Question

Show that in one time
constant 63% of the equilibrium charge
stored on the capictor?
charge​

Answer 1

Answer:

Capacitors are usually charged or discharged through a resistor, R. Capacitors charge and discharge following these formulae.

Discharge: V(t)=Vinitial × e^(-t/(R×C))

Charging: V(t)=Vb × {1- e^(-t/(R×C))}

Where R×C is the time constant and Vb is the battery voltage.

So, after a time period of T= R×C seconds, we have for a charging capacitor, V/Vb = {1-e^(-1)}=0.63 or 63% of the full voltage being supplied by the battery, Vb.

For a discharging capacitor, at t=R×C seconds, we get:

V/Vinitial = e^(-1)=0.37 or 37% of the voltage remains on the capacitor.

I hope this helps.