So changing the voltage in your equation doesn't change the wattage? Gotcha :becky:
Correct, because amperage will change
Power is work (energy) per unit of time, time isn't changing and since you can't just destroy or create energy from nothing that means your power is also constant.
If something is using 180 watts of power it is transferring 180 joules or 180 newton meters of energy every second, that's it's measure of work per unit of time. it doesn't create less or more work because you use a different voltage. if you increase both the voltage and amperage the power increases exponentially, so doubling both the voltage and amperage would increase the power 4X.
watts is joules per second (energy per time), volt is joules per coulomb (energy per charge, a potential), and amps is coulomb per second (charge per time)
if something can do a certain amount of energy per time -> if you decrease the amount of energy for each charge then you have to increase the amount of charges per time to still provide the same amount of energy per time. each charge has less energy, so you need to give it charges at a faster rate to keep the same amount of energy per time.
https://www.electronics-tutorials.ws/dccircuits/dcp_2.html
so if the 180W is at 19V
so P=VI; 180W = 19V * I --> 180W / 19V = I = 9.47 A
and 180W at 14V is 180W = 14V * I --> 180W / 14V = 12.86 A
or at 5V is 180W = 5V * I --> 180W / 5V = 36 A
you'll notice that as voltage decreases the amperage increases because of that relationship described by P=VI, one practical application of this related to vehicles is that you can use smaller wiring on 24V systems because of the lower amperage.
if you look at a power supply you will notice that the amperages are different on the different input and output voltages, if what you were saying is correct the amperages would be the same (the difference in the product of the voltage and amperage for input and output depends on the efficiency).
