If I were to connect an appliance to something other than an electrical wall outlet for power and that device produced more volts / watts than what my appliance needs to work, would it mess it upand if so how to I control this?
if V I R, you are effectively increasing V while keeping R constant; so I through the device must increase. Power dissipated by the device I^2 R : so by doubling the applied voltage, you'd quadruple the power dissipated by the device. If you're lucky and the components can handle the additional heat dissipation requirement, then nothing will happen except the device will become quite hot to the touch. If the components can't handle the additional heat being generated, the device will burn out. If you're lucky, there is a single fuse or component that will burn out and break the circult; stopping the flow of current. If you're not lucky, then the device could continue to heat up and eventually catch fire. you need to buy/acquire/build a voltage converter/power supply -- something that will take the input voltage and current and output the voltage'/current required by the device. Running 12V electronics from a car lighter socket, you encounter this all the time. The DC adapter basically takes the raw 12.6V and ensures that the voltage/current specification required by the device are met.
