# rate of change in a capacitor

I'm trying to understand something.
current in a capactior is given by: i= C dV/dt
if you take the deriv. of d/dt you get: i= C * (V / 1)
is this correct?
So if I have a 1mF capactior and 100 volts DC, then my instantanous current should be: 0.1A
Plugging this into MultiSim does not give me that value.
Can someone explain this to me?
Thanks in advance!
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On 4/5/07 8:41 PM, in article e7ednZTXfJzqXIjbnZ2dnUVZ snipped-for-privacy@comcast.com,

It is difficult to understand what you are asking because it is not a series of complete sentences. Of the sentence fragments present, "current in a capactior is given by: i= C dV/dt," probably makes the most sense. "if you take the deriv. of d/dt you get: i= C * (V / 1)" seems to be meaningless. Whatever "So if I have a 1mF capactior and 100 volts DC, then my instantanous current should be: 0.1A" is supposed to mean is incorrect.
Bill -- Fermez le Bush--about two years to go.
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If you connect 100VDC to a capacitor then it draws some current until it is charged up to 100VDC then draws no more current. Your calculus seems to have gone bad (it is wrong, or maybe 'not even wrong') and you seem to be expecting a capacitor to behave like a resistor.
V is the voltage across the capacitor terminals. dV/dt is the rate of change of the voltage across the capacitor terminals. i is the current into the capacitor. The setup you describe may result in some infinities (because a perfect voltage source and perfect capacitance are unphysical). These results will be an exaggeration of what would happen in reality.
Good luck.
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Hi there peter,

No i'm affraid it's not in most case. Depending on the rate of change of applied voltage across capacitor.

I dont understand the term 'instantaneous current'. The current in capacitor depends on the voltage applied to it. i= C dV/dt Given DC voltage the current will be zero (constant V, therefore i=0). Given AC, the current will change according to the frequency.
Regards,
Lukman
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It appears I've confused everyone because I don't understand the equation.
i = C dV/dt represents what?
I was under the assumption that it meant: i = C (delta V / delta t)
meaning: if I start off a circuit with zero volts and ramping up the voltage to 10 volts, that would equal my delta v. if I ramped it in 1 second, then my delta v / delta t would be equeal to 10.
So the current (i) after 1 second would be equal to C * 10
It appears dV/dt is NOT the deltas. It's more of: d/dt (V) where V is a voltage.
Hopefully my explanation about how confused I am will help someone understand what it is I'm trying to ask.
Thank you for all your replies, I look forward to more input.
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I think you have it. It's not about the delta v, that the voltage was 0 and now it's 10 volts. It's about the instantaneous slope.
The delta v and delta t stuff is just a approximate way to figure the slope. If the voltage was wiggling around it would be more obvious that one is dealing with instantaneous slopes ather than deltas.
dave y.
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On 4/7/07 1:02 PM, in article snipped-for-privacy@4ax.com,

As I read more, I realize that Dave's primary problem is that he is having trouble using language (English) correctly. I you believe, as I do, that your underlying language skill affects how you think, the problem is obvious. If you are talking detailed technical meanings, precise use of language is essential. The subject line reveals much. "change in a capacitor's voltage," would be more on point.
The problem shows up again in

Use of the word "that" is too ambiguous.
Dave may actually understand what is going on, but I cannot tell that from what he actually wrote.
The only thing I can add at this point is that current through the capacitor changes only while the voltage across the capacitor is changing. Before the change and after the change, current is zero.
Bill -- Fermez le Bush--about two years to go.
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Peter wrote:

No. dV refers to the change in voltage across the cap, when a fixed voltage is applied to charge it, or a fixed load is connected to discharge it. If you ramp up the voltage, you introduce a completely different variables.
Apply a fixed voltage to a cap through a resistance. Maximum current occurs when the cap has no charge on it. The current decreases exponentially and the voltage across the cap increases exponentially as the cap charges. Once the capacitor is fully charged there is no current. See http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html
Ed

