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Small question, I see that resistors are labeled in certain increments (e.g 5, 10 50 Ohms), although a resistor outside of a circuit is completely without voltage and therefore current. Does this just mean the resistor has a certain value when placed in an average circuit?

Also, I thought that, according to Ohm's Law, resistance was a variable along with current and voltage. Is this not true? I mean, if I speed up electrons through an ioninc matrix, through an increased voltage and current, shouldn't the matrix push back harder as well making more resistance?

I'm thinking of an ice cube falling through molasses. If the ice cube speeds up (due to increased push i.e voltage) shouldn't the molasses push back more as well? And yet it seems that resistors have a constant value, what the heck.

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  • \$\begingroup\$ Interesting premise, poor title. \$\endgroup\$
    – JYelton
    Commented Jul 14, 2014 at 22:19
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    \$\begingroup\$ "How are resistor values constant despite changing current and/or voltage?" \$\endgroup\$
    – JYelton
    Commented Jul 14, 2014 at 22:22
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    \$\begingroup\$ You can't change the resistance in steady state. I feel like you might enjoy reading about the memristor if you are curious about this kind of theory en.wikipedia.org/wiki/Memristor \$\endgroup\$
    – Funkyguy
    Commented Jul 14, 2014 at 22:31
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    \$\begingroup\$ You can rewrite your own molasses physics OR use the simplified models to help you as you get going OR learn "real" physics. No simple one line "law" describes what is believed to be an accurate model of anything. \$\endgroup\$
    – Russell McMahon
    Commented Jul 15, 2014 at 0:47
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    \$\begingroup\$ ... if you have never tried the following , try it. DO NOT LOOK THIS UP if you have not tried it. DO NOT. Just do it. THEN look it up :-). -> Get some "cornflour". A standard kitchen item. Put a few large spoonfuls in a cup and add enough water to make a wet but not runny paste. Now, stir it with a stick or your finger or something with a bit of cross section. stir it very slow/slow/medium/fast (if you can). | Here the resistance changes with pressure (Voltage) - but its the fluid not the resisting aspect that is changing. Very worth doing. \$\endgroup\$
    – Russell McMahon
    Commented Jul 16, 2014 at 7:06

6 Answers 6

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When the resistance value is constant (doesn't depend on I or V), we call that a linear resistor.

Common resistive materials (such as metals, carbon, etc.) are reasonably linear to a first order approximation. There is no physical mechanism that strongly causes them to "push back more" when a greater voltage is applied.

If you look more carefully, you will usually find that as you increase the current through a real resistor it heats itself up, and this does cause the resistance to change. However we try to choose materials for our resistors where this effect is small. TCR (temperature coefficient of resistance) values on the order of 100 or 200 ppm/C are easily available.

A diode (in a dc circuit) is an example of what we call a nonlinear resistor. A resistor whose resistance depends on the current through it. However, these tend to resist less, the more current they are passing.

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  • \$\begingroup\$ High voltage resistors typically suffer from voltage coefficient to a measurable degree. These ones brag about being 'only' 0.5% change at working voltage (2500V): digikey.ca/Web%20Export/Supplier%20Content/… \$\endgroup\$ Commented Jul 14, 2014 at 23:01
  • \$\begingroup\$ @SpehroPefhany, that kind of thing is why I weaseled around with "reasonably linear" and "to a first order approximation". \$\endgroup\$
    – The Photon
    Commented Jul 14, 2014 at 23:17
  • \$\begingroup\$ Isn't a diode just as linear in first approximation? \$\endgroup\$ Commented Jul 15, 2014 at 10:03
  • \$\begingroup\$ Hmm. I was just wondering, because it seems like electrons falling through metal is similar in circumstance to a skydiver falling through air. Since the 'pushback' of air is proportional to the diver's instantaneous velocity, shouldn't an increased current make for an increased pushback? \$\endgroup\$ Commented Jul 15, 2014 at 15:00
  • \$\begingroup\$ @MarcksThomas, that's true (because it's the definition of a "first order approximation" to be linear). But a 1st-order approximation is rarely useful (maybe in some rf circuits) for the ways we use diodes. \$\endgroup\$
    – The Photon
    Commented Jul 15, 2014 at 15:59
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A resistor retains it's value outside of what you might call a regular circuit. It can sit on a shelf with no perceivable external influences and still be characteristically a resistor of the same value.

