 3.33 amps

A 55W headlight that uses 12V would draw 55 /12 = 4.6 amps @ 12V

A 55 watt light bulb in a lamp at home would draw 55 / 120 = 0.46 amps @ 120V


As the previous post mentioned, inverters are not perfect when convertering 12V into 120V. If the converter consumes 1000W from the 12V battery, then a 90% effecient converter would generate 900W of 120V AC power best case. The other 100W is lost primarily as heat.

The other thing that gets tricky is that these ratings and the formula above are used for resistive loads, like light bulbs or hair dryers. Anything with a motor or transformer is considered an inductive load and can get much more tricky to calculate.

Consequently you need to give your self a safety margin when figuring out how big an inverter you need.

How does work in a practical sense?

Lets say you want an inverter for TV, DVD and Sat. Receiver. Look at the back of TV or in the manual. It should say how many watts it consumes. Lets say it is 400W. The DVD might be 100W and the Sat. receiver 50W - just as an example.

400 + 100 + 50 = 550 Watts. (just as an example)

You might think, well no problem, I'll use a 600 Watt inverter and have 50 watts left over. Depending on your inverter, that 600W might really be 600 x 90% effecient = 540 Watts of AC, less a 20% margin of error for the inductive transformers in the electronic of the TV, DVD and Sat. receiver 540 - 20% = 432 Watts.

Now you can see your 600 Watt inverter isn't big enough to do the job.

If we really need 550 watts of AC, add 10% to make up the effiency loss, then add a safety margin for inductive loads.

550 + 10% = 605 + 20% = 726 Watts.

Sounds more like an 800W inverter fits the job.

What does that mean in terms of wiring the 12V batteries to the inverter?

from the formula above:

current = power divided by voltage

In our example, we have an 800W inverter that runs on 12V

The current would thererfore be:

current = power divided by voltage
current = 800 watts divided by 12V
current = 66 amps.

That is important info because you can not use light gauge wire to carry 66 amps worth of 12V to the inverter nor could you use a 20A fuse to protect your inverter.

Now that's a lot of science for a guy who just wants to run a toaster on an inverter right?

800W / 120V = 6.66 amps

Using garryp's ratio 11:1, 6.66 x 11 = 73 amps.

That is a good ratio with a good safety margin.

This is all just MHO and should not taken as solid technical advise. In other words, don't blame me if you blow yourself up.