Answer:
[tex]t = \tau[/tex]
[tex]Q_{max} = CV_o[/tex]
Explanation:
during charging of the capacitor we know that the voltage across the capacitor is given as
[tex]V = V_o(1 - e^{-t/\tau})[/tex]
now the time to raise the voltage across the capacitor is given from 0 to 63%
so it is
[tex]V = 0.63 V_o[/tex]
[tex]0.63 V_o = V_o(1 - e^{-t/\tau})[/tex]
[tex]1 - e^{-t/\tau} = 0.63[/tex]
[tex]0.37 = e^{-t/\tau}[/tex]
[tex]t/\tau = 1[/tex]
[tex]t = \tau[/tex]
Similarly we know that during discharging we have
[tex]V = V_o e^{-t/\tau}[/tex]
so here this will decrease by 37 %
[tex]0.37 V_o = V_o e^{-t/\tau}[/tex]
again we have
[tex]t = \tau[/tex]
Also we know that during charging the maximum charge on the capacitor is
[tex]Q = CV[/tex]
[tex]Q = CV_o(1 - e^{-t/\tau})[/tex]
for maximum value of charge t = infinite
so we will have
[tex]Q_{max} = CV_o[/tex]