# Homework Help: Finding capacitance

1. May 7, 2013

### k31453

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

This is my data of Irms and angular frequency, and I have to plot the graph and teacher told me that the gradient will be capacitor's capacitance?

2. Relevant equations

3. The attempt at a solution

Did do anything wrong?

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2. May 7, 2013

### Staff: Mentor

Hi k31353. [Broken]

Have you worked out the value of the capacitance? (Can I presume that is the aim of the exercise?)

How many μF is it?

Last edited by a moderator: May 6, 2017
3. May 7, 2013

### CWatters

The general equation for a straight line is y=mx+c where m is the slope and c is a constant.

I suggest writing the equivalent equation for your experiment.

Is the slope C or 1/C or something else?

4. May 13, 2013

### k31453

Slope is C but how do i prove that the slope represents the capcitance of capacitor.
yes and the value is 0.1 uF

5. May 14, 2013

### Staff: Mentor

The slope is not exactly = C. The voltage you are applying must come into the equation somewhere, since if the voltage you are using is greater, the current must be greater.

Can you quote an equation relating all of those quantities, C, ω, I and V? That is what you need to examine to see what you are dealing with.

This information will be useful.

6. May 14, 2013

### CWatters

I asked because you plotted ω against I rather than I against ω. Perhaps I was awake too late last night but...

The imp of a capacitor is

Z = V/I = 1/jωC

rearrange to give ω in terms of I...

I/V = jωC

ω = I/(VjC)

ω = (1/VjC) I

which is now in the form

y = mx + constant

So is the slope m = C? 1/C ? or something else?

7. May 14, 2013

### k31453

Ok this is my circuit

[Broken]

Now and this are the data.

[Broken]

Then this is the question :
From the slope of the graph verify the value of the capacitor ? (Hint: if the system is sinusoidal AC, then I RMS = angular frequency(omega) * C * V RMS). Does it agree with the measure value? If not, why not?

So this is my answer :

[Broken]

So my capacitor answer is way different than it should be.

So this is the problem

Last edited by a moderator: May 6, 2017
8. May 14, 2013

### milesyoung

What does it say on your capacitor? Include any and all markings.

9. May 14, 2013

### k31453

Nothing actually thats all ive got

10. May 14, 2013

### milesyoung

Do you have a picture of it?

11. May 14, 2013

### k31453

12. May 14, 2013

### milesyoung

Ah doh, I missed an error of yours:

Vrms = Vpp/(2*sqrt(2))

The peak-to-peak amplitude Vpp is 2*Vp where Vp is the peak amplitude.

13. May 14, 2013

### k31453

14. May 14, 2013

### Staff: Mentor

Your table indicates at 1kHz and with a voltage of 7V the unknown capacitor drew 2.28mA.

Let's compare that with what a known capacitor would do. A capacitor of 0.1μF, at 1kHz and an applied voltage of 7V, would draw 4.4μA.

Big difference.

I wonder was your ammeter scale really μA where you recorded mA?

15. May 14, 2013

### k31453

Isnt it if u wanna get rms u just divide by squareroot of 2??

16. May 14, 2013

### k31453

I recorded ma.

17. May 14, 2013

### milesyoung

18. May 14, 2013

### milesyoung

His readings are fine, he just didn't calculate the RMS voltage correctly.

If Vpp = 10 V then:

Vrms = Vpp/(2*sqrt(2)) = 3.536 V
C = |Irms|/(2*pi*f*|Vrms|) = 112.5 nF ≈ 0.1 uF

for f = 100 Hz.

19. May 14, 2013

### k31453

Ahhhh gotchya..
U earn yourself a fortune cookie...
Thanks

20. May 14, 2013

### Staff: Mentor

No, I really do make it 4.4mA, so your figures are of the right order, just a factor of 2 difference. That factor of 2 might be accounted by that voltage conversion error milesyoung pointed out.

21. May 14, 2013

### CWatters

I agree. I checked the data for...

8V pk-pk, 1KHz, I=1.8mA

Vrms = 4/0.707 = 2.83 Vrms (eg not 5.65 Vrms)

C = Irms/(Vrms * 2∏f)
= 1.8 * 10-3/(2.83 * 2 * ∏ * 1000)
= 0.101 * 10-6F

Within 1% of 0.1uF

22. May 14, 2013

### k31453

[Broken]

https://www.dropbox.com/s/3d4x57necbe96lz/IMAG0301.jpg

So my calculation shows me 0.1UF but my graph slope shows me 0.649 F capacitor magntiude.

So From the slope of the graph verify the value of the capacitor ? . Does it agree with the measure value? If not, why not?

so it doesnt verify the value right !?? because the two answer are not same.

But what is the reason. because the graph should give me the capacitor value because c = q/ v

isnt it?

Last edited by a moderator: May 6, 2017
23. May 14, 2013

### Staff: Mentor

The slope of your I vs. V plot is I/V and this is not equal to C. In the text by CWatters which you quoted it is explained how to calculate C.

Besides, I/V has units of Ohms-1, so it could not ever equal a capacitance. (Examining the units of quantities on both sides is a handy technique you can use for confirming that equations you remember only vaguely may be correct.)

Q is not the symbol for current.

24. May 15, 2013

### k31453

Is it i m stupid or my professor because he wants us to draw Irms vs Vrms which is ridiculous because I is dependent variable so why he wants me draw on x axis ??

Then 2) i am doing another circuit which involves find capacitor from slop now they want me to plot graph Irms Vs angular frequency.. but it should be vice versa.

3) I have to identify low pass or high pass filter based on graph.

So they want me to plot Gain(db) vs frequency.. which should be vise versa as well.

I mean why they want me do like that??

25. May 15, 2013

### Staff: Mentor

I vs. V is okay. You vary V and plot the changing I. Sounds fine by me.

I vs. ω is correct. ω is the independent variable, so it's plotted on the horizontal axis.