- #1

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- 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

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