# Voltage divider data is a problem

1. Aug 30, 2014

### part

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

We did an Experiment on a voltage divider and a resistive voltage divider. we had to meassure the voltage in dependency of the resistence. we did divide every meassure by the main voltage or main resistance to fit it into one plot.
For evaluation i have attached the plot on our data. all four plots are absolutley correct. the points with errorbar is our data, the lines are the theoretical plots(just like in our task description pdf), which are at least all fine, and they show us that our data differs ALOT. why? the only possiblity for our mess is that we did not set up correctly the experiment somehow. the voltage was messured directly so the messurement errors are too little to have such a big effect. we really cant figure out what we could have done wrong.

#### Attached Files:

• ###### voltagedivider.pdf
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2. Aug 30, 2014

### milesyoung

I have no idea what's going on with the red data set - did you have a diode in there or something?

With regards to your blue data set, it looks spot on. I think you just plotted the line incorrectly. It should be a straight line between (0,0) and (1,1).

Edit: You should probably have placed this in 'Introductory Physics Homework' or 'Engineering, Comp Sci, & Technology Homework'.

Last edited: Aug 30, 2014
3. Aug 30, 2014

### part

You are right milesyoung thank you! i corrected it.

Maybe the resistave resistance( 240 Ohm) was to big in regards of the little main resistance( 1000 Ohm).

#### Attached Files:

• ###### corrected-plot-voltagedivider.pdf
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4. Aug 30, 2014

### milesyoung

It looks a bit like your red data set could fit if you had a very low value of $R_L$ and you just messed up a bit when plotting the data.

My German is extremely rusty, so to be of more help I'd need a schematic with resistor values on it and the code you used to plot your graph.

5. Aug 30, 2014

### part

values and data

Hello milesyoung!
the schematic is on page 1 in my attached "tasks experiment.pdf" (in 1st post). the values are 1000 Ohm for the overall resistor summ in the unresistive voltage divider and in the resistive voltage divider it is a 240 Ohm for the resistive resistor R_L. so yes i think You are right, maybe there was too little difference between R_L and the summ of R. But my point i dont get is: the voltage should be still proportional to the resistance. and i am really confused what resistance here in specific...

our voltage supply was a 10 Volt.
we meassured the voltage parallel to R_L.

here is my data and plot code attached in txt. for the plot i used GNUplot. of course we used direct current. Thank You :)

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• ###### scriptGNU.txt
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Last edited: Aug 30, 2014
6. Aug 30, 2014

### part

is it possible to link it or to replace my thread to there?

7. Aug 30, 2014

### milesyoung

When you have $R_L$ in your circuit, you're no longer measuring $R$ directly. You're measuring the equivalent resistance of the parallel connection of $R$ and $R_L$. I assume you've used an ohmmeter (multimeter on the Ω-setting) and just measured the resistance between the leads where you've measured $U$? That's why you're getting a much lower value for $R$ than you're expecting.

I have attached a plot of the ideal circuit compared with your measurements of $\frac{U}{U_0}$, but I've just used a vector of equally spaced points for the $\frac{R}{R_0}$ axis (Edit: I assumed you were using fixed increments when dialing your potentiometer). As you can see, your data looks fine.

If $R_m$ is your measured value, then you can determine $R$ from:
$$\frac{1}{R_m} = \frac{1}{R} + \frac{1}{R_L}$$
I think you'll have a problem correcting for it, though, since even a small measurement error in the equivalent resistance $R_m$ can give a large error in $R$.

#### Attached Files:

• ###### plot.png
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Last edited: Aug 30, 2014
8. Aug 30, 2014

### vanhees71

It looks everything right. Is the voltage source stabilized, i.e., really 10 V under all loads occuring in the measurement?

9. Aug 30, 2014

### vanhees71

From what you described, I also thought your column $R$ in your lab protocol is the resistence of the potentiometer only, and not the effective resistance including the load resistance in parallel. I think, milesyoung has solved the puzzle :-).

10. Aug 30, 2014

### part

Thank You very much milesyoung!
Do i need propagation of uncertainty to calculate the error for R? I see that if R and so the error of R is growing
the error will be overhelming due to the 1/R. Is that the reason you ment?

we did not check wether its stabillized. what could be the consequences if not? i guess if the voltage supply was lower there would have been no effect on the data of R because of R = U/I. am i wrong?

Last edited: Aug 30, 2014
11. Aug 30, 2014

### milesyoung

You're very welcome.

You have:
$$R = \frac{R_L R_m}{R_L - R_m}$$
Try plugging in your largest measured value for $R_m$ and see how much $R$ changes for small changes in $R_m$ (one ohm or such).