Stupid Question (I'm full of 'em) On pots

[Aurin]

Ok, So I know a potentiometer is a voltage divider when one of the outside lugs goes to ground. No problem understanding that.

What I want to know is, when you have let's say a 100k pot and you run input through lug 1 and lugs 2 and 3 are connected... How does that work?

I mean ...

IN:

--*
   |
  <
  ><--;
  <   /
   | /
   *
   +---- OUT

What does that actually accomplish? It seems to me like it limits current rather than voltage (The source having to go through the resistor, either way, to get to "out").

Am I right in that?

[markr04]

You're right. It's a variable resistor like that, limiting current.

[Aurin]

Awesome =) Thanks for verifying.... I'm trying to understand the circuits I'm working with rather than just blindly building them.... for two reasons 1) Easier to fix something you understand and 2) Eventually I'd like to try my hand at designing my own FX.

People like you answering these silly questions really helps and I appreciate it very much. Thank you.

[mjarus]

I have always wondered somethjing though... why bother connecting lugs 2&3?  It seems like if you just connect lug 2 and let 3 "float" it would be the same effect, right?

mjarus.

[niftydog]

it is usually done to reduce noise pickup on that floating lug. Otherwise it's just good practice to tie up all the loose ends, literally!

Like, you wouldn't hook up one end of a resistor to a circuit and just leave the other end floating, would you?

[mjarus]

I'll buy that for a dollar... thanks.  Makes sense, I guess if the other end accidently went to ground then you would get a divider rather than the intended result. 

I guess I shouldn't admit that I have left a floating resistor huh?

mjarus