# Voltage Divider

1. Sep 5, 2010

### JinFTW

1. The problem statement, all variables and given/known data

Consider the following voltage divider circuit with these values: R1= 10.0 kΩ, R3= 15.0 kΩ, E1= 13.0 V. What is the percent change in Vx when we connect a load resistor Rx = 10 R3?

http://a.imageshack.us/img25/1546/circuity.th.png [Broken]

2. Relevant equations

V1 = E * R3 / (R1 + R3)

R3x = 1/R3 + 1/Rx = 1/REq

V2 = E * R3x / (R3x + R1)

Use division with the voltage solutions to find the percentage change.

3. The attempt at a solution

I went ahead and found V1, which in this case was 7.8V

Then I found R3x, which was 13.6 kohms

After using R3x with the V2 equation it appears the percentage change is wrong. So I have to assume that 10*R3 isn't a simple multiplication of the resistance value, and that sounds right, except for the fact that I haven't taken a physics course for years and can't remember what you do differently when you multiply a resistance value.

Okay, I apologize I should have posted my results first:
V1: 13*0.6 = 7.8
R3x = 13.64

V2: 13*0.58 = 7.5

Percentage Change = 3.8 % (which is wrong)

I'm not sure why its wrong exactly, but any help would be greatly appreciated.

Last edited by a moderator: May 4, 2017
2. Sep 5, 2010

### Mindscrape

For the most part, it looks okay. V1 and V2 equations are right. The R3x equation is partly wrong but the value looks okay. Double check your numbers, and if you are still having trouble I can take another look.