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I have a question about the classic relation $V = RI.$ I know that is the voltage that creates the electrostatic force. But I do not understand how, in a circuit with just a current source and a resistance, I can "create" voltage. In other words why $V = RI\;?$ In an ideal circuit with just a resistance and a current generator, why the electrons move (thanks to what force) and why, when they pass through a resistance, create a potential difference. I tried to read some other answers in this sections but with poor results.. Many thanks.

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  • $\begingroup$ Voltage is not a force. $\endgroup$
    – user36790
    Commented Dec 7, 2015 at 2:14
  • $\begingroup$ I don't understand what you mean exactly, a current source produces current by creating a potential difference (voltage) between the two terminals. $\endgroup$
    – tmwilson26
    Commented Dec 7, 2015 at 2:22
  • $\begingroup$ @tmwilson26, that's not quite true. A current source produces a current when the voltage across its terminals is zero too. At least in the context of an ideal current source, the voltage across is entirely determined by the external circuit. $\endgroup$ Commented Dec 7, 2015 at 2:25
  • $\begingroup$ @AlfredCentauri - perhaps it would be better put as the ideal current source must provide whatever voltage is necessary to source the desired current? In reality, it seems that current sources are the least 'ideal' part of any set up, but that is another story... $\endgroup$
    – Jon Custer
    Commented Dec 7, 2015 at 16:08
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    $\begingroup$ I guess I'm too closely tied to trying to implement a current source in the real world... $\endgroup$
    – Jon Custer
    Commented Dec 7, 2015 at 17:22

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A current source can be thought of as a charge 'pump'.

If there were no voltage across the resistor, there would be no current through.

If there were no current through, charge 'pumped' by the current source would accumulate on one terminal of the resistor and be removed from the other.

But this would result in a voltage across the resistor which implies a current through.

Thus, there is just enough charge density difference on the terminals of the resistor such that the voltage across is just what is needed for the required current through.

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  • $\begingroup$ From your last answer to JonCuster, I think is much more clear. So is the change in charge density that produce a voltage. Am I right? $\endgroup$ Commented Dec 7, 2015 at 17:43
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A current source has a variable voltage source which is controlled by some circuit of active components like transistors which maintain a constant rated current in the load. This current is independent of the load. So, Even in this case a voltage creates a current.

A current is nothing but moving electrons. The electrons can be moved using any mechanism. The electrons in a conductor are moved using an electric field which is created by a voltage across the conductor.

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You are misinterpreting the relation & coming to an indeed wrong conclusion.

So, what does $\mathbf F= m\mathbf a$ tell you?

You: Oh! The mass when moves with acceleration creates force!!!!!

Now, this is ridiculous.

The actual relation from which Ohm's Law is derived is this:

$$\mathbf J= \sigma \mathbf E$$

Now, the physical implication of this relation is quite straight-forward:

The current density is proportional to the electric field that creates it.

From this you can derive Ohm's Law as:

$$\int \mathbf J \cdot d\mathbf a= \sigma \int \mathbf E\cdot d\mathbf s \\ \implies I= \sigma V \\ \implies \frac{1}{\sigma} \; I= V\\ \implies R\cdot I= V\;.$$

So, here it comes. It's just old wine in a new bottle.

The physical implication is always the same as above quoted; you've just flipped the constant of proportionality but that doesn't mean it changes the whole meaning of the statement.

If you've a constant voltage, then the current in the circuit is determined by the external elements like resistance $R$.

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See if you can relate to this analogy, we know that water flows from a higher level to a lower level similarly current flows from a higher level (high potential) to a lower level (lower potential). As it is almost natural, and we normally do not realize the presence of gravity (thanks Newton for this) but here, there is no concept of force. The relation is J=(sigma)*E(as stated in an earlier answer). For the matter of fact, if you want to know the force behind the phenomena, it is the electromagnetic force ( most of the phenomena occurring around us are due to this force, gravity and nuclear force are applicable only in specific places but if you wanna see the effect of a force an human level then this is the force).

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  • $\begingroup$ Isn't it true that the potential difference (voltage) in a closed circuit creates an electric field force that moves (repels) electrons (negatively charged particles) so that energy is propagated through a circuit & a resistor resists that energy flow due to the materials that are used in the resistor (which cause higher numbers of atomic collisions, dissipate energy, & drop voltage in the process or 'resisting') the current flow? $\endgroup$
    – DIYser
    Commented Dec 7, 2015 at 4:06
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What came first, the egg or the chicken?

It is very similar in this case: what came first, the current or the voltage? Ohms law is about steady state and direct current. So we will not delve inte dynamic stuff or alternating current or such things.

In the steady state that Ohms law describes, there is a resistance, a current and a voltage. They all exist at the same time. The current through the resistor requires a voltage. The voltage and the current requires a resistor. The Voltage and resistor requires a current. You might say that V=RxI! None of them comes first, they have to be there all at the same time.

So how do we reach this steady state. In your question you say that we start with a resistor and a current source. I imagine that you have a current source which wants to (actually is designed to or adjusted to) push a certain amount of Amperes through the resistor. The only way for that current source to do that is by modifying the Voltage. So it will increase (or perhaps decrease) the voltage until the current is exactly right. In effect, it is the current source that has to create the Voltage in order to reach enough of current. Once there, we have reached steady state and Ohms law applies. On the way there, before steady state, the relationstips may be different, or as they sometimes say left for the reader to calculate (hint: dynamic currents are much more complicated than Ohms law).

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