# Discharging circuit

1. Apr 1, 2009

### driedupfish

1. The problem statement, all variables and given/known data
http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys212/oldexams/exam2/fa06/fig17.gif [Broken]
After a very long time with both switches in the closed position, switch S2 is reopened. How long does it take for the charge on capacitor C2 to drop to 1/e (36.8%) of its fully-charged value (i.e. of the value it had just before S2 was reopened)?

2. Relevant equations
Time constant = RC

3. The attempt at a solution
I try to find the total resistance, but am not sure how. I am not even sure what it means by "drop to 1/e (36.8%)". Does it mean 36.8% of the total charge?

Last edited by a moderator: May 4, 2017
2. Apr 1, 2009

### Pythagorean

yes, since charge is a function of voltage here. As the voltage drops, less charge is accumulated on the plates of the capacitor.

Last edited by a moderator: May 4, 2017
3. Apr 1, 2009

### driedupfish

Okay, to find the time constant, RC, I have to find the equivalent resistor at that point. But I don't know how? With only S2 reopened, are all the resistors now in parallel?