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rate of change in a capacitor

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Peter

Jan 1, 1970
0
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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ehsjr

Jan 1, 1970
0
Peter said:
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

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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Peter

Jan 1, 1970
0
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

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.
 
P

Peter

Jan 1, 1970
0
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.

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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