In the circuit of the figure below, the switch S has been open for a long time. It is then suddenly closed. Take
(a) Determine the time constant before the switch is closed.
(b) Determine the time constant after the switch is closed.
(c) Let the switch be closed at t = 0. Determine the current in the switch as a function of time. (Assume I is in A and t is in s. Do not enter units in your expression. Use the following as necessary: t.)
The solved diff eqs for current over time
The Attempt at a Solution
(a) and (b) aren't problems for me but I do want to make sure my reasoning is accurate.
(a) Is pretty easy I guess. I just combined the two resistors in series and then multiplied by C, the reasoning being that the electrons that go through C also have to go through R1 and R2.
(b) The switch closing means that there is a short circuit, so all of the current from the voltage source goes through the middle branch and the right branch is ignored basically. So I guess you would multiply R2 by C because the discharging electrons have to go through R2?
(c) Is the problem here for me. At first it's kind of easy. The current from the left branch would be I1, the current in the middle branch can be Iswitch, and the current in the right branch can be I2. Using Kirchoff's junction rule, I1+I2=Iswitch.
The basic solution to the diff eq would be I(t)=I0e-t/RC. However, getting the initial current is kind of a problem for me. The only value for any sort of voltage we have is the ε, but how exactly can I use that?
I mean, what I'm thinking is that I could use Kirchoff's loop rule for before the switch is closed, but we don't have the charge on the capacitor, do we? Also, I believe the solution doesn't do it that way either. We could also use V=IR but it seems like that wouldn't work. Am I wrong?