# Potentiometer Problem

Discussion in 'Homework Help' started by yan500, Jul 12, 2011.

1. ### yan500 Thread Starter Member

Jul 12, 2011
48
0
Hello All,

I'm just dabbling in electronics for fun as a summer project but I really need some help on a problem. The problem is question 16 http://www.allaboutcircuits.com/worksheets/e_divide.html

I just can't figure out how to account for the 1k resistor. I've tried adding the two resistors together to figure out the voltage reading but that doesn't even make sense because the more resistance you have, the more voltage you will have.

I don't really know how to account for that resistor. Any help is much appreciated.

2. ### blah2222 Well-Known Member

May 3, 2010
573
36
I messed that up sorry, look below haha...

Last edited: Jul 12, 2011
yan500 likes this.
3. ### yan500 Thread Starter Member

Jul 12, 2011
48
0
Hmm...I'm getting 1.67V using your method but that is not the right answer. Or maybe I am doing something wrong.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
With the pot set at 25% then the value from wiper to ground is 1.25kΩ. The value from wiper to +10V is 3.75kΩ.

The 1.25kΩ (wiper to ground) is in parallel with 1kΩ. So from wiper to ground the effective resistance is 0.556kΩ.

The divider output is then Vout=10*(0.556/(3.75+0.556))=1.29V

yan500 likes this.
5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
So

$V_{out}=10X[\frac{%5k||1k}{(5k-%5k)+(%5k||1k)}]$

6. ### yan500 Thread Starter Member

Jul 12, 2011
48
0
Got it! I was visualizing it the wrong way, thanks a bunch!