why does sources with same voltage drive different current through same load?
There are 2 types of charging IC's used in cell phones. The 'dumb' IC uses a standard 1 amp charging current if the wall adapter can supply it. If you use your PC to charge the phone, it may only allow 1/2 amp of charging current, so that is all the charger IC has to work with. It will not use so much current that the source voltage drops below 4.0 volts. This includes a drop of 1/2 volt in the cable going to the wall-pack. The 'smart' IC in some smart phones will draw as much as 2 amps to charge the phone, if the charger can supply 2 amps and still maintain close to 4.5 to 5 volts under that load. It is generically called a 'fast' charge, and is usually controlled by a check box in the phones 'Battery' control directory. A phone that charges at 2 amps will come supplied with a 2 amp charger (wall-pack). My Samsung Note 2 has such a charger and I have it set for 'fast' charge. The charger IC in the phone makes all the decisions about how much current there is to charge with, is there a 'fast' charge setting for this phone, is the supply voltage at least 4.5 volts, plus monitoring the charging of the battery and reporting charge status to the CPU. It is a very busy IC when charging. NOTES:1)  The LM7805CT has a limit of 1 amp with heatsink, but there are simple booster circuits you can find on the web. A 12 volt 7.2 amp-hour battery will work, within the constraints of the LM7805CT IC.2) My answers in terms of current are in amp-hours or mA-hours.3) I am not implying a charge current > 2 amps or a charge voltage (at the phone) > 5.1vdc. I have not yet seen any phones capable of going out of these ranges.