pierlux said:
Someone better skilled than me in physics please chime in and explain better, I'm sure there is. But we're going a bit off-topic here, I think.
I think the point BruinBear was trying to make was that you are
conflating charge capacity with current flow. The term mAh is a measure of charge capacity over a period of time, and is not synonymous with mA, the actual current. The only time you would actually draw 27.195W is in the first moment of usage assuming you maximize the cameras' power draw for a moment. Actual voltage drops a little from the rated voltage, so on average you might, at full draw for some camera that actually needs 2.45amps, pull say 10.8V at 2.45A, for a power draw of 26.5W. I do not know of any reason you would be limited to 2450mA maximum current, however. Assuming you drained the battery in 30 minutes at 10.8v, you could draw ~5amps, or 53W!
Lithium battery voltage drops from the maximum rating to a slightly lower average during usage, peters off until it eventually drops off below a minimum safe level at which point a properly designed battery will usually shut off and stop supplying power. The math above is idealistic for constant power draw over a fixed period of time, and not actually representative of actual power draw by a camera in use. I honestly not sure what the actual power draw of a Canon 1D X is, however it is not continuous at a constant level...it bursts when the shutter is pressed, then drops to a lower ambient level.
Burst power draw in a 1D X, assuming max shutter speed, full AF drive of a 600mm f/4 L II lens, while tracking a moving subject, at full-size RAW+JPEG writing to two separate cards concurrently, could likely draw more than 26W. Assuming you actually draw 3400mA for a period of 8 seconds of continuous shooting like that, followed by idle draw of 20mA for 5 seconds, that would be say 10.8v times 3.4A for 0.002222 hours (36.7W over 8 seconds), 10.8v time 0.02A for 0.0014 hours (0.22W over 5 seconds), so ~37W, or 0.82Wh. (I've completely ignored resistance here...I don't know what kind of resistance you'ed have in something like the 1D X.)