For the circuit below, the switch has been closed for a period of time long enough for steady state conditions to have been reached.
a: Determine the time constant before the switch is opened
b: Determine the time constant after the switch is closed
c: Determine the voltage across the capacitor before the switch is opened
d: Determine the voltage across the capacitor after the switch is closed
e: Draw a graph of the voltage vs time, with sufficient time duration that steady state conditions are shown
v(t) = V(0)(1-e^(-t/τ)
The Attempt at a Solution
Here is what I got
a: τ = 33.3Ω * 90mF, τ = 3
note: 33.3Ω is a result of (50*100/150), or (r1 + r2) || r3 as the circuit is flowing through the switch
b: τ = 33.3Ω * 90mF, τ = 3
note: 33.3Ω is a result of (50*100/150), or (r3 + r2) || r1 as the circuit is flowing through the capacitor (until the capacitor becomes an "open circuit")
c: v(t) = ?, not sure what to do here, or how to go about it. I have a formula, but every approach I can think of I either end up with 0, or infinity. Speaking in terms of what happens with the e function, -t/τ. Because when the switch is closed, we can think of t as infinity I think (because the voltage\current relationship will be the same for ever in ideal conditions), so you would get e^ (-infinity/3) --> 0), e^0 = 1, so 24(1-1) = 0.
d: Don't think you even need to mathematically prove that the voltage across the capacitor (as the only "exit" route, as the capacitor charges, current goes to 0 as voltage goes to its limit), so voltage across the open switch circuit would be 24V.
Could use some help here, am I on the right path with my time constants? If not, what am I doing wrong, and what can I do to find c? (probably also what can I do for d, I can't imagine a verbal explanation would satisfy the instructor)