cyclone.1983

Just bought a new Rubicon, comes in Monday. Was curious, it has a plug in for electronics, or whatever and I was curious as to what it could handle? I assume it will handle my motor for my air mattress but can it handle larger things like my 20 inch LCD TV? I know nothing about power and inverters ect so I was curious.

Dronac

150?! That about $25 worth of inverter...

mvnolan

Wattage isn't even your real issue with most inverters. Check it's amperage rating. Most cheapo inverters are anywhere from 5-10 amps max. Not enough to run a vacuum cleaner, but maybe a 20" lcd. Check the ratings of the inverter, then check your lcd. Remeber, most electronics draw more amperage on startup than when they are actually up and running.

gundybears

If you know volts and watts, amps are easy to figure out. Watts = Amps x Volts.

Dronac

So in this case:

150 = Amps x 12
150/12 = Amps
Amps = 150/12
Amps = 12.5

gundybears

Yep. A 300 watt would handle 25 Amps and so on.

H315M4N

Don't forget on the 110VAC side it's only 1.3 amps. I=P/V= 150W/110VAC=1.36A. That's not even enough to charge a lot of laptops.

gundybears

True. Odd that vehicles that include 110V outlets are usually rated under 200W.

lacavol

You also have to check what it will run. Some inverters put out a waveform that will run a switching power supply but won't run motors and transformers. They are beginning to get better in quality though. Motors are really tough to start because of the inrush.

Mark Doiron

I wouldn't use the word "most", but you're right that some devices require more power at start-up. This is especially true for devices with motors, especially capacitor-start motors (common in refrigerators and such). However, power inverters also have an initial surge current that they permit without harm or malfunction to the device. They don't necessarily offset--some capacitance start devices may force that draw to exceed the surge time limits of the inverter.

This is true for DC circuits. However, with AC circuits you're actually calculating the apparent power, which is measured in volt-amperes (or kilovolt-amperes, often abbreviated KVA). In a simple analysis, in order to calculate the real power draw you'd need to know the power factor (which is the cosine of the phase shift between the current and voltage). What's going on is that perfect AC current has a sinusoidal waveform (and perfect AC current likely isn't coming from your inverter, but that's a story will defer for the moment). If the powered device is a perfect resistor with no inductance or capacitance (essentially, though not exactly, a light bulb), then the current and voltage stay in phase. The power factor is 1 (close to it for a light bulb), and power always flows to the load (light bulb). But, make the circuit have a little capacitative or inductive loading, and the voltage and current shift away from each other. It is this shift that is measured in degrees to determine a theoretical power factor.

It's actually a bit more complicated than that, but an important consideration is that a power factor of around 0.7 is commonly used by engineers as a "WAG" for complex AC circuits (please don't go using that number--it can be anywhere from 0-1 in the real world). So, while the rudimentary calculation you propose would give a 120-watt value for a 10-amp draw on a 12-volt circuit (12-volts from the battery times 10-amps on the device label), that is not what's really needed. The actual power draw is 84-watts (the preceding times the 0.7 power factor).

Unfortunately power factor is not explicitly stated on devices. If you measure with a wattmeter and ammeter, you can calculate backwards into the power factor, but what's the point (since you already know the watts required from measuring it). Well, if you combine multiple devices with different power factors, it quickly becomes complex. But if you want to dive right into math with imaginary numbers (remember √-1=i, or j to electrical engineers?), you can calculate the expected, real power draw.

Yes. It's easier and cheaper to make an inverter that puts out a square wave than a modified sine wave, or, best yet, a true sinusoidal wave. That's why some inverters are much more expensive: They have the additional components and engineering effort to put out a proper waveform. The shape of the waveform is also another factor in determining how much power a device will actually draw: A true sinusoidal waveform is the most efficient.

Bottom line: If you use the rudimentary calculation of power draw outlined by Dunemobbin, you'll have an inverter that's oversized for your needs. So, you should be fine, except you may have spent a little more money perhaps than was necessary.