Calculate the capacitance of the capacitor

[TRE]

1.Question: A charged capacitor is connected across an 8.0 kΩ resistor and allowed to discharge. The capacitor loses 63.2% of it's original charge in a time of 6.2 seconds. Calculate the capacitance of the capacitor. (This is everything they give and nothing extra, I have checked and made sure).

What they give me is:

R = 8000 Ω
R*C = 8000Ω*C
t=6.2 sec)
Percentage value of : 63.2%

Homework Equations

: None of the equations I have at my disposal can make use of the above values without a voltage or current value given. This is why I am asking for help because I don'tΩ know what equation to use.

I have looked at one equation that may help, but it uses voltage,:
V=V₀ * e^(-t/RC)

The Attempt at a Solution

: R = 8000 Ω
RC = 8000Ω*C
t=6.2 sec)
Percentage value of : 63.2%

I can use:

63.2% for V ∴ = 0.632

100% for V₀ ∴ = 1

e = 2.718 (Value is given like this in my textbook)

RC= 8000Ω*C

t = 6.2 sec

C = ? (value I am asked to calculate)

Solution:

0.632 = 1 * e^(-t/R*C)

ln(0.623) = ln(e) * (-6.2/8000*C)

C = 1.688946652 * 10^-3 F

My answer does not seem correct as it is a very low value and besides I don't know whether I used the correct value for V and V₀ or should they have been switched ?

I would really appreciate some help.

Thank You
Regards
Trevor

 

[Jony130]

You have all information needed to solve this problem. It's seems to me that you don't full understand what time constant is or maybe you don't read the Question too carefully.

This part is very important
The capacitor loses 63.2% of it's original charge in a time of 6.2 seconds.
And all you need is this equation t = R*C