RLC circuits conceptual help!!!

Discussion in 'Homework Help' started by TheSpArK505, Jun 27, 2014.

  1. TheSpArK505

    Thread Starter Member

    Sep 25, 2013
    Hi everyone.
    I,m studing RLC circuits and i have two confusions i would like to understand.

    1.in series RLC circuit the book says that the final voltage at the

    capacitor (at t=∞) is same as the source...why? is it because C behave

    as Open ?!! why don't R dissipate some power and thus draws voltage

    so that Vc not equal to Vs and thus current flows.

    2.For the circuit attached..the switch has been closed for a long time ..OK.

    before it is opened..are theree two short circuits one is the switch and the other is due to L because it behave as SC at t=∞..how can i calculate value of current"do I simply divide by two"?

    how do C and L behave after the switch opens?

    Waiting your help:confused:
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  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
    Regarding 1.
    Yes. In circuit with DC source, after large amount of time passes, the capacitor is an open circuit. Since it is series RLC circuit, once capacitor acts as an open, the pass for current is broken, there is no current flowing though the resistor, by Ohm's Law (V=I*R) you have V=0*R=0 volts.
  3. TheSpArK505

    Thread Starter Member

    Sep 25, 2013
    but if voltage of the capacitor eaquals to the voltage of the sourse that means no power dissipation in R...Why is that?
    and please if any one can help with the second partof the Q.
  4. nDever

    Active Member

    Jan 13, 2011
    In response to the first question, if the voltage across the cap is the same as the source, then there cannot be any voltage across the resistor because then, that would be a violation of Kirchoff's voltage law. If V = 0, then P = 0.

    For the second question, before the switch is opened, the circuit has reached steady state, therefore the cap is an open circuit and the inductor is a short circuit. No current will flow through the resistor if it is shorted and no current goes to the cap, so what must the current be?

    After the switch is opened, the source is essentially "turned off". Since you know that the current through an inductor cannot instantly change, then the current through it is the same just before and just after the switch opens. What is the resulting circuit and how do you think the inductor and capacitor should respond?
    Last edited: Jun 27, 2014
  5. shteii01

    AAC Fanatic!

    Feb 19, 2010
    Because in order to have voltage across R you MUST have current through R. The current through the R is zero, therefore voltage across R=0*R=0 volts.