# Voltage Drop Explanation. Is it correct?

When the battery is turned on, a wave propagates within the circuit to determine the current of the sytem.At this point, the current is inhomogeneous withing the circuit(but this dissapears almost at the speed of light). Let's suppose there are also resistances of different resistivity connected to this circuit. The electrons within the circuit are a part of the same system, thus, if one slows down, the rest will slow down too. The "voltage drop", at this point affects the whole system at the same time. The loss of potential is calculated in the initial turn on, and the current is thus established. The reason why this drop of voltage equals to the V of the battery is because all is calculated according to this wave which is launched from the battery.

• The voltage drop equals the one of the battery because that's the mover of electrons. For the rest it seems rather accurate. For all these questions I find the water flow analogies very helpful Commented Jan 25, 2019 at 9:27
• Can you please explain your first sentence in more detail? Commented Jan 27, 2019 at 2:06

It is difficult to understand the question as it is more a series of statements. Since the question is a bit answer-like, the answer may be a bit question-like:

1/ If the wave propagates (with velocity $$c$$) through the system on the switch-on, why not also during the switch-off = voltage drop? These are quite symetrical situations.

2/ How important is the speed of electrons and assumption about slowing down with resistance?

Look at e.g. https://physics.stackexchange.com/a/145422/75057 and https://en.wikipedia.org/wiki/Drift_velocity and you find an example that a drift may be even like 20 $$\mu$$m/s for large currents.

Accept the symetry, forget about speed of electrons, you may get your picture more clear.

Make an experiment. This wave ON and OFF is actually possible to view on an osciloscope. Just take a long cable (30 cm means 1 ns delay, 30 m makes 0.1 $$\mu$$s) and a short cable. Make a circuit that automatically switches ON and OFF and you will see the same delay between the two cables, no matter what resistance you put. The resistance will reduce the signal. Your picture may look like here - where the delay corresponds to 30+ meters of a cable (blue).:

I see now that yellow and blue are not the identical signals, the blue is slightly broader, sorry.

When the battery is turned on, a wave propagates within the circuit to determine the current of the system.

When the battery is turned on it establishes an electric field almost instantaneously (speed of light) throughout the conductor.

At this point, the current is inhomogeneous within the circuit (but this disappears almost at the speed of light).

The current exists almost instantaneously as the free electrons in the conductor (typically metal) all simultaneously (within the speed of light) experience a force from the electric field.

Let's suppose there are also resistances of different resistivity connected to this circuit. The electrons within the circuit are a part of the same system, thus, if one slows down, the rest will slow down too. The "voltage drop", at this point affects the whole system at the same time.

No problem with this statement.

The loss of potential is calculated in the initial turn on, and the current is thus established. The reason why this drop of voltage equals to the V of the battery is because all is calculated according to this wave which is launched from the battery.

The voltage drops around a circuit equals the drops in electrical potential energy of the electrons in the circuit. The reason why the sum of these voltage drops equals the battery voltage (Kirchoff’s voltage law, KVL) is because the sum of the drops in potential energy equals the rise in potential energy from the battery. Therefore KVL is simply a consequence of the conservation of energy.

Now you may ask, what happens to the potential energy the electrons acquire from the battery? The answer is it is first converted into kinetic energy and then the kinetic energy is lost and is converted to heat due to the collisions between the electrons and the fixed particles of the conductor as the electrons move through the conductor. You can think of it as a series of brief increases (prior to a collision) and decreases (following a collision) in the kinetic energy of the electrons, so that the average velocity of the electrons, called the average drift velocity, will be slow.

Finally, the electric current through a cross sectional area of the conductor is defined as the rate of charge transport through that surface, or

$$i(t)=\frac{dq(t)}{dt}$$

And this reflects is the average drift velocity of the electrons in the conductor.

Hope this helps.