Charging Capacitor Calculation

's' means seconds.

RxC yields the time constant or the time it takes for the capacitor to discharge to .368 of its volt level at the beginning of a run.

.368 just happens to be 1/e.

Thus the discharge curve in its barebones form:

V = (1/e) ^t

(We pretend TC is 1 to keep it simple.)

Charge curve:

V = (1-1/e) ^t

0.001 s / RC = 0.001/(6500*1e-9) = 153.8

exp(153.8) = 1.53e-67

The app doesn't manage to display the exponent because it shows ridicolous number of digits.

It's saying at the time, t = 0.001 sec, which is equals to 153.846153 time constants.
I_max = V_b/R = 125/6500 = 0.019230769 A.
e^{-0.001/(6500*1e-9)} = 1.5327252873438256e-67
So the current at 153.846153 time-constants is I_max*1.5327252873438256e-67 = 2.947548629507357e-69 A.

If you scroll to the right when you're in the current fields, you'd see the exponent symbols.

One of the first things you learn when studying physics is to use adequate number of digits for empirical data. I really wonder why an app apparently maintained by physicians prints full double precision. Most likely lack of coding skills.

FvM said:

0.001 s / RC = 0.001/(6500*1e-9) = 153.8

exp(153.8) = 1.53e-67

The app doesn't manage to display the exponent because it shows ridicolous number of digits.

Click to expand...

Most of your math checks out. Perhaps I'm missing a variable?

Akanimo said:

It's saying at the time, t = 0.001 sec, which is equals to 153.846153 time constants.
I_max = V_b/R = 125/6500 = 0.019230769 A.
e^{-0.001/(6500*1e-9)} = 1.5327252873438256e-67
So the current at 153.846153 time-constants is I_max*1.5327252873438256e-67 = 2.947548629507357e-69 A.

If you scroll to the right when you're in the current fields, you'd see the exponent symbols.

Click to expand...

This seems to check out. Thank you.

assuming the capacitor is discharged when the switch is closed,
Vcap = Vbat(1 - e^(-t/RC))
Qcap = CVcap
RC is the time constant, in seconds (hence the s) which is usually denoted by Greek lower case tau

both Vcap and Qcap follow the red charge curve
with different values

If the capacitor is charged to Vcap, open the switch, remove the battery and then close the switch,
the capacitor discharges according to
Vcap = Vbat(e^(-t/RC))
and as above:
Qcap = CVcap
RC is the time constant, in seconds (hence the s) which is usually denoted by Greek lower case tau

general rule of thumb:
if time is 5 time constants (5*R*C) or more, the capacitor is essentially fully charged or fully discharged
note in the example, 153 time constants yields 10^-69, a very small number
so the capacitor is either fully charged or fully discharged, as the situation demands

build a table in your favorite spreadsheet:
x e^-x 1-e^-x where x is 0 to 5, step by 1
then plot the two curves
e^-x is the general discharge curve and 1-e^-x is the general charge curve.

lots of people make lots of silly mistakes, so the double precision printing of numbers is not the biggest issue

chiques said:

This seems to check out. Thank you.
View attachment 180226

Click to expand...

Take a look at it again. You should use Imax rather than Qmax. With Qmax, your answer should be charge (in Coulombs) and not current (in Amperes) which you indicated. Actually, for a charging capacitor, the stored charge is accumulated over time and so can only increase with time. The more the charge accumulated, the less the current due to lesser difference in potential between the driving voltage and the voltage across the capacitor. So, even if you were to mean charge with your last expression, it would still be incorrect as your expression would imply that charge decays with time.