- #1

FOIWATER

Gold Member

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## Main Question or Discussion Point

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.

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.