|
Home > Archive > Electrical Engineering > April 2007 > rate of change in a capacitor
You are viewing an archived Text-only version of the thread.
To view this thread in it's original format and/or if you want to reply to
this thread please [click here]
| Author |
rate of change in a capacitor
|
|
|
| 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.
| |
| dave y. 2007-04-07, 5:25 pm |
| On Sat, 07 Apr 2007 10:16:32 -0500, Peter <private@private.com> wrote:
>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.
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.
| |
| Salmon Egg 2007-04-07, 5:25 pm |
| On 4/7/07 1:02 PM, in article 3etf13ho03kl8vhfhgvir9pc812ujnu0hd@4ax.com,
"dave y." <nospam@myhouse.com> wrote:
> On Sat, 07 Apr 2007 10:16:32 -0500, Peter <private@private.com> wrote:
>
>
>
> 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.
>
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[color=darkred]
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.
| |
|
| Peter wrote:
> 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.ed...ric/capchg.html
Ed
>
>
>
>
> 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.
>
| |
| operator jay 2007-04-09, 3:25 am |
|
"Peter" <private@private.com> wrote in message
news:VuqdnVk2aNtNKIrbnZ2dnUVZ_uiknZ2d@comcast.com...
> 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.
>
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
| |
| daestrom 2007-04-09, 8:25 pm |
|
"Peter" <private@private.com> wrote in message
news:VuqdnVk2aNtNKIrbnZ2dnUVZ_uiknZ2d@comcast.com...
> 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.
>
>
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
| |
|
| "daestrom" <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote in
news:461acf0f$0$4883$4c368faf@roadrunner.com:
>
> "Peter" <private@private.com> wrote in message
> news:VuqdnVk2aNtNKIrbnZ2dnUVZ_uiknZ2d@comcast.com...
>
> 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.
| |
| dave y. 2007-04-11, 3:25 am |
| On Mon, 09 Apr 2007 22:16:23 -0500, Peter <private@private.com> wrote:
>"daestrom" <daestrom@NO_SPAM_HEREtwcny.rr.com> wrote in
>news:461acf0f$0$4883$4c368faf@roadrunner.com:
>
>
>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.
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.
| |
| Salmon Egg 2007-04-11, 3:25 am |
| On 4/10/07 8:14 PM, in article fhko13ldj170olemt2mddd45uapjon7ahm@4ax.com,
"dave y." <nospam@myhouse.com> wrote:
> 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.
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.
| |
|
| Salmon Egg <salmonegg@sbcglobal.net> wrote in
news:C241C1D9.6EDBC%salmonegg@sbcglobal.net:
> On 4/10/07 8:14 PM, in article
> fhko13ldj170olemt2mddd45uapjon7ahm@4ax.com, "dave y."
> <nospam@myhouse.com> wrote:
>
>
> 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.
| |
| Salmon Egg 2007-04-14, 3:25 am |
| On 4/13/07 4:39 PM, in article ucOdnS7aVsFbib3bnZ2dnUVZ_tTinZ2d@comcast.com,
"Peter" <private@private.com> wrote:
> 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.
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.
| |
| dave y. 2007-04-17, 3:25 am |
| 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.
|
|
|
|
|