Is My Approach to Solving a Source-Free RLC Series Circuit Correct?

[wcjy]

Homework Statement

Given the following circuit with the source voltage V1=60(V). The switch in the following circuit has been connected to A for a long time and is switched to B at t=0. The current i(t) through the capacitor C for t>0 has the following expression:

$$i(t) = Ae^{xt} + Be^{yt}$$

Relevant Equations

$$α = \frac{R}{2L}$$
$$ω_o = \frac{1}{\sqrt{LC}}$$
$$ S_{1,2} = - α +- \sqrt{ α^2 - w_o^2}$$

Hello, this is my working. My professor did not give any answer key, and thus can I check if I approach the question correctly, and also check if my answer is correct at the same time.

for t < 0,
V(0-) = V(0+) = 60V
I(0) = 60 / 50 = 1.2A

When t > 0,
$$α = \frac{R}{2L}$$
$$α = \frac{30}{2(10)}$$
$$α = 1.5 $$

$$ω_o = \frac{1}{\sqrt{LC}}$$
$$ω_o = \frac{1}{\sqrt{10*50*10^{-3}}}$$
$$ω_o = \sqrt{2}$$

$$ S_{1,2} = - α +- \sqrt{ α^2 - w_o^2}$$
$$ S_{1,2} = - 1.5 +- \sqrt{ 1.5^2 - \sqrt{2}^2}$$
$$ S_{1,2} = -1.5 +- j0.5$$

$$i(t) = e^{-1.5t}[Acos(0.5t) + Bsin(0.5t)]$$
When t = 0, i = 1.2
$$ A = 1.2$$

when t = 0
$$L\frac{di}{dt} + Ri + V= 0$$
$$\frac{di}{dt} =-\frac{1}{L} (Ri + V)$$
$$\frac{di}{dt} = -\frac{1}{10} ((50)\frac{6}{5} + 60)$$
$$\frac{di}{dt} = -\frac{1}{10} ((50)\frac{6}{5} + 60)$$
$$\frac{di}{dt} = -12$$

when t = 0
$$\frac{di}{dt} = -1.5[ e^{1.5t} [ Acos(0.5t) + Bsin(0.5t)]] + 0.5e^{-1.5t}[-Asin(0.5t)+Bcos(0.5t)]$$
$$-12 = -1.5A + 0.5B$$
$$ B = -20.4$$

Therefore
A = 1.2
B = -20.4
x = 1.5
y = 1.5

 

[Gordianus]

The statement says the switch has been in in the A position for a long time (the capacitor id "fully" charged). No current flows in the "initial" condition.

 

[wcjy]

oh so if I(0) = 0,
A = 0
di/dt = -6
B = -12
the rest remains the same?