- #1

sliperyfrog

- 27

- 0

## Homework Statement

In the circuit below (drew it the best I can) the switch is opened, how much time elapses before the capacitor loses half of its initial stored energy.

## Homework Equations

[/B]

q(t) = Qe

^{-t/RC}

U = q

^{2}/2C

## The Attempt at a Solution

[/B]

So first thing I did was I set (1/2)Q

^{2}/2C = q

^{2}/2C then i replaced q

^{2}for Q

^{2}e

^{-2t/RC}so I get (1/2)Q

^{2}/2C = Q

^{2}e

^{-2t/RC}/2C I cancel out the like terms and I am left with 1/2 = e

^{-2t/RC}I than get rid of the e so I get ln(1/2) =-2t/RC then i get -ln(2) = -2t/RC getting me t = RCln(2)/2 and now this is were i got confused I know C = 200E-6 F but i don't know what R is suppose to be, I think it is suppose to be 90Ω since the 30Ω and the 60Ω resistors connect together at where the capacitor is.