- #1

horsedeg

- 39

- 1

## Homework Statement

In the circuit of the figure below, the switch S has been open for a long time. It is then suddenly closed. Take

*R*1 = 56.0 kΩ,

*R*2 = 200 kΩ, and

*C*= 14.5 μF.

(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*.)

## Homework Equations

V=IR

The solved diff eqs for current over time

Kirchoff's rules

## 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 I

_{1}, the current in the middle branch can be I

_{switch}, and the current in the right branch can be I

_{2}. Using Kirchoff's junction rule, I

_{1}+I

_{2}=I

_{switch}.

The basic solution to the diff eq would be I(t)=I

_{0}e

^{-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?