Before I ask my question, it's important that you know how I understand the underlying physics of circuits, so you could point out any misunderstandings or incomplete thoughts. A battery creates an electric potential difference between its two terminals, changing the electric field around it (the field already exists in space, but the battery gives it values other than 0). When the battery is connected a circuit, this electric field change is channeled through the circuit. The potential difference created by the battery can then push the electrons (the force is exerted through the electric field), giving them a forward motion in addition to their already random motion. However, this flow of charge isn't the same as the flow of energy, because the Poynting vector must be perpendicular to the the flow of charge. Thus, the energy must flow out of the battery into the field, where it can then flow back into the circuit as needed (ie. it powers a lightbulb).

However, my concern with this understanding is that the energy might not be conserved since energy is needed both to get the charge to flow and to power the lightbulb. How is it possible that the battery generates enough energy to cause an electromotive force to cause the electrons to move (this is a force applied over a distance, so it must use energy), but then also enough for the energy to flow to power the lightbulb?

Furthermore, how is it possible for the applied voltage of a battery to resistors running in parallel to be the same as the total voltage for the circuit if the energy outputted by the battery is finite?


2 Answers 2


How is it possible that the battery generates enough energy to cause an electromotive force to cause the electrons to move (this is a force applied over a distance, so it must use energy), but then also enough for the energy to flow to power the lightbulb?

Imagine you are powering machinery by turning a wheel that is connected with a chain. Besides the energy going into the machinery, we have to accelerate the chain as well. That energy is lost (we don't recover the energy from the chain when it slows down). But under most circumstances, the energy from that is so small compared to the energy that goes into the load that we can ignore it.

Same with the battery. The energy that goes into moving the electrons (and setting up the magnetic field around the wires) can be modeled as the inductance of the circuit. For most simple circuits, the inductance is very small and the energy it takes to get to steady-state is correspondingly tiny. Therefore when analyzing the circuit, we only consider the energy that goes into the load (the lightbulb).

This is especially true since the acceleration loss is a one-time energy loss, while most other losses (resistance in the wires, load) are continual losses over time. So ignoring the loss prior to steady-state is usually completely valid.


For circuit analysis, we don't really need fields. A simple one-dimensional description is perfectly sufficient.

Let's first lock at the simple case of a battery connected to a (moderate) conductor. A conductor is a material where the electrons are "loose enough" so that they can move without too much force required.

When the battery touches the ends of the conductor a force proportional to the voltage is applied to electrons and they start to accelerate. However, there is always friction inside the material (ignoring superconductors) so the electrons accelerate until the force of voltage and frictions balance each other. If they go any slower, the voltage will accelerate them and if they go any faster the friction will slow them down.

Friction turns kinetic energy into heat and that's exactly where the energy goes. The friction of the electron flow generates heat. The amount of energy is a function of the "electric friction" (resistance), the driving force (voltage) and the amount of electron flow (current). So we have

$$P = V*I = V^2/R = I^2*R$$

where $I$ is current, $V$ is Voltage , $R$ is the resistance, and $P$ is the dissipated power.

Now let's look at a battery, two wires, and a light bulb. A "wire" is a conductor with a very, very low friction. Electrons can accelerate freely and nothing slows them down. Hence it's a REALLY BAD IDEA to put a wire across a batter since things tend to melt, explode, burn or do other nasty things.

The light bulb has a much higher resistance than the wires so the light bulb determines the speed limit for the electrons. All electrons go at the same speed since there are sitting "one behind each other" and the electrons in the wire can't go any faster. That means that almost all the energy is dissipated in the light bulb. The bulb and the wires see the same current but the resistance is vastly different since. If the resistance of the bulb is 1000 times higher than that of the wire, it will consume 99.9% of the entire energy.

In short: for regular resistive materials all the energy goes into heat that's generated by electrons bumping into things.


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