First order circuits

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FOIWATER
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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 kirchoff's 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.
 

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  • #2
anorlunda
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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.
 

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