I understand voltage to be a potential for electrons to be pushed through a circuit.
However, in a battery, you have an electron build-up that creates the voltage. Once current begins to flow, electrons are now moving through the circuit.
Does this mean that the voltage actually begins to decrease as a direct result of current flow? Specifically are electrons "used up" or do they simply lose energy (dissipated as heat in circuit) which leads to a lower voltage potential?
Or another scenario, do the electrons, upon returning to the battery, still act as voltage potential to continue to push a consistent amount of current through a circuit despite the loss of energy?
Or is there something else I'm missing entirely?
Thanks.
Best Answer
The voltage of the battery depends on the chemistry of the cell it is based on. For ex, a Lithium-Polymer cell has a nominal voltage of 3.7V and that of a lead-acid cell is 2V. For cells belonging to a particular chemistry, the voltage depends on many factors, the prominent one being the concentrations of the electrolyte, electrodes etc. The increase and decrease of cell voltages while charging and discharging is due to the changes in the these values. However, the number of electrons inside of a battery doesn't change in normal operation no matter what. Applying Kirchhoff's current law, you can check it for yourselves. No matter your circuit and its operating conditions, the current going out of the battery should be equal to the current going in. The voltage only changes because the chemicals inside the cell are changed slightly and not because of a change in the number of electrons. Coming to the heat part, the heat generated in the circuit is compensated by the loss in potential energy of the battery.