Calculating Capacitance: Irms and Angular Frequency Data Analysis for Homework

[k31453]

Homework Statement

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

Homework Equations

The Attempt at a Solution

Did do anything wrong?

 

[NascentOxygen]

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

Hi k31353.

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

How many μF is it?

 

[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?

 

[k31453]

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?

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

 

[NascentOxygen]

Slope is C but how do i prove that the slope represents the capcitance of capacitor.

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.

yes and the value is 0.1 uF

This information will be useful.

 

[CWatters]

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

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?

 

[k31453]

Ok this is my circuit



Now and this are the data.



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 :



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

So this is the problem

 

[milesyoung]

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

 

[k31453]

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

Nothing actually that's all I've got

 

[milesyoung]

Do you have a picture of it?

 

[k31453]

Do you have a picture of it?

ok that's all i got

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

 

[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.

 

[k31453]

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.[/QUOTE
I don't get that part

 

[NascentOxygen]

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?

 

[k31453]

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

 

[k31453]

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?

I recorded ma.

 

[milesyoung]

Have a look at the figure included under 'Root mean square amplitude' here:
http://en.wikipedia.org/wiki/Amplitude

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

 

[milesyoung]

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?

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.

 

[k31453]

Have a look at the figure included under 'Root mean square amplitude' here:
http://en.wikipedia.org/wiki/Amplitude

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

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

 

[NascentOxygen]

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[strike] 4.4μA.[/strike] EDIT: 4.4mA[/color]

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.

 

[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

 

[k31453]

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

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.

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 doesn't 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?

 

[NascentOxygen]

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.

 

[k31453]

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.

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??

 

[NascentOxygen]

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 ??

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

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.

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

 

[k31453]

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.

exactly so am i stupid or is it trick question ? now as above post v/i slope represents resistor.
Now how the hell i can find resistor / impedendence of the circuit where only C is given.

 

[NascentOxygen]

C = I / (ω V)

⇔[/size]

C = ( I / V ) / ω

.......

 

[k31453]

.......

whats your point ?

 

[milesyoung]

You know the magnitude of the voltage and current are related by:
V/I = 1/(ω*C) ⇔ I = (C*ω)*V + 0 = (C*V)*ω + 0

Doesn't that look similar to the form of your line fit: y = a*x + b?

Now how the *beep* i can find resistor / impedendence of the circuit where only C is given.

What is the magnitude of the impedance in terms of V and I?

 

[CWatters]

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 ??

Neither.

Start from...

Z = 1/jwC = V/I

then rearrage it to be the same form as y = mx + c. You can do that two ways depending on which you want on the y axis...

1) With V on the y axis...

V = (1/jωC) I

In which case the slope is = 1/jωC

2) With I on the y axis..

I = (jωC) V

In which case the slope is = jωC

Note that both cases the slope can be used to calculate C but in neither case is the slope exactly = C.

PS: You could always plot I vs jωV. The the slope would be C.