Solving RLC Circuit Homework | Help Needed with Finding A and B

[jesuslovesu]

Homework Statement

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)
http://img503.imageshack.us/img503/448/rlcwo0.th.png

Homework Equations

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?

 

[CEL]

Homework Statement

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)
http://img503.imageshack.us/img503/448/rlcwo0.th.png

Homework Equations

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?

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 $$v_o(t)$$.

 

[piercas]

It looks like you have the right idea, but there may be a small mistake in your calculations. First, let's review the equations you are using:

1. Kirchhoff's voltage law (KVL): This states that the sum of the voltages around a closed loop in a circuit must equal zero. In your circuit, the loop consists of the voltage source, the resistor, and the capacitor. So, we can write:

V_0 - i_R(t)R - v_C(t) = 0

2. Kirchhoff's current law (KCL): This states that the sum of the currents entering a node in a circuit must equal the sum of the currents leaving that node. In your circuit, the node is the one connecting the resistor and the capacitor. So, we can write:

i_R(t) = i_C(t) + i_L(t)

3. Ohm's law: This states that the voltage across a resistor is equal to the current through it multiplied by its resistance. So, we can write:

v_R(t) = i_R(t)R

4. Capacitor equation: This states that the current through a capacitor is equal to the capacitance multiplied by the derivative of the voltage across it. So, we can write:

i_C(t) = C(dv_C(t)/dt)

5. Inductor equation: This states that the voltage across an inductor is equal to the inductance multiplied by the derivative of the current through it. So, we can write:

v_L(t) = L(di_L(t)/dt)

Now, let's apply these equations to your circuit. Since we are given the initial conditions at t = 0, we can use them to find the values of A and B. First, we can use KVL to write:

V_0 - i_R(0^+)R - v_C(0^+) = 0

Substituting in the given values, we get:

V_0 - 2(2) - 0 = 0
V_0 = 4

So, we have found the value of A to be 4. Now, we can use KCL to write:

i_R(0^+) = i_C(0^+) + i_L(0^+)

Substituting in the given values, we get:

2 = C(dv_C(0^+)/dt) + L(di_L