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

In summary, the conversation discusses the concept of capacitance and how it operates in a circuit. Specifically, the conversation focuses on the voltage across a capacitor and how it cannot change instantaneously. The conversation also touches on the role of an inductor in determining the voltage across a capacitor and the importance of having a load in the circuit. The conversation ends with a clarification on the understanding of the capacitor's operation between t(0-) and t(0+).
  • #1
dwn
165
2

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.
Code:
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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  • #2
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
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
Why do you think the charge on C is 9V before the switch is opened?
 
  • #5
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.
 
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  • #6
You should focus on the inductor. What is the voltage across an inductor when current thru it is constant?
 
  • #7
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
Please go back to my post #2 and think about it.
 
  • #9
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
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!
 
  • #11
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
dwn said:
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.
 
  • #13
well, that's really not helpful.
 
  • #14
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
rude man said:
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+).
 
  • #16
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+.
 

1. What is an RLC circuit?

An RLC circuit is a type of electrical circuit that contains a resistor (R), inductor (L), and capacitor (C) in series. These components work together to create a resonant frequency that can be used for various purposes, such as filtering, amplifying, or oscillating electrical signals.

2. How does an RLC circuit work?

An RLC circuit works by combining the properties of the resistor, inductor, and capacitor to create a resonant frequency. When an electrical current flows through the circuit, the capacitor stores energy, while the inductor resists changes in the current. This results in the circuit oscillating at the resonant frequency, which can be used for various applications.

3. What are the applications of RLC circuits?

RLC circuits have various applications in electronics, including filtering, amplifying, and tuning radio and television signals. They are also used in power supplies, voltage regulators, and electronic filters in audio equipment.

4. How does an RLC circuit differ from other types of circuits?

An RLC circuit differs from other types of circuits in that it contains all three passive components (resistor, inductor, and capacitor) in series. This allows for the creation of a resonant frequency and has unique properties that make it useful for specific applications.

5. What are the advantages of using RLC circuits?

Some advantages of using RLC circuits include their ability to filter unwanted frequencies, their high stability, and their low cost. They are also versatile and can be used for various purposes, making them a valuable tool in electronics.

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