# Why is all the voltage used up in a circuit?

We always assume that all the voltage, in no matter what type of (basic, I guess) circuit we are dealing with, is used up. I can see why this is true. Imagine a circuit with low resistance - and correct me if I am wrong, but from electrostatics, we know that an electric field, similar to magnetic fields, accelerates charges.

Now mustn't it be that some charges arrive at the end of the circuit (into the positive charge reservoir) with some velocity, higher than that with which they began.

Essentially, if seems to me what we always assume that the electrons always do the maximum that they could, no matter what the conditions.

Is it my misunderstanding that such an assumption exists or are there some ideas I am completely oblivious of?

• Voltage is simply the difference between two potentials, which are nothing else than the ability of the system to perform work. In a resistive DC circuit all the work that this potential difference does on charges is completely converted to heat by charges loosing energy to the atoms, molecules or ions of the conductors. In a circuit that is not completely resistive, it may accelerate charges, but in that case one has to analyze the circuit as an AC circuit including all electromagnetic fields that are being created. In the end, energy conservation guarantees, that no energy is lost. – CuriousOne Aug 28 '14 at 7:31

You make a good point. In a battery a chemical reaction (a redox reaction) creates the potential difference, and the potential difference is calculated assuming the electrons start and finish in well defined energy states. If electrons returning to the cathode have some residual kinetic energy then the could affect the reaction and change the EMF. However we can easily show this effect is negligable.

Suppose we take a $1.5$V battery and let an electron flow unimpeded from the anode to the cathode. The energy gained by the electron will be $1.5e$, where $e$ is the charge of the electron ($1.602 \times 10^{-19}$C). The kinetic energy of the electron is going to be equal to this energy, so we have:

$$\frac{1}{2}m_ev^2 = 1.5e$$

and rearranging gives:

$$v = \sqrt{\frac{3e}{m_e}}$$

Feeding in the mass and charge of the electron we find $v \approx 7 \times 10^5$ m/s.

The actual velocity of electrons in a wire is called the drift velocity, and in most cases the drift velocity is less tham a millimetre per second. The collisions of the electron with the atoms of the wire it's flowing through are so numerous that the electron sheds all but one part in $10^9$ of its velocity, or one part in $10^{18}$ of its energy. The remaining energy is so small that it can be ignored.