(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: Voltage Divider

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