Homework Help: I love rlc circuits : Series

1. May 9, 2014

dwn

1. The problem statement, all variables and given/known data

Image

2. Relevant equations

Critical damping : e-αt(At + B)
R = 1.5

3. The attempt at a solution
I'd like to post this before anything else to make sure that I've analyzed the circuit correctly.
Code (Text):
t < 0:    v[SUB]c[/SUB](0) = 9V  ,  i[SUB]L[/SUB](0) = 9/4 = 2.25A

i = e[SUP]-t/3[/SUP](At + B)
i(0) = B = 2.25A
di/dt = -1/3e[SUP]-t/3[/SUP](At + 2.25) + Ae[SUP]-t/3[/SUP]
di(0)/dt = -1/3*2.25+ A = V[SUB]c[/SUB](0) = 9V

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Last edited: May 9, 2014
2. May 9, 2014

rude man

Your v_c(0+) is wrong. What is v_c(0-) and what does the capacitor have to say about changing to v_c(0+)?

3. May 9, 2014

dwn

v_c(0-) = 0 V Is the reason for this bc it is not supplying any charge to the circuit? I'm having a difficult time distinguishing when I should have a v_c for 0- and when there is not...I see it as a sort of battery that stores charge in it until there is not longer a source present.

4. May 9, 2014

rude man

Why do you think the charge on C is 9V before the switch is opened?

5. May 9, 2014

dwn

I know that during the steady state the cap is an open circuit, but after reading in a book that cap's are similar to batteries along came the idea of it holding a charge (9V in this case). Essentially the battery will hold the charge that it contains if it is removed from a circuit, until it has somewhere to dissipate the charge (aside from internal resistance of course). Then once the switch is flipped, the voltage in the capacitor decreases by V(0)e^(-t/RC) or using the characteristic equations in RLC circuits.

Last edited: May 9, 2014
6. May 9, 2014

rude man

You should focus on the inductor. What is the voltage across an inductor when current thru it is constant?

7. May 9, 2014

dwn

0 V --- current is not changing so di/dt = 0. (VL = L di/dt).

I don't see how that helps me with understanding whether v(c) is 0 or some other constant.

8. May 9, 2014

9. May 9, 2014

dwn

So its safe to say that v_c(0-) with any current/voltage source will never "hold" a charge of that source? In which case, why are there instances when v_c(0-) has a value. Just to be clear.

10. May 9, 2014

rude man

Again: what is the voltage across the inductor at t = 0-? What is the voltage across the capacitor at t = 0-? What must therefore be the voltage across the capacitor at t = 0+?

Hint: you can have all sorts of voltages across a capacitor at t = 0-, but not if there's an inductor across it!

11. May 9, 2014

dwn

It's t(0-) = t(0+) because the cap and inductor cannot change at suddenly at t=0.

Why does my statement mystify you?

12. May 9, 2014

rude man

Dunno, it just does.

13. May 9, 2014

dwn

14. May 9, 2014

rude man

The hint of post #10 was not helpful?

A capacitor holds any charge sitting on it if there is no load across it. I mean an L or an R or both. In your case what sits across the capacitor?

15. May 9, 2014

dwn

Thank you. That made sense to me. As long as the current has another path to take, the capacitor doesn't hold a charge (hence the open circuit). I had a difficult time understanding how the capacitor operated between t(0-) and t(0+).

16. May 9, 2014

rude man

The basic idea is that the voltage across a capacitor cannot change instantaneously. So if that voltage was zero at t=0- it must also be zero at t=0+.