DC Circuit#1: I have a 10V battery connected to a 10 Ohm Resistor. Therefore, the current will be 10/10 = 1 Amps
DC Circuit#2: I have the same 10V battery connected to a 5 Ohm Resistor. Therefore, the current will be 10/5 = 2 Amps
(Assume the conductor wire does not have any resistance at all)
What I do know: The negative terminal of the battery produces a repulsive force on the electrons. This force travels at near the speed of light. When the electrons enter the resistor, they start colliding with the lattice. The kinetic energy of the electrons are converted to heat energy which is dissipated. The drift velocity of the electrons slow down inside the resistor. Now this slowing down effect is propagated backwards because of which the entire drift velocity through the whole of the circuit attains a steady state. In Circuit#1, this steady state drift causes a charge flow per unit time, of 1A. In Circuit#2, the resistance is less, so the steady state drift is more, so the charge flow per unit time is 2A. Because the whole of the circuit has the same steady state, an ammeter connected anywhere in the circuit shows the same Ampere reading. I completely get this 'same current everywhere thing'. My question is more about the potential drop.
My question: In Circuit#1, the resistance is more (10Ohm). This means that the electrons encounter more collisions while travelling through the resistor. So the electrons lose much energy by the time they come out of the resistor. Compare this to Circuit#2, the resistance is less. This means that the electrons collide less. Agreed that the electrons lose energy while travelling through the resistor, in Circuit#2 also. But my point is that if you compare the electrons at the point of exit point of the resistors, between Circuit#1 and Circuit#2, the electrons in Circuit#1 have lost much more of their ability to do work because they were slowed down more by the collisions, than the electrons in Circuit#2. Then why ,in the whole world, is the potential difference measured across the resistor the same(10V), in both the circuits?
If you keep on increasing the resistance, I understand that the current will keep on decreasing. But why doesn't the potential difference between both ends of the resistor change? The energy difference of the electrons at the entry point of the resistor, and the electrons existing the resistor will definitely depend on what was the amount of collisions that happened in the journey through the resistor. So why doesn't the potential difference also depend on this?