Q&A

# Critically damped oscillation issue

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For a critically damped oscillation for a series RLC circuit the equation of current has this form I(t)=D1te^(-at)+D2e^(-at) where D2 =I(0+) and D1-aD2 = dI(0+)/dt=VL/L.

Due to L1 :I(0)=I(0+)=0A and by applying KVL in VL1=-VC1=-2V

And by substituting the values we get D1=-2A/s and D2 = 0A and we end up with an equation of I(t)=-2te^(-0.5t) but this cant be correct.What am I doing wrong?

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Your derivation is correct, you just missed the sign: $V_L=-V_C=2;\mathrm{V}$, because the capacitor charges with +2 V, and the discharge accounts for the negative sign on the inductor. The notation of $V_L=-V_C$ may be confusing, so think of it as $V_C=-V_L$, maybe it makes more sense.

Verifying your results is never a bad idea, so here it is:

I(R2) is plotted in blue while in black it's the test voltage, modelled as a function with the same format as your result, except for taking care of the delays (the uramp() part, such that it's zero until the event).

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I still dont get it VL=-VC = -2V. (5 comments)
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I don't know how to attach images to a comment thread, so I will just open up an answer for this. I redrew the schematic for you to reference.

I think that symbol at the top of your drawing probably represents a switch. We are likely looking at damping when the switch closes. We want to see what happens when the capacitor has a 2V difference across it (fully charged), and then the switch suddenly opens.

I myself am terrible at math and derivations, but I think I found the goal formula for you to aim for: the current $I1$ in a critically-damped, series RLC circuit. I used MathJax as recommended by Olin.

$$I(t)=\frac{V_0}{L}te^\frac{-Rt}{2L}$$

$$\frac{R}{2L} = \alpha$$

$\alpha$ is the attenuation for this particular circuit configuration (series RLC).

Someone who is actually good at math should be able to guide you to the derivation, but hoped this helped a little bit.

UPDATE Dave Tweed helpfully pointed out that my schematic above is probably wrong in terms of what the OP wanted. It definitely is wrong with my own interpretation of what the OP wanted, though... Lol. Dave understands me better than I do.

I fixed the circuit schematic below:

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