Well,
It's not so much Ohm's law but how far you push the wall wart beyond it's design parameters.
Generally speaking, say your wall wart is rated at 6V, 175mA output. Charging a single cell or a number of cells whose voltage is much less than the wart's rating will yield a higher mA output.
If charging beyond the wart's voltage rating, for instance a six cell pack, (7.2V nominal), the mA output of the wart will decrease.

I mention the wart's design parameters because I have fried one or two warts by exceeding them.

The math may say one thing but the actual results may yield another. Your best bet is to use your meter.

Using your example and the equation: W=E*I, it can be shown what your wart's output would be at 7.2V.

W=15V*.200A
W=3 , 3 watts
To find the current, I, at a lower voltage one must rearrange the equation to read: I=W/E.
I=3W/7.2V
I=.417A or 417mA

From what I remember, that is how one would do it. If I had my new scanner operational I'd post a sheet of Ohm's law and derivations of it.

Hope this helps,
Jeff