# SHowing direct proportion

1. Apr 13, 2010

### Catkill

First post so sorry if its not exactly as it should be.

My problem is that for my physics investigation - a graded piece of coursework which goes towards my exam grade - we have to carry out a few experiments and do write ups with uncertainties etc.

In my investigation this relationship is ment to be directly proportional (Impedance with frequency in an AC circuit) yet my graph looks like this:

http://img690.imageshack.us/img690/9779/helpfizz.jpg [Broken]

Which is hard to see if its directly proportional at all. Anyoneone know a way i can say that - the trend is roughly linear but not through the origin.

Thank You
Blair

Last edited by a moderator: May 4, 2017
2. Apr 13, 2010

### Stonebridge

What were the values of L and R in this experiment?

3. Apr 13, 2010

### Catkill

L i dont have the value in henry's but it was a 500 coil iron cored inductor (I belive) and R was 1100 ohms - these were kept constant

4. Apr 13, 2010

### Stonebridge

There's something not quite right, then.
I'm assuming your vertical axis is ohms. If so, then as the frequency gets lower and approaches zero, the impedance of the circuit should get closer to the value of the resistor. (1100 ohms) This is because the reactance of the inductor falls to zero as the frequency gets nearer zero, just leaving R.
From you graph it looks like R is about 100 ohms. The graph does not go through the origin because there will always be the purely resistive component, R, there.
There will also be some other resistance due to the circuitry and also some in the inductor itself.
Now the value of L in your circuit is critical.
The total impedance of your circuit is √(R2 + X2)
R is the resistance and X is the reactance of the inductor, L.
X=2πfL which means it is directly proportional to f.
However, if R is large enough, or L is small enough, your graph will not be a straight line.
It depends on the relative sizes of these two values.
Is it possible to check the value of R you used, and find out the value of L.
The main point I'm making is that the graph isn't necessarily going to be a straight line.

5. Apr 14, 2010

### Catkill

Apologies it was 110 ohms, I dont have the equipment anymore to re-do the experiment so will have to still do the write up with these. How would I go about explaining this in my analysis?

Thanks
Blair

6. Apr 14, 2010

### Stonebridge

I've plugged some numbers into a spreadsheet and arrived at this.
Vertical axis is impedance in ohms, horizontal is frequency in Hz.

R is 110 ohms and I've used L=0.007Henry
This gives the best fit I could get for your results.
You will see that the line curves to meet the vertical axis at R=110ohms, but it is straight for most of the way above about 3 or 4 thousand Hz.
I can't explain why your results look the way they do. With appropriate values of L and R, you should be able to get a reasonably straight section of line above 5000Hz.
Firstly, there is nothing "wrong" with getting a curve. The relationship between impedance and frequency isn't linear. However, it becomes virtually linear as the value of the reactance of the inductor gets greater as the frequency increases, and the value of the R in the circuit becomes proportionally less.
Your graph is fine in that it levels out to give the value of R=110 ohms. However, it doesn't seem to become linear for higher frequencies.

What were you going to do with this graph? When you measured the gradient, what were you going to do with the value? Or were you just out to "prove" that you get a straight line.
It would be unwise to claim that these results give a straight line when they clearly don't, and would bring into question what the value of any gradient you drew actually represents.
The gradient of the "straight" portion, by the way, should approximate to 2 Pi times the inductance. If you measure this on mine it should give (roughly) the value of the inductor I used.

The graph should have looked something like my one above, but I am guessing in the dark a little here.

7. Apr 14, 2010

### Catkill

It was more to prove the relationship so i could say reactance is directly proportional to frequency then in the "second section" ive got experiments with filters showing how this can be used.

Dont know how i would factor the gradient into it ? If you have msn or e-mail i coudl send you the rest of the document showing my other experiements and how ive tried to explain it.

Thanks
Blair

8. Apr 14, 2010

### Stonebridge

Well, the next step is to look at the actual measurements you took. I'm guessing the frequency was read directly from the output dial of a signal generator.
How did you measure impedance? This, I presume, was calculated from other readings you took. Again, I'm guessing you measured ac current and voltage. But what did you measure current through and, more importantly, the voltage across? Can you show or describe the circuit you set up?
I think it's possible to send docs via the private messages on here. Try this. If it doesn't work, PM me anyway and let me know.