Hey guys,

my question is actually no homework, but I thought it would fit in here best.
a) DC Voltage: charge can be calculated via Q = C * U - right? (U = voltage)
b) AC Voltage: should be the same except for U beeing not a constant value but time dependent?
C) How do you calculate the charge on a capacitor in a circuit as described below?

Circuit: AC Voltage Source (f.e sin with an amplitude of +/-10V) ---> goes into a Capacitor (f.e 1nF) ---> goes through a resistor (f.e 10MOhm) --> ground.

--> Im trying to catch the charge between the capacitor and the resistor

Cheers,
BR
Alex

 

What you have is called an RC circuit.

Since the reactance of a capacitor is a function of frequency, you need to specify the frequency of the input sine wave.

Then you can use complex math, phasors or Laplace Transforms to analyze the current and voltages.

Google RC circuit or look at this:

http://en.wikipedia.org/wiki/RC_circuit

 

you cant charge a capacitor with ac. the only way charge could be specified is instataneous charge at a point of time.,and at the next point of time it would be different

 

So I possibly could calculate the charge if i had given the function of the input voltage like ui = 10*sin(2*pi*50)
with a given time for example 1 second
?

 

The question is why do you want to calculate charge?

The charge is constantly changing.

You can calculate the current through the R and C as well as the voltage across R and C. These are not constants and will follow a sine wave, same as the input sine wave. There will be phase shifts observed in the results.

This is standard course material in a Year I or Year II college or university physics or electrical engineering program.

If you wish we can show you how to do the calculations using simple mathematics.

 

Should be easy to do, shouldn't it? They have changed the definition of voltage many times. Being this isn't homework....think of it this way. One Coulomb of charge will give you one volt of potential. This is directly proportional. So if you know the voltage at a certain time and place......you know exactly the amount of charge at that place and time.

 

One Coulomb of charge will give you one volt of potential.

Where did you get that? That is incorrect.

 

Hi,

Just a couple quick notes...

Current is defined as:
I=dQ/dt

Charge that is transferred from one place to another is simply the integral of that over the time period, and that is what allows us to test a battery for the equivalent 'charge' it contains.

The energy in a capacitor is E=C*V^2/2 and also E=Q*V/2. Equate the two to find Q.

And yes, if you know the function of the driving voltage you can find the function of the change in charge over time. It will most likely be a function as well (not a constant) if the driving function is always changing.

 

But [Q] = C = A*s so where do you get the integral from?
In my understanding this would be

Q(t) = i(t) * t

so in this respect i need to get the i(t). If i only know the u(t) it should be easy to calculate i(t) right? So lets get started with this. Given the C-->R in the circuit this should be:

i(t) = u(t) / (R + 1/(jωC))

"simplified" you get:

i = û / (√(R²+(-1/(ωC))) ) * e^(jωt - arctan((-1/(ωC))*1/R)

which means i got a phaseshift (the 2nd term) and something which is dependent on the frequenzy. This would be only correct if u(t) = û * sin(ωt).
Please correct me if there is sth. wrong in my calculations so far.
(If you come to the same conclusion it would be nice if you say so aswell)
How would you get i(t) for a random function (not just a single sin)



Cheers
BR Alex

 

Q(t) = i(t) * t

Doesn't that relationship only hold if the current is constant?

Why not use the relationship Q=C*V, where V is the voltage across the capacitor terminals? In the case of a series RC circuit excited by a sinusoidal source one can readily find the steady state AC capacitor voltage using the voltage divider relationship.

For a complex arbitrary voltage source one would need to be more specific about the function involved & as a consequence, the means of obtaining a particular solution.

Is there a point to this investigation?

 

Hi again,

Charge accumulates over time therefore it does make sense to integrate, and also we have:
I=dQ/dt

and turning that around we simply have:
dQ/dt=I

and so again it makes sense to integrate to get Q. Since there is a time frame involved we then have to know the time period too:
Q=Integral[t1 to t2] I dt

If we had a conveyor belt with apples on it we might see 20 apples pass per second, but the total number of apples that passed would be the total sum which is the integral. But if the rate slowed down to 10 apples per second we would not loose any charge that already passed, we'd just be accumulating at a slower rate now.

Q(t)=i(t)*t doesnt work and as a simple example if we have i(t)=1 we have:
Q(t)=1*t
so after 1 second Q=1, after 2 seconds Q=2, etc. but then i(t) goes down to 0.1 amp and now after 3 seconds we have Q=0.3, and 2-0.3=1.7 so where did all that charge go? The only way to remove any charge is to reverse the current.

