Engineering What is the critical damping equation for this series RLC circuit with R = 1.5?

[dwn]

Homework Statement

Image

Homework Equations

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

The Attempt at a Solution

I'd like to post this before anything else to make sure that I've analyzed the circuit correctly.

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

 

[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+)?

 

[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.

 

[rude man]

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

 

[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.

 

[rude man]

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

 

[dwn]

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

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

 

[rude man]

Please go back to my post #2 and think about it.

 

[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.

 

[rude man]

Your statement mystifies me.

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!

 

[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?

 

[rude man]

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

It's not "t(0-) = t(0+)". It's v(0-) = v(0+).

Why does my statement mystify you?

Dunno, it just does.

 

[dwn]

well, that's really not helpful.

 

[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?

 

[dwn]

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?

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+).

 

[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+.