Capacitor Charging/discharging problem

[Learnphysics]

Homework Statement

In the circuit shown below the switch has been opened for a very long time and closes at
t = 0. Calculate the output voltage VO at:
1: t = 0−; 2: t = 0+; 3: t = t1 = 2.7 µs; 4: t → ∞.

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Homework Equations

Vc = Vc - Vc(e^(-t/RC) (Charging)

Vc = Vc(e^-t/RC) (Discharging)

The Attempt at a Solution

I found the correct answer for questions 1,2 and 4.
-18, -18, and -8.18182

however question 3 is giving me a problem.
I reasoned that when the switch closes the capacitor will begin discharging into the two resistors, and so I used:
Vc = Vc0(e^-t/RC)

With Vc = -18,and R = R1||R2. However the answer i got was incorrect.

I also don't see how R1 is || to R2, but the book's example assumed they were.

The example's solution is to first assume Vc is the voltage across R1, and then say that the voltage at the Vo node is Vc0(e^-t/RC) + Voltage across R2.

Why is this.

 

[lewando]

so I used:
Vc = Vc0(e^-t/RC)

If you use this equation, what is Vc when t = ∞? Is this consistent with your result from part 4?

 

[Learnphysics]

If you use this equation, what is Vc when t = ∞? Is this consistent with your result from part 4?

I'm still a little sketchy on this kind of stuff, how i got that answer is, i assumed at infinite time the capacitor is fully charged, thus acting as a open circuit, and i could use simple voltage division principle to find voltage drop over the 1k, and that'd be the voltage at Vo.

If i use the discharge formula, what would my V0 be? -18? or the voltage across the 2.4k resistor?

There's Voltage drop over the 2.4k resistor is -9.8181818.

either way assuming an infinite time, it should be 0. using Vc = Vc0(e^-t/RC)

 

[lewando]

I'm still a little sketchy on this kind of stuff, how i got that answer is, i assumed at infinite time the capacitor is fully charged, thus acting as a open circuit, and i could use simple voltage division principle to find voltage drop over the 1k, and that'd be the voltage at Vo.

Right, at time t = ∞.

If i use the discharge formula, what would my V0 be? -18? or the voltage across the 2.4k resistor?

There's Voltage drop over the 2.4k resistor is -9.8181818.

Which 2.4K resistor?

either way assuming an infinite time, it should be 0. using Vc = Vc0(e^-t/RC)

Right, if you use that formula, which is wrong.

Vo will start at -18 (Vc0) and decay to -8.2 (Vo @ t = ∞), so your equation should support this concept. You need 2 terms: a variable (decaying) term and a constant (final value) term. Come up with one and graph it until you are comfortable with it. Hint: you will have to adjust the "Vc0" coefficient of the variable term to account for the constant term.

BTW, the 2 resistors can be considered parallel because the ideal voltage source has no internal resistance.

 

[Learnphysics]

Oh right, there is no 2.4k resistor, I mean the voltage drop across the 1.2k.

Ahhh, so they're in parallel i didn't know ideal voltage sources had no internal resistance.. Or i knew it but i hadn't internalized it yet.the constant final term would be -8.2, and the variable decaying term would be -18(e^(-t/RC))

Vo = -18(e^(-t/RC)) -8.2?

So that, at t = infinite, we have -8.2 volts...cool.

and when t = 0(Initial condition) we have -26.2 volts. But that doesn't sound right... it should only be -18 initially.

(-18 + 8.2)(e^(-t/RC)) - 8.2 = Vo?

This looks a little more correct? right?

 

[lewando]

Right. Evaluate at 2.7 µs and you should be done.