Series RLC

1. Mar 31, 2008

jesuslovesu

1. The problem statement, all variables and given/known data
I'm having some problems solving this RLC circuit, if anyone could help.

R = 2 ohms
C = 2/3 F
L = 1/2 H

The top picture is when t < 0
The bottom picture is when t > 0
Find $$v_o(t)$$ (notice V_0 is the defined +- voltage over the resistor)

2. Relevant equations

3. The attempt at a solution

Well I found that:
$$i_L(0^+) = 2 A$$
$$v_c(0^+) = 0$$
$$v_o(0^+) = -4 V$$
Which I believe are correct,
I found the general equation to be $$v_0(t) = Ae^{-t} + Be^{-3t}$$
(My main problem is finding A and B, they should be 2 and -6 but I just can't get that)

A + B = -4
$$dv(0^+)/dt = -A + -3B$$
This is where I get a little sketchy, but since Cdv/dt = i_c then I was thinking that since $$i_L(0^+) = i_C(0^+)$$ for series circuits I could just say
$$-A + -3B = 2/C$$
However when I solve those two equations I get A = -4.5, B = 1/2
Does anyone know what I did wrong?

2. Apr 2, 2008

CEL

You should always solve a series circuit for $$v_c(t)$$ and a parallel circuit for $$i_L(t)$$.
Once you have $$v_c(t)$$ , you can differentiate it in order to get $$\frac{dv_C}{dt}$$. Since the current in a series circuit is the same for all elements, you have $$v_c(0)$$ and $$\frac{dv_C}{dt}(0)$$.
Knowing $$v_c(t)$$ you can calculate $$v_o(t)$$.