RLC Series Circuit Analysis: Questions and Answers

Click For Summary
SUMMARY

The discussion centers on the analysis of linear RLC series circuits, specifically the differences in Laplace transform applications when dealing with an uncharged capacitor versus a charged capacitor. The user employs Kirchhoff's voltage law and transforms voltage and current equations accordingly, achieving correct results for uncharged scenarios. However, when inserting a charged capacitor, the user encounters discrepancies in signs, indicating a need to adjust the transform for the capacitor to account for its role as a power supplier rather than a consumer. The professor's method of representing the capacitor and inductor with their own transforms is confirmed as a valid approach.

PREREQUISITES
  • Understanding of RLC circuit theory
  • Proficiency in Laplace transforms
  • Familiarity with Kirchhoff's voltage law
  • Knowledge of initial conditions in circuit analysis
NEXT STEPS
  • Study the impact of initial conditions on Laplace transforms in circuit analysis
  • Learn about the representation of capacitors and inductors in the frequency domain
  • Explore advanced applications of Kirchhoff's laws in RLC circuits
  • Investigate the role of energy storage elements in dynamic circuit behavior
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing RLC circuits and their dynamic responses to changes in initial conditions.

FOIWATER
Gold Member
Messages
434
Reaction score
12
Just a quick question for you people - If I have a linear RLC series circuit where there is an uncharged capacitor at time t=0, and the switch is closed, is it an entirely different analysis (in terms of laplace transforms) than If, say, the some battery charges a capacitor, then I put it into the circuit without any external voltage source?

I Have been analyzing these circuits and getting correct answers.

I use kirchhoffs voltage law, when I have a resistor I use v(t)=Ri(t) and transform it to V(s) = rI(s) , the voltage drop across the inductor Ldi/di as L(SI(s)-i(0)) and the voltage drop across the capacitor (uncharged scenerio) as 1/c integral i(t)dt transforms to (I(s)/SC + v(0)/S) once I solve for the volt drop. I re arrange 1/c integral i(t)dt to i(t) = Cdv(t)/dt and transform it and solve for V(s).

Now I make an equation as V(s)/S = RI(s) + L(SI(s)-i(0) + (I(s)/SC + v(0)/S) and this has been working for these circuits.

However, there is a scenerio my professor has given us where the charged capacitor is inserted. No longer I can use this transform for the capacitor certainly? I get similar answers but signs incorrect. It is as if I must use (I(s)/SC - v(0)/S), not plus. My professor showed me I can represent the capacitor, and inductor, as two components with their own transform. For example, the capacitor impedance represented as 1/SC in complex frequency domain, and in series with v(0)/S and it makes sense to me that if the capacitor is now charged and thus supplying power as opposes to consuming it, the transform I initially stated for the capacitor cannot be correct. However his methods seem to contradict each other, its friday and I cannot wait for the class to pick back up I need to know.

Any information appreciated greatly.
 
Engineering news on Phys.org
The Laplace transforms describe a system, and the external stimulus to the system. Your problem is that you are changing the system, by switching in a capacitor. Be sure to write equations only for the post switching time.

The time domain solutions will contain an arbitrary constant of integration. You can solve for the value of that constant that also matches the initial charge on the capacitor.
 

Similar threads

  • · Replies 32 ·
2
Replies
32
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
Replies
44
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
12
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K