As tnk pointed out, if i(t) is changing with time then we have to integrate. A simple example is when we have a battery that was charged and we want to know the ampere hour rating or at least the ampere hours that had really been stored in the battery.
If we use a constant current load, then we can do:
Q=I*t
but if it varies as with a resistive load we might have 2 amps for 3 hours, then 1 amp for 5 hours, so we'd have to add up all the products of i(t)*dt:
Total=2*3+1*5=6+5=11 ampere hours

If we then let this run for another 10 hours with 0.1 amps then we'd have to add that on too:
11+0.1*10=12 AHr total.

So it's basically a sum of products, not just one product. Using the integral means we allow the time delta dt to approach zero, and this gives us an exact figure because we sum the product i*dt at every possible instant of time.

Since the voltage across a capacitor is indicative of the charge contained in that cap, knowing the voltage across a cap leads to a much simpler solution.

 

Cheers you two,

i got that now..makes totaly sense to me now. I m not convince that its easier to use the u relationship since you need to know the voltage over your capacitor uc right? So in that case you might have to use the

Uc(t) = Uin(t) * Zc/(Zc+Zr) ... well ok its about the same effort isnt it?

 

Cheers you two,

i got that now..makes totaly sense to me now. I m not convince that its easier to use the u relationship since you need to know the voltage over your capacitor uc right? So in that case you might have to use the

Uc(t) = Uin(t) * Zc/(Zc+Zr) ... well ok its about the same effort isnt it?

I guess it boils down to how efficient the chosen method proves to be. If one finds the current first then this has to be integrated to determine the unknown charge - an operation which may require more effort and care.

 

Cheers you two,

i got that now..makes totaly sense to me now. I m not convince that its easier to use the u relationship since you need to know the voltage over your capacitor uc right? So in that case you might have to use the

Uc(t) = Uin(t) * Zc/(Zc+Zr) ... well ok its about the same effort isnt it?

Hi,

Either way you must know the capacitor voltage. Since you must calculate this or something like this anyway, might as well use that.

Note that to calculate the current you need Vc(t) and to get that you have to find the time solution of an RC circuit driven by a sine or cosine wave. To get the current you then:
I(t)=(sin(wt)-Vc(t))/R
then integrate over all time to get the charge.

But since you already know Vc(t) then why not just multiply by C:
Q=C*Vc(t)

The cap voltage time solution for a driving sine wave is:
VcSin(t)=(w*C*R*e^(-t/(C*R))-w*cos(t*w)*C*R+sin(t*w))/(w^2*C^2*R^2+1)

where w=2*pi*f

and the cap voltage time solution for a driving cosine wave is just the time derivative of that divided by w:
VcCos(t)=(1/w)*d(VcSin(t))/dt

I dont think it is any easier if you instead do:
I(t)=E(t)/(R+Zc)
either.

 

A long time ago in a land not too far away, grown men started rubbing cats with amber and glass. It was quite the buzz. Well, like most things, competitions started. A guy in France would say....hey I rubbed my cat 10 times and I was able to do this. And a guy in Italy would say hey...I rubbed 15 times and was able to do that. Finally they all got together and decided how many rubs to agree on. When all this started....they thought static discharge was like a fluid...they really did. And they really did catch it in a jar.....really. The jar couldn't hold too much, so they connected a bunch of jars together and they called it a battery. This was the first power supply for circuit experimentation. So they had to make a standard for electric or charge potential. You can look it up....a certain amount of charge in a certain area at a certain distance away and certain effect on a charge or some such stuff. Later, this standard was called a volt. These standards and relationships are built into every capacitor....that’s what a capacitor is. A capacitor is a voltage and current standard. These power supplies had to be equal or at least comparable for these experiments. The standard reference was one farad capacitance. It will take one coulomb of charge to give you one volt. When that volt discharges....it becomes one amp. When you charge with one amp.......you have one volt potential.
This is the standard electron flow and circuit operation theory today. That is...if you use capacitors.
The last I heard, voltage is now determined by frequency at certain energy level of some atom.......I’m sure it is some “relativistic quantum effect” by now.
In my younger days I must admit, I have rubbed a cat or two. And I know that an accumulation of charge will definitely produce a potential, but I don’t think that’s what happens in a circuit. I believe that in the future we will find that the potential in a circuit comes from alignment, not accumulation. I also believe that alignment not flow, causes the charge on a capacitor.

 

Placing several cats nose to tail in a circle (aligned) and rubbing them sequentially might produce a synchrotron. Persians might be the best breed. One could extract the line beams from the cat's eyes.