When a current flows through a component such as a lightbulb, how is work done?
I was thinking it was the speed of the electrons and that after a current goes through a lightbulb the electrons in the circuit would lose some kinetic energy, but drift velocity is dependent on current.
How is the potential energy of the electrons used up in a circuit?
Work is a force applied over a distance. So when charges move through an E field the field exerts a force while the charge moves over a distance, thus doing work. This is quantified by $\vec E \cdot \vec J$ which gives the power density at any location, which is the rate of doing work per unit volume.
The kinetic energy of the electrons is rarely relevant. The electrons do not carry energy through the circuit, the fields carry the energy. This is described quantitatively by Poynting’s theorem. The role of the charges is to set up the fields and to allow the transfer of energy between the fields and the matter through $\vec E \cdot \vec J$
If you wish to think about electrons traveling through a circuit then you should use $V=IR$, where $V$ is the potential, $I$ is the current and $R$ is the resistance.
When you place a component, such as a flashlight bulb, into the circuit you increase the resistance. Since the voltage (potential) stays the same, the current must decrease. Yes, the current is constant throughout the circuit, but the current after the flashlight bulb is introduced is reduced throughout the circuit. In this way you can see that as energy is used to light the bulb, the flow of electrons decreases. Note: with no component, you have a short circuit and current races through the wires of the circuit generally creating a dangerous situation. As you add more components in series (along the same line) the current further decreases with each additional component with the total resistance being the sum of all the individual component's resistances.
