1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find iC(t) of RLC circuit

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data
    find iC(t) in the circuit and plot its waveform for t>=0.

    2. Relevant equations
    D1= i(0)-i(infinity)

    wd=sqrt(W0^2 - a^2)
    s1= -a+sqrt(a^2 - w0^2)
    s2= -a-sqrt(a^2 - w0^2)

    3. The attempt at a solution
    just a bit confused would this be a parallel RLC after all? doesn't seem like it, cause after switch moves it seems to be in series, except for current srouce. then I would calculate R(eq) by adding those two resistance? 0.1+0.52=0.62ohm?? i also get iL(0-)=iL(0)=4mA and iC(0)=0 because no current through open circuit?
    Last edited: Apr 5, 2013
  2. jcsd
  3. Apr 5, 2013 #2


    User Avatar

    Staff: Mentor

    You could change the 4mA current source and its parallel 0.1 Ω resistor into its Thevenin equivalent. Then you'd definitely have a series circuit for t > 0.
  4. Apr 9, 2013 #3
    i used source transformation and brought the 0.1ohm on top, and got an 0.4mV source. then i divided by the R(eq) of the two 0.1ohm resistors in series to get iL(0)=iL(0-)=2mA. then Vc(0) would be the voltage across the 0.1ohm resistors, 0.2mV. since iC(0)=iL(0) then using equaiton I=C(dv/dt) i get v'C(0)= iL(0)/C= 2mA*1.8=3.6mV. also, vC(infinity)=V(s)=0.4mV.

    a (for series)= R/2L= 0.62ohm/(2*0.05)=6.2 np/s
    w0 = LC^(-1/2)= 6 rad/s

    since a>w0 it would be overdamped. so using the formula i listed i get s1=−4.64 and s2=−7.76
    A1= (v'(0)-s2[v(0)-v(infnity)])/ s1-s2 = (3.6mV-(-7.76)(0.2mV-0.4mV))/3.12= 0.66mV
    A2= -[(v'(0)-s1[v(0)-v(infnity)])/ s1-s2] = (3.6mV-(-4.64)(0.2mV-0.4mV))/3.12= -0.86mV

    vC(t) =vC(infnity) +A1*e^(s1t)+A2*e^(s2t)
    vC(t) = 0.4mV+0.66mV*e^(-4.64t) - 0.86mV*e^(-7.76t)

    so far i did this right?
    Last edited: Apr 9, 2013
  5. Apr 9, 2013 #4


    User Avatar

    Staff: Mentor

    Yup. Looks good so far.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted