# verifying equality in both equations that gives the voltage across a capacitor in an AC RC series

Discussion in 'Physics' started by kiroma, May 28, 2015.

1. ### kiroma Thread Starter Member

Apr 30, 2014
40
0
Hello.
In a AC RC circuit, this is the Capacitive Reactance, as you are reading should know (if you're going to answer to this question, haha).
(1)$image=http://latex.codecogs.com/gif.latex?%5Cfrac%7B1%7D%7B2%5Cpi%20fC%7D%3DXc&hash=b73196030e3770ad7935b16bb7c03979$
And this is the total impedance, given a R which is the resistance in series.
(2)$image=http://latex.codecogs.com/gif.latex?%5Csqrt%20%7B%5Cfrac%7B1%7D%7B%282%5Cpi%20fC%29%5E2%7D+R%5E2%7D%3DZ&hash=a7d0bcefe36a106fbd65227e08ef38e8$
Then the rms current in this circuit can be written by Vrms/Z
And it times the Xc will give us the voltage across the capacitor, as follows:
(3)$image=http://latex.codecogs.com/gif.latex?%5Cfrac%7BVrms%7D%7B%5Csqrt%20%7B%5Cfrac%7B1%7D%7B%282%5Cpi%20fC%29%5E2%7D+R%5E2%7D%7D*%5Cfrac%7B1%7D%7B2%5Cpi%20fC%7D%3DVc&hash=ffea6f68264ad149351330ec5098e005$
Until here I know it's all true, and I don't have any doubts about it.
But I did want to integrate an half period of the sinusoidal current to see if the voltage is equal as calculated before.
(4)$image=http://latex.codecogs.com/gif.latex?%5Cint_%7B0%7D%5E%7BT/2%7D%5Cfrac%7BVrms%7D%7B%5Csqrt%20%7B%5Cfrac%7B1%7D%7B%282%5Cpi%20fC%29%5E2%7D+R%5E2%7D%7D*%5Cfrac%7Bsin%28%5Comega%20t%29%7D%7BC%7Ddt&hash=2bddc0b1e8994fa32347ef8299523c22$
Integrating the currrent (Vrms/Z) will give us the charge "q".
By the definition of capacitance, we have that: V=q/C
That's why I put a divided by C.
Ah, omega = 2*pi*f
The integral I showed before (4) is not the voltage across the capacitor. It's simply exactly the double voltage.
Which means it should be divided by 2 before, to give us the Vc.
Solving the integral (4) give us
(5)$image=http://latex.codecogs.com/gif.latex?%5Cfrac%7BVrms%7D%7B%5Csqrt%20%7B%5Cfrac%7B1%7D%7B%282%5Cpi%20fC%29%5E2%7D+R%5E2%7D%7D*%5Cfrac%7B1%7D%7B%5Cpi%20fC%7D&hash=6027707c408dd46651b0857649ab6753$
This is what I showed you in (4).
Multiplying by 1/2 will give us exaclty the equation number (3), which is the Vc.
Here comes the question
Why multiplied by 1/2?
Where does it come from?

Solution I thought hours later:
The second half of the first half period charges the capacitor with Vc.
Then it requires half of the second half period (which is negative) to discharge the capacitor, and another half of the second half period to charge negatively the capacitor.
And it goes on.
IF this is the answer, to where it goes the first half of the half period?

Last edited: May 28, 2015
2. ### WBahn Moderator

Mar 31, 2012
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You are confusing instantaneous and aggregate quantities. Vrms/Z gives you the aggregate magnitude of a waveform, not the instantaneous values of that waveform. When you integrate a waveform, you need to use the instantaneous values.

3. ### kiroma Thread Starter Member

Apr 30, 2014
40
0
So, how would it be after the correction?
Because, sorry, but I didn't understand what you meant.

Edit:
Ok, I think I understood what you meant.
Vrms is the root mean square value. But if I take the maximum value, which is Vrms*sqrt(2) and integrate, then I'll have the maximum value of the capacitor voltage. So I'll have to divide by sqrt(2) and this is just a constant in the integral which I didn't mention because it cancels itself.

Last edited: May 29, 2015
4. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
I think your confusion runs deeper than WBahn suspects.

Your equation (2) is incorrect, despite your declaration that we all agree.

Equation (3) is a strange way to go about calculating things.
WBahn is right.

Work with instantaneous values.

Last edited: May 29, 2015
5. ### WBahn Moderator

Mar 31, 2012
17,757
4,800
If you take the maximum value and integrate that, then you will get a value that corresponds to what you would get if your waveform was a DC voltage at that maximum value. But that is not the waveform you are working with.

6. ### WBahn Moderator

Mar 31, 2012
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I suspect you are correct.

7. ### MrAl Well-Known Member

Jun 17, 2014
2,438
492

Hi,

Interesting idea, but i dont think that will work because the voltage across the capacitor is not sin(w*t) it just so happens that when the R and C is excited with an AC voltage of 1v it is really:
Vc(t)=(w*C*R*e^(-t/(C*R)))/(w^2*C^2*R^2+1)-(w*cos(t*w)*C*R)/(w^2*C^2*R^2+1)+sin(t*w)/(w^2*C^2*R^2+1)

(and multiply that by whatever amplitude the source voltage really is)

and for your purpose you can set the exponential term to zero to get just the sinusoidal part which would exist after say 10 time constants. You can use this to check any results you may come up with when trying other methods.

