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Basic Electronics » Electric field in a resistor

April 15, 2010
by bretm
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I'm getting confused by conflicting descriptions on the web of voltage, potential difference, emf, and so forth.

When current flows through a resistor, there is a voltage drop across the resistor. I get that.

What makes the current flow through the resistor? I read that current flows because an electric field is applied to it, and the electric field pushes electrons through the material.

But it seems to me that the resistor is responsible for the electric field: current flows through the resistor, the electrons bump into atoms and lose some of their energy in conversion into heat, this loss of energy across the length of the resistor sets up a drop in electric potential, and this "volts per meter" represents an electric field. So the resistance causes the electric field.

How to reconcile those two? 1) current flows because you apply an electric field (or a voltage), and 2) the dissipation of power produces an electric field and voltage drop.

It seems like #1 has to be wrong because of the superconductor example: no resistance, hence no voltage drop, hence no electric field. And yet #1 seems to be the common explanation for how current is formed in a resistor.

For capacitors, same dilemma and confusion of explanations. Negative charge on one side, positive charge on the other side, electric field between the two sides. But once again, those charges didn't get separated because you applied an electric field--the electric field was formed because the charges became separated.

So why does one say that you "apply a voltage" across a resistor or a capacitor? Isn't it more correct to say that you "shove electrons into one end and pull electrons out of the other end" of a resistor or capacitor? The voltage drop becomes a consequence, not a cause.

How you do that shoving is really a separate issue, right? It could be a chemical reaction (battery), a varying magnetic flux (generator), thermal energy (thermcouple), photon collisions (photovoltaic cell), or who-knows-what.

Is this just a confusion between volt as the unit of electromotive force (actually force per charge per meter) and as the unit of electric potential difference across an electric field?

I see this conflict everywhere, sometimes in almost the same breath. See this for example. In one sentence it quotes a source that seems to make sense:

"When a metal wire is subjected to electric force applied on its opposite 
ends, these free electrons rush in the direction of the force, thus forming 
what we call an electric current."

In the very next sentences it gets turned around:

When a metal wire is connected across the two terminals of a DC voltage 
source such as a battery, the source places an electric field across the 
conductor. The moment contact is made, the free electrons of the
conductor are forced to drift toward the positive terminal under the
influence of this field.

Isn't that wrong? Isn't it the voltage drop caused by resistance in the conductor that sets up an electric field across the length of the conductor? Isn't it the electromotive force on the electrons that cause them to drift?

April 15, 2010
by bretm
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Now to play devil's advocate against myself...

In a pinball machine, balls roll down a ramp under the influence of a gravitational field. They bump into things and dissipate kinetic energy as heat. It would be silly to say that the dissipation of gravitatonal potential energy causes the gravitational field.

But the superconductor example throws me off. If it's the voltage source that creates the field that moves the electrons, and a field strength is measured in volts per meter, and there is no voltage drop in the superconductor because there's no resistance, then there can't be an electric field. And yet there's current.

April 15, 2010
by wayward
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Hey bretm,

with my naive understanding, I'd guess it's like this: since there is no ideal voltage source, each battery actually has some internal resistance which can (perhaps?) be modeled as a resistor in parallel with an ideal voltage source. This internal resistance should be very small, to the order of milliohms, so we don't normally factor it in our calculations.

Now, when a resistor is connected to a battery, the entire setup is really two resistors in series—the resistor we observe and the internal battery resistance. Those two would act as a voltage divider, normally to the benefit of the "big" resistor, since it's orders of magnitude greater than the internal resistance of the battery. But if our "big" resistor becomes a superconductor, then the entire voltage drop falls upon the poor internal resistance of the battery, causing it to overheat and go blam. Much the same as shorting a battery.

Does this make any sense? Am I on the right track?

April 15, 2010
by wayward
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Edit: meant to say "...internal resistance which can (perhaps?) be modeled as a resistor in series with an ideal voltage source".

April 15, 2010
by bretm
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I think my point still applies if you're not talking about batteries or superconductors. But I think it's starting to come together in my mind. The key is understanding that a voltage source such as a battery is generating an electric field (a static field for a DC source) even if no current is flowing. The series of pages starting here put the pieces together for me.

Answer: #1 is correct. And #2 is correct if worded better. When you "apply a voltage" across a resistor, you are exposing the resistor to an externally-generated electric field. This does induce the charges to move. There is a voltage drop whether or not there is a complete circuit. The magnitude of the voltage drop across the resistor depends on the resistance. If the resistance of the load goes to zero, the voltage drop in that part of the circuit is zero.

April 15, 2010
by bretm
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And this lecture (Word doc) clarifies it even more. There are conservative fields in which the work done moving a test charge between two points doesn't depend on the path taken. Such as field has the concept of "electric potential". There are non-conservative fields in which the work done does depend on the path taken. Electric potential makes no sense for those fields.

There is a non-conservative field inside the battery. The charge carriers inside the battery in an open circuit experience no net force because the field is counter-acted by the field caused by the build up of charges at opposite terminals of the battery. It's the conservative external field that drives current in the external circuit.

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