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I have seen the explanations about the energy lost in from of heat, but that energy comes from the net reduction of kinetic energy of the electrons (reduction of current). More over voltage is a state function (depends only on the distance) by formula $$V = kQ/r$$ In the analogy of gravity. When we make a ball roll on an inclined plane the difference in potential energy is only depending upon the initial and final height, no matter whether we roll it over a smooth plane (no resistance) or a rough surface (resistance). The heat released due to friction is compensated by the loss of kinetic energy at the bottom of the inclined plane (the ball going in the smooth plane has more kinetic energy at the bottom due to total conversion of potential energy compared to the one on the rouge surface due to loss in heat).

My question is keeping the above explanations in mind voltage should not drop across a resistor but why does it do so.

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  • \$\begingroup\$ But it's the voltage drop between the two ends of the resistor that makes the current flow through it. \$\endgroup\$
    – Finbarr
    Apr 5, 2022 at 15:54
  • \$\begingroup\$ place a resistor across battery terminals ... where else could the battery voltage drop, if not across the resistor? \$\endgroup\$
    – jsotola
    Apr 5, 2022 at 16:03
  • \$\begingroup\$ Conduction is inverse parameter to resistance. When current flows the power is dissipated in heat and the energy from time causes temperature to rise according to thermal Rth. Consider friction like resistance. A change in slope between altitudes like V potential difference may induce a force of current depending on R. \$\endgroup\$ Apr 5, 2022 at 16:09

2 Answers 2

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The theoretical model behind Ohm's law (voltage drop beeing proportional to current and resistance) is the Drude Model.

One crucial point in this model is, that electrons don't fly or drift unimpeded, but bounce very often against crystal ions (see image; from Wikipedia):

enter image description here

After each bounce the electrons direction is totally random. The electric field (voltage) has only little time to act on the electrons between bounces to create a net motion (current) in opposite direction of the field.
The result of the model is that current density \$\bf j\$ is proportional to the electric field \$\bf E\$. The proportionality constant \$\sigma\$ is called conductivity (inverse of resistance):

\$\bf{j} = \sigma \bf{E}\$

In lumped-element-model this yields Ohm's law:

\$I = GV\$ or \$ I = \frac{V}{R}\$

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  • \$\begingroup\$ Curd, In the first line of your contribution you write "voltage drop". May I ask "why"? Isn`t simply the voltage applied to both ends of the part creating inside the E-Field? Is there any "drop"? \$\endgroup\$
    – LvW
    Apr 5, 2022 at 19:17
  • \$\begingroup\$ .. because OP asked for the reason of voltage drop, i.e. current through resistor is the precondition. If a relationship between physical quantities is a physical law (like Ohm's law resulting from the Drude Model) you can solve it for any quantity if the others are given. \$\endgroup\$
    – Curd
    Apr 6, 2022 at 6:32
  • \$\begingroup\$ Of course the current will be caused again by an external E field, but it must satisfy Ohm's law (within the limits where the model is valid) \$\endgroup\$
    – Curd
    Apr 6, 2022 at 6:39
  • \$\begingroup\$ Curd, I am aware of the fact that - from the mathematical point of view - any physical law can be solved for any quantity. However, when somebody asks "why does voltage drop..." I think, he wants a physical explanation of the term "drop". That was the background of my comment - nothing else. \$\endgroup\$
    – LvW
    Apr 6, 2022 at 7:49
  • \$\begingroup\$ Then the simple answer is: because the external current source creates a potential difference (=voltage drop) in order to maintain the current; otherwise there wouldn't be any current \$\endgroup\$
    – Curd
    Apr 6, 2022 at 10:59
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The term "voltage drop" has taken on a new (and somewhat misleading) meaning.

Originally, the term was intended to describe the fact that the voltage avalaible at the battery terminals "drops" when a current is drawn (due to an internal source resistance).

However, today it is very often used as a synonym for "a voltage is created" across a resistor. Such a view "works" and can be used for calculating purposes, but - physically spoken - it is not correct.

The voltage across a resistor is a precondition for the current to flow. No current through the resistor without a driving voltage

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