Question 1182991
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  V\ =\ 5\,\text{volts}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  C\ =\ 80\,\times\,10^{-9}\,\text{farads}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  R\ =\ 40\,\times\,10^3\,\text{ohms}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  T\ =\ RC\ =\ 3.2\,\times\,10^{-3}\,\text{sec}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  v_c(t) =\ V\(1\,-\,e^{\frac{-t}{T}}\)]


where *[tex \Large t] is elapsed time in seconds.


<pre>
(msec)    1       2       3       4       5       6       7       8       9       10
        1.342	2.324	3.042	3.567	3.952	4.233	4.439	4.590	4.700   4.780	

          11      12      13      14      15      16      17      18      19      20 
	4.839	4.882	4.914	4.937	4.954	4.966	4.975	4.982	4.987	4.990
</pre>


     *[illustration Charging_Voltage_RC_Circuit.png]

I don't have a way to measure the gradient at a point on an excel graph, so I approximated by calculating *[tex \Large \frac{v_c(0.005998)-v_c(0.006002)}{0.000004}\ \approx\ 239.6172] and this differs from the value of the first derivative at 0.006 by only one digit in the fourth decimal place.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dv_c}{dt}\ =\ \frac{Ve^{-t/T}}{T}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dv_c}{dt}\|_{t=0.006}\ =\ \frac{5e^{-0.006/3200}}{3200}\ \approx\ 239.6171]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ v_d(t)\ =\ \frac{Ve^{-t/T}}{T}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \frac{dv_d}{dt}\ =\ \frac{Ve^{-t/T}}{T^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \frac{dv_d}{dt}\|_{t=T}\ =\ \frac{5e^{-1}}{3200^2}\ \approx\ 1.8\times\10^{-7}]


Essentially, zero.

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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