Find iC(t) of RLC circuit

1. Apr 4, 2013

asdf12312

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
i(t)=i(infinity)+e^(-at)[D1cos(wdt)+D2sin(wdt)
D1= i(0)-i(infinity)
D2=i'(0)+a[i(0)-i(infinity)]

aux.:
w0=(LC)^(-1/2)
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. Apr 5, 2013

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.

3. Apr 9, 2013

asdf12312

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

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
4. Apr 9, 2013

Staff: Mentor

Yup. Looks good so far.