can someone solve this capacitor discharge formula?

Question

i was thinking if someonesolved it for me (at least once) then i will get an idea how it works and know how to work the problem so heres the problem, i was trying to find the voltage discharge in a cpapcitor and i found the formula and i couldn't figure it out so here you go

V=v(e^t/T)
the instintanious voltage is equal to the initial voltage times exponent to the power of time over tau

and now for the details
the capacitor has the capacitance of 80uf and has a rated voltage of 330v and the battery i am charging the capacitor with has 1.5 volts and 5.0 amps and i wanted to see how much the capacitor discharges in 1 second

and there was a marking on the capacitor thet read 3(2) om i think that was the resistance

Answer 1

C=80uF
V=Voc1.5 volts
I=Isc=5.0 amps (assuming short circuit current)

So thevnin's theorem implies the
Rth = Voc/Isc = 1.5/5 = 0.3ohms

So the circuit is an RL circuit,

Time constant T = RC = 0.3*80uF = 24us

Now you can use that formula for charging

Answer 2

Capacitor Discharge Formula

Answer 3

The formula you're using has the Greek letter Tau in it which is called the time constant. Tau = time constant = resistance X capacitance. Without knowing what resistance is in series with the capacitor, there is no way to predict the charge/discharge rate.

Answer 4

in maximum utility they think approximately (4 or 5)*(RC) because of the fact the time it takes for discharging.r is total resistor of circuit seen via capacitor or in useful attitude u can calculate the time that LED is on,its the time for discharging of C