Ok I didn't understand, I studied the constant long ago, from my notes: after a time constant, the capacitor voltage will be 27% of the starting voltage (example: if the capacitor voltage is 10v, after a time constant it will be 2.7v across it)

we can say that a capacitor is almost totally discharged after 10 time constants
After a time constant, the capacitor should have 63% of the applied voltage. We usually consider the capacitor charged in 5 or 6 time constants.
For time constants of 6 down to 1
1-e
(-6) = 0.997 or 99.7% charged
1-e
(-5) = 0.993 or 99.3% charged
1-e
(-4) = 0.982 or 98.2% charged
1-e
(-3) = 0.950 or 95.0% charged
1-e
(-2) = 0.864 or 86.4% charged
1-e
(-1) = 0.632 or 63.2% charged
The old HP48GX is getting a workout this morning!
If you play with Octave (free) or MATLAB, here is a simple script that prints a table of values
ETA: And plots the charge and discharge curves...
V0 = 1; % assume a supply voltage
R = 10000; %10k Ohms
C = 0.1*10^-6; % 0.1 ufd
Tau = R*C;
t = linspace(0,8*Tau);
Vchg = V0 * (1 - exp((-t/Tau)));
Vdis = V0 * (exp((-t/Tau)));
plot(t, Vchg)
grid minor
grid on
hold on
plot(t,Vdis)
ylabel("Voltage")
xlabel("Time - seconds")
%there is a known bug in 'legend', the [h,~] is a kludge to work around it
[h,~]=legend("Charge Voltage","Discharge Voltage");
figure(gcf) % or shg command - pull plot to top
for Tau = 0:6
pct_charge = V0 * (1 - exp((-Tau))); %percent charge
fprintf("%% Tau = %.0f Percent Charge = %5.2f%%,\n", Tau, 100*pct_charge)
end
% Results appended to source
% Tau = 0 Percent Charge = 0.00%,
% Tau = 1 Percent Charge = 63.21%,
% Tau = 2 Percent Charge = 86.47%,
% Tau = 3 Percent Charge = 95.02%,
% Tau = 4 Percent Charge = 98.17%,
% Tau = 5 Percent Charge = 99.33%,
% Tau = 6 Percent Charge = 99.75%,
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