8. ### kiroma Thread Starter Member

Apr 30, 2014
40
0
I'm not integrating just the maximum value. I'm integrating like Imax*sin(2pi*t)dt which is the charge "q".

I know to admit when I'm wrong, but when it's always the same constant that is missing... I don't think I'm wrong. By the way I'll talk to my electric circuits I professor about it.

Voltage is not sin(2pi*t), but the current is! I don't think you understood me what I did there.
What I did there was to integrate the current to get the charge in the capacitor and then divide by the capacitance to get the voltage across the capacitor.

To all: how do, then, I calculate voltage across a capacitor in an ac series circuit?
Because every value I put to get a voltage, the same exact value I find in the falstad circuit.
If you don't know what is fans had, search on Google, it's a very trusty app made in java.

9. ### WBahn Moderator

Mar 31, 2012
17,757
4,800
It's still a bit hard to tell exactly what it is you are trying to do -- I suspect that's an "English as second language" issue that we just need to struggle through.

Why are you wanting to integrate the current in the capacitor to find the voltage on it at a particular time, namely T/2 where T is the period of the driving frequency. But what is it that you think this will tell you? In general this value will not have much meaning, if for no other reason than you have a phase shift and so the current integrated over a one half period interval is going to depend on when you start that interval and, because of the phase shift, there is nothing special about starting the interval at time t=0. You also need to take into account the initial charge on the capacitor, which, in general, will not be zero.

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10. ### MrAl Well-Known Member

Jun 17, 2014
2,438
492
Hi again,

Ok i'll take another look at what you are talking about and see if i can sort it out.
Perhaps if you explain in words what you are trying to do that would help too.

LATER:
Ok, it looks like the problem is still with your sin(wt) for the voltage OR current.
For the network RC excited by a sine wave the current is really:
I(t)=(w^2*sin(t*w)*C^2*R)/(w^2*C^2*R^2+1)+(w*cos(t*w)*C)/(w^2*C^2*R^2+1)
-(w*C*e^(-t/(C*R)))/(w^2*C^2*R^2+1)

and if we allow an appropriately long time to pass we would have simply:
I(t)=(w^2*sin(t*w)*C^2*R)/(w^2*C^2*R^2+1)+(w*cos(t*w)*C)/(w^2*C^2*R^2+1)

and you can see right away that this is more complicated than just sin(wt). This can be equated to a form of A*sin(wt+ph) but you'd have to do that too then, and that is still much different than sin(wt) because it has both a new amplitude and an added phase shift.

So you'd have to try using that current expression and see if you can get it to work.
You may have to solve for the zero crossing or something first too.
Try that and see what happens.

Here are two plots of the current with different component values and different frequencies...

• ###### RC_Network-1.gif
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Last edited: May 31, 2015
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11. ### kiroma Thread Starter Member

Apr 30, 2014
40
0
Hi again. Sorry, I was busy, that's why I didn't answer.

It's funny the way you find English is my second language. I wonder where did you find it.
By the way I'm Brazilian

That's the point. It's not at a particular time. The same way you integrate the power from 0 to 2*pi so in the end you have the average power, I integrate current from 0 to T/2 to have the voltage peak at the end of the cicle, like it's done with the power I mentioned above. Then the peak value I divide by sqrt(2) to have the rms value as I said before it cancels itself because it's from rms to peak and then peak to rms.

For those who didn't understand what equations (2) and (3) meant:
1/(2pi*f*C)=Xc
sqrt(Xc²+R²)=Z (impedance)
Because when you're dealing with vectors, I learnt I need to do that because the voltage in the capacitor is delayed 90° in relation to the voltage of the source.
Then we apply pythagoras theorem which is a²=b²+c²
Where a=Z, b=Xc and c=R.
Then we have the Z, the impedance.
Dividing the Vrms by the impedance we'll find the current through the circuit, which is the same for all elements.
And it multiplied by Xc we'll give us the voltage across the capacitor. Like we do with resistors, Ir*R=Vr.
Which is Vc=Vrms*Xc/Z
I tested and it works.

The other manner to do that is by integrating the current, which we know, over the sinusoidal signal.
Knowing the current is a kind of sen(2*pi*f*t) and supposing a instant t after a long time has passed, the quantity of charge that passed through the resistor and went to the capacitor would be the integral of the current, isn't it?
C=q/Vc then
Vc=q/C
So
Vc=integral(i(t))dt/C
That's what I did.

And the question should be why the voltage across the capacitor is just half of Vc. (laughs)

12. ### WBahn Moderator

Mar 31, 2012
17,757
4,800
If I spoke Portuguese, even at a modest level of fluency, and posted on a Portuguese-language forum, you would probably spot pretty quickly that Portuguese was not my native language. You would find that the way I expressed many things was at least a bit off and, therefore, hard to understand, particularly if the topic was technical.

