The Question States:

The switch is in position x for a long time before moving to position y at t=0.
Find the expression for V1(t) at t≥0.

My Answer:

First I found v0(t) using voltage divider. at t<0 where the capacitor is an open circuit.

Then at t>0 I used T= RC and did 1/T
Therefore:
vo(t) = vo(0)*e^-t/T

Then I just said that vo(t) = v1(t) at t>o where the capacitor beneath the switch is now a short circuit because they are in parallel is that correct?

Thanks

 

First you find the voltage across the capacitor at \(t=0^-\) just before the switching occurs

\(
v_C(0^-) = \frac{10\text{k}\Omega}{5\text{k}\Omega + 10\text{k}\Omega} 20\text{V}= \frac{40}{3} \text{V}
\)

Right after the switching occurs at \(t=0^+\), the capacitor voltage stays that same, that is

\(
v_C(0^+) = v_C(0^-) = \frac{40}{3} \,\,\,\,\,\, \text{V}
\)

The capacitor voltage for \(t \ge 0^+\) is given by

\(
v_C(t) = v_C(0^+) \text{e}^{-t/\tau}
\)

where \(\tau = C R_{th}\) where \(R_{th}\) is the Thevenin resistance "seen" by the capacitor for \(t \ge o^+\); therefore, \(\tau=(10\mu\text{F})(100\text{k}\Omega)=1\text{s}\).

The final solution for \(t \ge 0^+\) is
\(
v_C(t) = \frac{40}{3} \text{e}^{-t} \text{V}
\)

I think this is pretty much what you had but you did not provide any numerical results. For example, if you just write \(RC\) for the time constant it is not clear which resistance you are referring to. So just be a little more clear with that. I know which one you meant but most graders don't want to assume anything and will not try to read your mind.

Best regards,
Vahe

 

at vo(t) = v1(t) at t>o where the capacitor beneath the switch is now a short circuit because they are in parallel is that correct?

Thanks

your process seems right. but can you explain why do you think it is shorted ? as i understand, at t >0, the capacitor acts like a DC voltage source dissipating its stored energy through 100 kΩ resistor. it is not shorted. and it can be viewed both as parallel and series.

let me know if my understanding is wrong.