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Yes, the d/dt are kinda like deltas and in this case they would happen give the exact same answer as deltas because the ramp function is a nice straight line. The d/dt are actually the 'derivative', a calculus operation that essentially finds the slope at any / all instants. However the slope of your ramp is constant so deltas give the same result.
Note that the current is C * 10 for the entire time:
i(t) = C * dV(t)/dt
since your dV(t)/dt (the slope of the graph of V vs t) equals 10 [volts/second] at any time t between 0 and 1 s, the formula
i(t) = C * 10
is valid for t between 0 and 1 s.
j
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It may be more meaningful if you re-arrange the equation a bit.
Thanks to Liebnitz notation, we can do some things like....
dt * i / C = dV
This says if you apply a fixed current for a short time, then the change in voltage on the capacitor terminals is equal to the current times the time (in seconds) divided by the capacitance. This is charging the capacitor with a constant current source.
In your example, (i=C dV/dt), if you want to figure out the current, you could. Suppose you are ramping up the voltage applied from 0V to 10V over 10 seconds. Now, for this to work with deltas instead of derivatives, you must ramp the voltage at a constant rate for the entire 10 seconds. If you do that, then....
i = C * dV / dt = C * 10V / 10s = C * 1 V/s
If you don't ramp the voltage 'evenly', then the current won't be constant. So we could take shorter, and shorter intervals. Down to as short a time span as you want, and the 'instantaneous current' would be what you get if you measure the amount of voltage change over an 'instant' of time. Suppose you measure the voltage change from one particular nanosecond to the next nanosecond and found that the applied voltage changed by exactly one nanoVolt. Then the current during that nanosecond would be...
i = C * 1e-9 V / 1e-9 s = C * 1 V/s
Same number as before.
daestrom
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So how do I attempt to solve for the derivative? In calculus class, you're always given something like x d/dx and you solve the derivative in that fashion which turns out to be 1 (correct?). Now I'm given somethign not so straight forward such as: C dV/dt
The deriv. of d/dt V = 0, correct?
This is why I'm so confused about this equation.
Thanks for everyone's help so far.
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The essential problem is you're not far enough along in your studies. After calculus comes a class in differential equations and that's where you'll learn to solve these types of equations. Patience.
dave y.
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On 4/10/07 8:14 PM, in article snipped-for-privacy@4ax.com,

The differential equations (DEs) involved here are so simple that if you do not know how to solve them with simple integration of simple functions, you should try another endeavor.
Bill -- Fermez le Bush--about two years to go.
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wrote:

I asked a question, got a few answers, but then as we came to the conclusion you insulted me.
I agree, it's all basic stuff, however, I don't understand and came here for an answer - that's why we have newsgroups. If we could close this issue and explain how to apply the basic capacitor calculus formula (i = C dV/dt) instead of considering them deltas, we could pass on wisdom and advance technology instead of throwing subtle insults.
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On 4/13/07 4:39 PM, in article ucOdnS7aVsFbib3bnZ2dnUVZ snipped-for-privacy@comcast.com,

In truth, you just provided a better formulation of of a question than the formulations of your previous question.
In what follows, I hope a "�" shows up as a delta as it does on my computer.
i = C dV/dt) and I = C �V/�t) state almost the same property, except i represents instantaneous current while I represents average current over the time �t. You go from I to I by taking the limit as �t -> 0. That is how you fo from finite changes to derivatives.
Bill -- Fermez le Bush--about two years to go.
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Ok, why not do it. Here's an example of how this is used.
Assume your circuit is a battery E charging a capacator C thru a resistor R.
Let the voltage across the capacitor be V
The current in the capacitor is i = C*dV/dt
The voltage across the resistor is R*i
If you sum the loop voltages you get,
E = R*i + V
Substituting in the formula for current in a capacitor,
E = RC*dV/dt + V
Rearranging to the normal form,
dV/dt + V/RC = E/RC
The above is a first order linear differential equation. This simplest of examples can be solved by separation of variables, but in real life the equations get harder and you'd want to use Laplace transforms if possible.
dV/dt = E/RC - V/RC = (E-V)/RC
dV/(E-V) = dt/RC
Integrating both sides,
ln(E-V) = t/RC + constant1 E-V = exp(t/RC + constant1) = exp(t/RC) * exp(constant1) E-V = exp(t/RC) * constant2
We have to determine the integration constant. At t=0, the capacitor wouldn't have had time to charge, so it's initial value is zero. Therefore at t=0,
E-0 = exp(0) * constant2 constant2 = E
Therefore
E-V = exp(t/RC) * E
V = E*(1 - exp(-t/RC) )
So the voltage across the capacitor has an exponential rise up to the battery voltage with time constant RC.
There ya go,
dave y.
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