Let's take a look at what you are trying to claim.

You are claiming that if you integrate the current into a capacitor over half a cycle that that will give you the peak total charge stored on the capacitor. From that you can get the peak voltage on the capacitor and, from that (and for a sinusoidal signal) you can get the RMS voltage on the capacitor.

Let's examine the first part of that claim -- that integrating the current into the capacitor over half a cycle will give you the peak total charge stored on that capacitor. Go back and carefully read my Post #9, particularly the last sentence.

In sinusoidal steady state, what is the voltage on the capacitor at the moment that the current changes direction from positive to negative. You are blindly assuming that it is zero. It's not. If you want to find out the voltage on the capacitor at the end of the integration period, then you MUST take into account the voltage at the beginning of the integration period. You can't just assume that it's zero.

Draw the voltage and current waveforms for the capacitor. You want to start your integration when the capacitor has zero charge on it (i.e., zero voltage) and end your integration when the capacitor has max charge on it (i.e., max voltage).

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13. ### studiot AAC Fanatic!

Nov 9, 2007
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Is the integral of power from 0 to 2pi not zero?

14. ### kiroma Thread Starter Member

Apr 30, 2014
40
0
Hmm. I think that's why that constant was missing, because I didn't consider the initial charge.
I can't upload a image right now because I'm working, but ASAP I'll upload one showing lower current and higher at the first half of the first cycle. I just can't understand why Ic=C*dv/dt doesn't hold true within the first semi-cycle, because in t=0, dv/dt=max but Ic=0.

Quoted it again because it blown my mind. How could I didn't see things simple like this?!
What you mean is
$image=http://latex.codecogs.com/gif.latex?%5Cint_%7B0%7D%5E%7BImax%7DI%28t%29dt%3D%5Cint_%7B0%7D%5E%7BVmax%7DCdv&hash=737369b4c32224223dc318619a83197a$
From q=CV
dq/dt=C*dv/dt
I(t)*dt=C*dv/dt

With this, yes, we can set the upper limits to Vrms and Irms
Which will be still equal and will let us get the Vrms across the capacitor
And Vmax is acquired in half a semicycle

Now, knowing that half semicycle is used charge the capacitor, half of the negative semicycle is used to discharge then half negative semicycle for it to charge and the half positive semicycle to discharge the negative voltage (and goes on), as time is a lot higher than 0, I make the following affirmation:
The first half positive cycle worth modulus of half negative semicycle, speaking about current.
ASAP I'll post a image of it.
I don't know how to prove this and I'd like to.

In a pure resistive circuit with a sinusoidal, this integral is power/2. That was only an exemple, not an accurate analogy.

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15. ### kiroma Thread Starter Member

Apr 30, 2014
40
0
What is the problem? Did I commit an terrible error?

16. ### WBahn Moderator

Mar 31, 2012
17,757
4,800
Don't be so sensitive. My first post here took a couple weeks to get a response. Members do not live and breath for the privilege of making super timely responses to your posts.

Also, in your last post you said that you would post some images ASAP. Perhaps we are holding off to see what you post.

17. ### kiroma Thread Starter Member

Apr 30, 2014
40
0
Ok, sorry. I'm just paranoid, thinking I did something wrong, always. I know, but seeing the period of time to get some reply, I thought at this time I'd have another. But no problem if it's just an issue of time.

By the way I think you understood my question. But anyways here it goes a bit more easy to understand.
I am stating that the first semicycle is equal to half of the negative semicycle, in modulus, as image shows.

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18. ### WBahn Moderator

Mar 31, 2012
17,757
4,800
Actually, I didn't see a question at all in that post.

The image you just posted is largely unrelated to your question because it is not the sinusoidal steady state response, which is what all of your work has been dealing with. Do you see how the waveform is not symmetric and how adjacent peaks are changing. This is the transient response portion of the total response. Your general thought that the total charge in the first semicycle is equal to the total charge transferred in the first half of the second semicycle is largely correct. How close it is to being correct is going to depend on the time constant of the circuit relative to the period of the applied signal. In your circuit you have a time constant of about 5 ms, which is about 1/4 of a cycle. So you can expect transient effects to play a role across that first semicycle boundary. Make both the resistor and the capacitor, say, four times the size they are currently and this will probably be more apparent.

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19. ### kiroma Thread Starter Member

Apr 30, 2014
40
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Yes, I saw that. Yeah.
My question now is how do I achieve the equation which draws this transient effect?
I tried to simulate 4x what they were, but didn't find anything visible
There's just a small variation from the first sinusoid to the last, considering from 0 to 500 ms.

Last edited: Jun 3, 2015
20. ### WBahn Moderator

Mar 31, 2012
17,757
4,800
You need to solve for the transient response. The best (?) way to do that is using what are known as transform methods. Specifically, using either the Laplace or Fourier transforms to solve the differential equation that describes the system. There's a pretty good mathematical gap that you have to cross to go from the formulas you are using and transform methods. Do you have any calculus behind you?

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