Looking at your spreadsheet, this may be the problem.

A sine wave with a peak to peak voltage of 2V has an amplitude of 1V. The rms voltage is therefore \(\frac{1}{\sqrt{2}}\) or 0.71 V.

Dude if i am using 0.71 rms value. it does give me correct answer but why i do that rather than divide peak to peak value by squareroot 2?

 

To get the rms value of a sinusoid, you do NOT divide the peak-peak value by the sqrt(2), you divide the AMPLITUDE by the sqrt(2). You might remember that, back in Post #5, I warned you about not mixing these together.

 

Dude if i am using 0.71 rms value. it does give me correct answer but why i do that rather than divide peak to peak value by squareroot 2?

I think WBahn has explained the relationship between amplitude and rms values well enough.

To be fair, there are inconsistencies in the spreadsheets so it was not obvious: sometimes it was \(V_{peak}\), sometimes \(V_{p-p}\).

Glad to know I am a 'dude' - although not entirely sure whether that's a good thing.

 

I think WBahn has explained the relationship between amplitude and rms values well enough.

To be fair, there are inconsistencies in the spreadsheets so it was not obvious: sometimes it was \(V_{peak}\), sometimes \(V_{p-p}\).

Glad to know I am a 'dude' - although not entirely sure whether that's a good thing.

Thats good thing bruv. and thanks for the support I aint like physics !!!

 

and it ain't cool to call your teachers 'dude' and 'bruv'.

 

and it ain't cool to call your teachers 'dude' and 'bruv'.

hahahaha

Okay

mm so

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?

 

Units, units, units!!!

You're making the same mistakes all over again.

Don't take raw numbers and ignore units.

C = Q/V = charge/voltage = current x time / voltage

What happened to time?

What happened to the scale units of Vrms and Irms?

Just eyeballing your graph, your slope of 0.649 is not correct.

Btw - Didn't they teach you dimensional analysis in school?

 

Units, units, units!!!

You're making the same mistakes all over again.

Don't take raw numbers and ignore units.

C = Q/V = charge/voltage = current x time / voltage

What happened to time?

What happened to the scale units of Vrms and Irms?

Just eyeballing your graph, your slope of 0.649 is not correct.

Btw - Didn't they teach you dimensional analysis in school?

yeah i did change it to mA to a but still i got wrong answer..

i think i got the answer ...

isit c= q/v only apply for dc circuit right?

 

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??

 

Because you are still not getting it and you are not paying attention.

I have repeatedly asked you to do two things:

1) Write out the proper equation.
2) Write out the units (dimensions) and check for equality.


The generic equation is y = mx where m is the slope.

m = y/x

The units of y/x must match the units of m.

The equation is I = V/Xc where Xc = 1/ωC.

I = VωC

Which is y, which is x and which is m?

The slope of I vs V is ωC.
The slope of I vs ω is VC.

ALWAYS check for equality of units on both sides of the equation:

Units of I/V must be equal to units of ωC.
Units of I/ω must be equal to units of VC.


Note: This is going to be my last post on this matter. I cannot help you further than what I have written here.

 

Right. That means plot rms current on the y-axis and angular frequency on the x-axis.

You have also been told that plotting \(I_{rms}\) vs \(V_{rms}\) means I goes on the Y axis and V goes on the X axis.

What your professor wants is not ridiculous. He is saying the same thing as we are saying. And we are all correct!