I am looking at hi-jacking an USB cable to supply to supply > 2.1A to the target. What is the physical current limitations of a standard USB cable?
Basically, that resistor is used to tell the device what kind of interface it's connected to.
If the resistor is 56K, it indicates to the connected device that it's actually connected to a USB-2.0 port, and it has to negotiate with the remote host to learn the amount of power it can safely draw.
If the resistor is 10K, it should indicate that it's connected to a host that is able to provide 3 amps of power without any negotiation being required.
The concern here is that even the highest-power USB-2.0 interfaces were only specified up to 1.5A (higher for dedicated chargers without data).
Basically, the concern is that there is no way to safely predict what will happen to every piece of hardware out there when the connected device draws 2X the current it was originally designed for. Depending on the hardware design, it could fail in a multitude of ways, some of which are capable of potentially damaging any connected devices.
If you're lucky and have a properly designed host, the failure would just be the open-circuit disconnection of a PTC which would prevent charging, but reset once the overload is disconnected. .
However, the above implies you have a host-device which has a PTC, which is often not true in many ultra-cheap hubs. If your device does not have any protection facilities, you instead get to discover the failure mode of it's power supply for yourself.
Again, ideally, the power-supply would just shut down on overload. However, again, this implies proper safety considerations have been taken into account in the power supply, which generally raise it's price. On the other hand, it's possible (for example) the host uses A DC-DC converter to convert a local 12V rail to 5V for the USB interface, and the overload causes the switching MOSFET to fail short-circuit, resulting in 12V on the USB 5V line, which would almost certainly damage both the host's internal circuitry, and the electronics of the device.
The absolute, pathological worst case would be the above, only the connected device/charger then catches on fire, and burns down your house, killing your cat/cat-substitute (significant-other, kids, dog, ferret, small angry squirrel etc...).
This is a unlikely scenario, and the failing system described above is certainly not well designed, but it would work normally otherwise, and is standards-compliant. The adapter manufacturers cannot know that there are not devices like that out there, and as such selling a device which intentionally violates the USB specifications just to make a connected device charge faster is highly irresponsible.
In effect, the 10KΩ resistor doesn't "prevent" it, it's actually used as part of a system for informing the connected device about the host it's connected to. It's effectively lying to the connected device, and claiming the host can do things it was likely never designed to do.
Yes. It would be like they were never connected. You do compromise the effectiveness of the shield around the cables by cutting it though, so try to minimize that or ensure you tightly tape it back up.
But couldn't you use a 2.0 cable in the first place? The usb 3.0 micro b connector has two parts, a 2.0 micro b five pin part and a newer 3.0 only five pin part. A usb 2.0 micro b cable can plug into the corresponding side of the 3.0 micro b jack for 2.0 and power only access.