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RC time constant

  1. May 18, 2009 #1
    Hi, I have a simple question: if I have a circuit with a capacitor, how do I calculate the voltage on the capacitor in function of time, I mean I know that tau(τ)=RC and there is the general formula X(t)=X(0)+[X(0)-X(∞)]-t/τ. So here comes my question should I consider the capacitor as the load and calculate the Thevenin equivalent voltage at t=0 and t=∞ and use the X(t)=X(0)+[X(0)-X(∞)]-t/τ forumla ,where τ=RThC.
     
  2. jcsd
  3. May 19, 2009 #2

    Dale

    Staff: Mentor

    You can do it that way. A simple "rule of thumb" is that a capacitor is a short for high frequencies (t=0) and an open for DC (t=infinity).
     
  4. May 20, 2009 #3
    Ok, I have some circuits and I don't know if I resolved the problems right so here are they:
    and I have another question: files
    1. The voltage of an ideal a.c. source has the expression:
    ug = 100 *20.5sin (2*104 ∏*t +∏/3) V.
    1.1. Find the value of the frequency and the value of the voltage at the moment t=0
    1.2. Find the complex expression and the rms value of the current if the source supplies the impedance consisting in the resistance of 80 Ω connected in series with the inductance of 3/∏ mH (0.955 mH).

    I don't really know what does it mean at the question 1.1. The frequency at the moment t=0. The voltage is 100 *20.5sin (∏/3).
    The second question(1.2) I've solved like this:
    URMS=100V ; R=80Ω ; L=3mH => XL=j*2*104∏*3/∏*10-3=j*60Ω.
    IRMS=URMS/|Z|, where |Z|=(802+602)1/2=100Ω => IRMS=100/100=1A.
    And the complex expression of the current is Ucomplex/Rcomplex=100j∏/3/j60=100(cos∏/3+jsin∏/3)/j60=100(1/2+j31/2/2)/j60=50+j50*(31/2)/j60.

    Is this correct?
     

    Attached Files:

    Last edited: May 20, 2009
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