1. Oct 28, 2009

### Apteronotus

Hi,

I've asked this one question in one form or another a couple of times now and no one seems to have a good idea as to how to solve it. So I'll ask it again, in the hopes that the right person may come across it.

Suppose you have a simple capacitor, where the potential on one side is held constant and the other varies randomly.

What is the capacitive current?

thanks,

2. Oct 28, 2009

### Bob S

Use the relation Q=CV, where Q is the charge, C = capacitance, and V = voltage across the capacitor. Now differentiate, yielding

dQ/dt = I = C dV/dt

Bob S

3. Oct 28, 2009

### Apteronotus

The problem is that since V is a random function, its derivative is not defined, (to my knowledge).

4. Oct 28, 2009

### Staff: Mentor

I'm not sure what kind of answer you want. Do you understand the relationship between the voltage on a capacitor and the current flowing through the capacitor? It's a direct relationship. Just because you only have a random representation for the voltage, that doesn't change the fundamental capacitor equations.

What is the context of your question? What is the physical setup? Maybe that will help us understand why you are asking this.

5. Oct 28, 2009

### Apteronotus

Hi,
So perhaps I should make myself more clear. Though I have no background in physics, I do understand the relationship $$\left(I_C=C\frac{dV}{dt}\right)$$ between voltage on a capacitor and the current flowing through the capacitor.

What I dont understand is, given that V is random with some distribution, how do you calculate this equation?
It seems that to calculate, one would have to take the derivative of a random signal, V. To my knowledge this is not possible.

To put it another way, if all we know is that V~N(0,1) then what is $$\left(\frac{dV}{dt}\right)$$?

The context is the current across a biological cell, due to a voltage gradient across the cell membrane. Some of the current will pass through the cell membrane, some will cause a build up of charge on the cell. What are the equations describing this when the voltage gradient is noise? (Before, I tried to keep my question as general as possible, so that other forum members could also benefit, but this is the context of my question)

Thanks

6. Oct 28, 2009

### Staff: Mentor

Ah, much more clear now,thanks. THe random voltage fluctuations will have a finite bandwidth, which makes them differentiable. Does that help? I'm not real strong on differentiations of random functions, but I'm pretty sure that you can do it after you define the bandwidth characteristics of the random voltage signal.

7. Oct 28, 2009

Your biggest problem is trying to state "varying randomly", while accepting the relationship between voltage and current on a capacitor. In particular, **WHY** would the voltage be "randomly"? If what you are seeing is that you MEASURE it "varying randomly", then something must be CAUSING it to do so. What you CAN then say is that you KNOW the current flowing into/out of the capacitor follows the relationship above. Absent something CAUSING the changes (i.e. injecting current into/out of the capacitor), the voltage would be steady - and related to the charge on the Cap.

8. Oct 28, 2009

### Staff: Mentor

Welcome to the PF, LeadDreamer. Your style of posting is making me dizzy....

9. Dec 2, 2009

:) I apologize for the style; I tend to write the way I speak (generally fairly animated), and text doesn't lend itself well to that...

10. Dec 3, 2009

### Drill

As far as I know

since the voltage is Random function ,the capacitive current will be a random function also ,then the best way to solve this is in the frequency domain and
use 1/jwc as the impedance

but first we need to know the randomness of the voltage

11. Dec 4, 2009

### Redbelly98

Staff Emeritus
A capacitor resists changes in voltage. So, it's impossible for a real power source to produce a voltage that changes randomly from one instant to the next.

12. Dec 5, 2009

### MaxwellsDemon

Personally if I had a bunch of random data that isn't a nice function of time, I would look for statistical quantities and determine their behavior. For example, maybe the average potential is a nice function of time and this would allow you to find the average current per unit time.

13. Dec 5, 2009

### omkar13

As we know q=cv and i=dq/dt==>dq/dt=d(cv)/dt but since derivatives are only defined for continous functions probably we can take like this
i=cv/t where i=current at required time
c=capacitance at that instant
v=potential at that instant
if anyone think that this is wrong they can mail me at [email address deleted] and even in the forum

Last edited by a moderator: Dec 5, 2009
14. Dec 5, 2009

### Redbelly98

Staff Emeritus
This is saying that i=q/t or in other words q=i*t. That is not true in general, though it would be true when i is constant and q=0 initially.

15. Dec 6, 2009

### omkar13

sorry my idea is that i=Dq/Dt but not q/t However i think no current passes through a capacitor since there will be no electrical connection between the plates

Last edited: Dec 6, 2009
16. Dec 14, 2009

Sigh. A few issues and confusion remains.

The basic relationship for a capacitor is simply Q = CV ; that is to say, the *Charge* on the (opposite plates of the) capacitor is related to the Capacitance times the Voltage - regardless of the rates that they change.

All other relationships derive from this. For example, the common I = C*(dv/dt) derives from the fact that I (current) is *defined* as the rate of change of Charge - i.e. flow.

The continuous nature (others have stated) of the derivative function is not enforced by the fact that derivatives/integrals are not defined at singularities - it's defined by the physics. Step (instantaneous) functions in voltage on a capacitor require Step (instantaneous) changes in the *Charge* on the capacitor - i.e. charge would have to somehow instantaneously appear/disappear - violating basic conservation principles.

As far as "charge not flowing through a capacitor", well... no, not exactly, but...

The point of a capacitor is that there is "equal and opposite charge on the opposing faces" - so if there is a current (charge flow) "into" one plate, it will attract the opposite charge on the opposing face - which causes a charge flow - which is a current. So **semantically** current doesn't flow "through" a capacitor - but as a "black box" device, current flows in one lead, and out another, so it might as well be treated as "flowing through".

Net of all of this - if you have a "random voltage" - all you can say is what the charge is on the capacitor at the moments of measure. You can also say that *if* the voltage has changed, then the *charge* has changed - which can be netted as a current over the measurement time period.

Much of this stems from the original contention of "voltage varies randomly" - are you asserting this as a thought experiment? A series of measurements? What kind of "random"? Way too many things unexpressed to make a real answer.

17. Dec 15, 2009

### omkar13

sorry i didnot get you i didn't think of all that random voltage and all but i tried to explain him that no current passes through capacitor