It can even theoretically be proven that this is the case by a little thought experiment but you'd have to understand the physics to a much deeper level. For instance, if you had a sensitive voltage measuring device, you would be able to measure the noise voltage that the resistor creates and, if you had an accurate measurement of temperature you could precisely say what the resistance was.

Other than that ohms law defines three variables but, for a perfect resistor, voltage is perfectly proportional to current.

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  • \$\begingroup\$ A good comment. I like this \$\endgroup\$ Commented Jul 15, 2014 at 14:51
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Ohm's law is an equation on three variables: (\$R=\frac{V}{I}\$). This means that for given values of two of the variables, you can solve for the third. Or if you set one variable and vary a second, you can see the change in the third.

For example:

  • Constant resistance, varying the voltage, see the current vary. This is the case where you have a variable voltage source into a fixed resistor. The current through the resistor will vary proportionally to the voltage.

  • Constant resistance, varying the current, see the voltage vary. This is the case where you have a variable current source into a fixed resistor. The voltage across the resistor will vary proportionally to the current.

  • Constant voltage, varying the resistance, see see the current vary. This is the case where you have a constant voltage source connected to a potentiometer (between the wiper and one end). As you vary the resistance, the current will change as the inverse.

Does this help?

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    \$\begingroup\$ The answer was useful, but the answer to my question has been resolved in the comments of The Photon's response. You may find it hilarious \$\endgroup\$ Commented Jul 15, 2014 at 16:59
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How can a resistor not be constant?

If you have a power source that is both a constant current \$I\$ and a constant voltage \$V\$, then surely that current must flow between the two terminals of the power source, and the voltage across that power source must always equal \$V\$.

So, say we have a 1A constant current and a 1V constant voltage, but nothing connected to it, then that 1A must be flowing through what - open space? For that to be the case the resistance of open space must equal 1Ω (\$R=\frac{V}{I}\$). If that were the case, then every battery in the world would pretty much be instantly flat as it is discharged by a 1Ω resistor. All the power lines would explode and melt down, and we would never have got out of the stone age.

So clearly air cannot be 1Ω, which means that logically a constant current and constant voltage together do not make any sense.

The only way you could get 1V at 1A is by using a 1Ω resistor to complete the circuit. Any other value resistor just wouldn't work and something would have to give, usually with the release of the Magic Smoke™.

although a resistor outside of a circuit is completely without voltage and therefore current

So, a battery with nothing attached has no resistance and no current, so the potential difference between its terminals is 0V? But the potential difference is not 0V, it's the voltage of the battery.

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  • \$\begingroup\$ I don't think this actually clarifies anything. \$\endgroup\$
    – pjc50
    Commented Jul 15, 2014 at 10:55
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If a resistor obeys Ohm's law the resistance is constant (i.e. independent of voltage and current) by definition.

Ohm's law is an idealisation that is a more or less good approximation of reality depending on the particular case. The most important reason why a device doesn't obey Ohm's law is because its resitance depends on temperature, which in turn is affected by the current flowing through the device.

E.g. Ohm's law it is a good approximation if the resistor is a wire of Constantan. It is a bad approximation if the resistor is a light bulb (has low resistance at low temperature/current, has high resistance at high temperature/current; That's why it can be used as a simple current stabilizing element).

A physical model that explains linearity between current density and electrical field (i.e. constant resistivity) is the Drude model.

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  • \$\begingroup\$ Thank you for your comment. I've been reading the Drude model and am glad you brought it up, thanks \$\endgroup\$ Commented Jul 15, 2014 at 14:52
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Ohm's law is only valid in steady conditions and the variable parts of the equation are V and I. R is a constant. Resistor values are constant because it's a physical property due to the way they are made.

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