photonphysicist Loc: Texas

I like that it has USB and a "wall plug".
I can use my Canon battery charger with this.
https://mycharge.com/collections/adventure-series/products/adventureultra

Harvey Loc: Pioneer, CA

Great tip and advise - I have kept one of this ($40) type inverters in my car for years to use for my laptop, battery chargers and lighting for several years- nothing like having it when miles and hours from a wall plug.

Rongnongno Loc: FL

That can be useful when shooting long time lapse animation outdoor, in the middle of nowhere. Small size ans seemingly powerful. (45w) Would have to check for the power requirement of a Nikon camera brick first - using the 110v output.

Harvey Loc: Pioneer, CA

I have not thought of using it as a power source for my 110 converter to my DLSR out in the field - I just bought one for my long indoor macro and close-up test/practice.
Thanks for the tip. - it works well for the items I mentioned.
Harvey

JD750 Loc: SoCal

I would be wary. Read user reviews.

45 watts is the peak output but for how long? Notably missing was a watt-hr or amp-hr rating. That is a pretty important specification for a battery. Why is it missing?

Then there is the matter that annoying web site pop up offering 10% off and no way to dismiss it. Personally, I will never buy from a company that pulls that stunt on their web site. But that is just me.

photonphysicist Loc: Texas

Thanks for the assessment. you've presented a few things I had not considered - but will.

aphelps Loc: Central Ohio

They claim the battery has a capacity ov 13400mAh. That's13.4 Ah. Input to the inverter. The inverter converts battery dc volt to ac 110 volts but there are losses. If you estimate 80% efficiency (typical) that means you will get around 10 Ah at output. If peak power is 45 watts, running power will be less....perhaps on the order of 38 watts. Output of 38 watts at 110 volts is about .35 amps. Dividing output of 10 Ah by .35 A gives you about 26 hours of run time at that load. Now I've made a bunch of assumptions to get that....your actual mileage may vary!

photonphysicist Loc: Texas

V=IR, P=IV

burkphoto Loc: High Point, NC

In English, please define the variables...

Rongnongno Loc: FL

Voltage = Resestivity (Ohms) * Intensity (Amp) (V=RI)
Power = Volt * Amp (P=VI)

Therefore P = Resestivity * amp squared as Volt = RI so RI*I = RI^2

'Ohms law' This is the base for all electrical circuit, even the more advanced.

I do not understand why the US is still in 110v as so much power is lost in the wires it is not even funny, this and the heat/safety problem.

Think of a bulb that is 100w at 100v it needs 1a. Nothing with wrong with that, right?
Now use 100w=100v*1a This tells us the R = 100 Ohms.
Raise the voltage to 200v. To get 100w Ohm must be 0.5A

So far no saving, right?

Use formula P=RI^2 Still the same power, right.

How about the wires in the house? They have a restivity calculated on their section and their length. Let's say for simplicity's sake that they are 1 ohm.
How much power is lost in the first case? P=RI^2 so 1w is lost. Small number right?
Now try with 200v A is equal to .5 so P=RI^2 = 0.25W lost.
Since there is less power lost there is also less heat in the wire, safer and more economical.
How much power is your house losing it the wiring????

The US wants to save energy? Start with that.

(Sorry venting)

In french we use U.

JD750 Loc: SoCal

Thank you for that.

“Caveat Emptor”.

JD750 Loc: SoCal

Where was that claim of 13400 mAH? I did not see that.

Ok your math is good. I agree with your analysis.

burkphoto Loc: High Point, NC

Thanks!

I agree, 240-volt service is more efficient...