What does the statement have to do with the question? There is nothing confusing about I = E / R.
Your question has all the trappings of an x-y problem as any value will work so long as the element resistivity is linear and there is no output load current.
Telling us what you’re actually atempting to do will probably get you a far more useful answer then the above response.
I've got a circuit which is 5v with a 1k pull-up resistor.
Is that the only thing in the "circuit"? If not, give us a schematic or tell us what you know. That fixed resistor may make things harder. Can you get rid of it?
That example uses a [u]potentiometer[/u] (AKA "pot") which is a standard 3-terminal variable resistor used as a variable [u]voltage divider[/u]. A standard linear pot will give you a linear voltage (2.5V at the center position, within tolerances).
I'm trying to use the variable resistor to give a linear voltage output from 0-5,
You won't find a "plain" variable resistor. You CAN use just 2-terminals of a pot to make a variable resistor, and you CAN put that in series with your 1K resistor to make a variable voltage divider. But, it's NOT LINEAR and you can't get the full 0-5V range because there is always some voltage drop across the fixed 1K resistor.
Thanks for the help, I've been doing it wrong, I've been using only 2 legs of the pot, and using a pull-up resistor. I'll do it without a pull-up, and connect:
first leg - 5v
2nd leg - output /wiper
3rd leg - ground.
Jimster:
I'll do it without a pull-up, and connect:
first leg - 5v
2nd leg - output /wiper
3rd leg - ground.
What are you attaching to the wiper? You don't want it to draw more than 10% of the current in the pot (stiff voltage divider rule of thumb); less is better.
I know it's a pain to attach images on this forum, but when discussing circuits, it's the best way to convey information. That avoids having to ask so many clarifying questions.