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So according to the ideal voltage loss equation V = IR if I double I (per se) I will double the loss in voltage. This is a bit odd and I just want to make sure my intuitive explanation is correct.
I am assuming voltage loss is mainly due to some electrons bouncing off the resisting material and thereby lowering the kinetic energy of the electrons that come later from the collisions that ensue (this would lower current again but less so until it reaches a sort of equilibrium). If I increase the current, the speed with which an electron bounces off the material will increase and therefore the KE loss of electrons further down the line (related to voltage loss) would increase as well. This doesn't seem exactly proportional but close enough where I hope it is the right explanation.
I am only now worried about the connection with KE loss and both current and voltage, so I may have to rethink some things (sorry I may have posted too soon).
I am assuming voltage loss is mainly due to some electrons bouncing off the resisting material and thereby lowering the kinetic energy of the electrons that come later from the collisions that ensue (this would lower current again but less so until it reaches a sort of equilibrium). If I increase the current, the speed with which an electron bounces off the material will increase and therefore the KE loss of electrons further down the line (related to voltage loss) would increase as well. This doesn't seem exactly proportional but close enough where I hope it is the right explanation.
I am only now worried about the connection with KE loss and both current and voltage, so I may have to rethink some things (sorry I may have posted too soon).
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