Asked by Katie

How long will it take a charged 80-ìF capacitor to lose 20 percent of its initial energy when it is allowed to discharge through a 45 ohm resistor?
Answered by Katie
i mean 80 microfarads
Answered by drwls
The time constant for charge decay is
RC = 3.6*10^-3 s, if your ì symbol means "micro"

For 80% of the initial energy, the charge must decrease to sqrt(0.8) = 89.4% of the initial value.

exp(-t/RC) = 0.894
t/RC = .1121
t = 4*10^-4 seconds