Energy and Speed of Electrons in a Circuit

Question

I really wonder about the relationship between potential and kinetic energy in a circuit. At the positive terminal of the battery, there will be high potential energy. The difference in electric potential between two terminals will cause the electrons to move from positive to negative terminal. The electrons thus move with an increasing speed as its potential energy decreases, converted to kinetic energy. When it passes the bulb, its decrease in potential energy will be converted to heat, leading to even less kinetic energy.

But this seems wrong to me since as the electrons move closer to positive terminals, there will be the force from the positive terminal increasing the speed of electrons.

However, I learn that the current in the series circuit is constant. So, the drift speed of electrons in this circuit is the same, contradicting what I previously said.

I am totally confused with different concepts. Could you explain what exactly happens to the electrons and the energy of them in the circuit?

Answer 1

As a result of the conservation of charge the current is always constant in a circuit (continuity equation).

We know by Ohm’s law that the current density is proportional to the applied electric field $$\mathbf j=\sigma \mathbf E$$ where $\sigma$ is called the conductivity. Since we apply a constant voltage, the electric field is the same across the circuit. (Think of it as if the electron is nearer to the positive terminal it feels a greater attraction but also a lesser repulsion from the negative one). But if the conductivity changes the electric field need not to be the same as we see later. Due to scattering at atoms the electrons get slowed down making their speed constant (this makes up the resistance). This explains the proportionality in Ohm’s law. From this we get that the drift velocity is proportional to the applied electric field.

We can write the current through the circuit as $$I=nev_dA$$ where $n$ is the charge carrier density per volume, $e$ the elementary charge, $v_d$ the drift velocity and $A$ the cross-sectional area. As the current is the same in the circuit the drift velocity can indeed be different by changing the cross-sectional area $A$ and charge carrier density $n$. For example if you half the cross-sectional area the drift velocity doubles. You can think analogously of a water circuit. The analogon of a resistor (e.g. the bulb) is a narrowing. If the area is halved in the narrowing the water will flows double as fast than before. But after the narrowing it will flow with the same velocity as before it.

As mentioned above the electric field need not to stay the same across the circuit. If we revisit Ohm’s law (using absolute values instead of vectors) and by definition $j=\frac I A$, we get $$\frac I A =\sigma E.$$ For simplicity let’s assume the area $A$ is the same for the wires and the bulb. Since the bulb is a resistor it will have a smaller conductivity $\sigma$. (Note that the conductivity is a material constant and does not depend on the area or the length). But this means the electric field inside the bulb has to be greater because the current is the same across the circuit. This complies with Kirchoff’s first law stating that the voltage across a loop sums up to zero. Usually we treat the wires as perfect conductors with infinite conductivity and assume that the resistance can be modeled as one discrete element. Hence Ohm’s law gives us that all the battery’s voltage drops at the bulb. But Ohm’s law gives us more, if there are two or more resistors in a circuit, we can calculate the voltage drops based on their individual conductivities (at the smaller resistor will be the smaller electric field).

You are right, the electrons lose kinetic energy and convert it to heat in the resistor but in the same time the bigger electric field accelerates them again and compensate their energy loss by raising more the potential energy. From an energy point of view, you have a constant voltage source making sure there is always the same potential energy. This means if you lose heat at the bulb the voltage source will compensate for that.

There is another way to look how energy gets transferred from the battery to the resistor which comes handy when considering high frequencies. (I guess it is also more correct). We can look at the created electric and magnetic fields by the current. The Poynting vector $\mathbf S$ measures the energy flux and is defined as $$\mathbf S = \mathbf E \times \mathbf H.$$ If you do this for your circuit you will see that the Poynting vectors points from the battery to the lamp as in the picture below from Wikipedia. That is to say the energy flows from the battery to the bulb. Look also at this great post by @wbeaty.

Source: https://en.m.wikipedia.org/wiki/File:Poynting_vectors_of_DC_circuit.svg

Answer 2

It is true that the potential energy created by the battery is converted to the kinetic energy of electrons and then to the heat or other forms of energy depending on the type of a load.

But individual electrons do not speed up in a wire as they do in vacuum.

In a vacuum tube, for instance, the kinetic energy of electrons increases as they fly from the cathode to the anode and, when they reach the anode, this energy is turned into heat.

In a wire, individual electrons collide with atoms very frequently, so they keep gaining and losing their kinetic energy between consecutive collisions.

As a result, given the same wire type and, therefore, the same field strength, the average speed of electrons in a wire near the negative terminal should be about the same as their average speed near the positive terminal.

If the wire diameter was greater, the average field would be weaker, so more electrons would contribute to the current, but their average speed would be lower to yield the same current. Still, the nature of the movement would be the same stop-and-go regardless of the wire size.