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Q&A

Voltage of tank circuit is invalid

+0
−2

In a previous question we found that for a critically damped oscillation:

$$ I(t) = 2 t e^{-0.5 t} $$

But if we integrate to find the voltage through the capacitor:

$$ V_{CO} + \frac{1}{C} \cdot \int_0^t{I(t)}dt $$

the integral is always a positive value which doesn't make any sense since the capacitor is being discharged.

What am I doing wrong here? Help appreciated!

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1 comment thread

When it is critically damped, there is not oscillation. (1 comment)

1 answer

+3
−1

You need to be more careful.

First, you should have been able to see for yourself that the first equation isn't written right. I'll assume the whole "-0.5t" is the exponent of e. You were too sloppy to notice, or too lazy to fix it. Either way, it's really rude to the volunteers here. That's what the downvote is for.

Second, pay attention to the sign. The top equation always yields a positive result for positive values of t. Obviously integrating that will also yield a positive result. You have to decide which way is positive current flow, and therefore positive voltage, and stick to it. In this case, a diagram of the circuit with I and V clearly labeled would have probably caused you to see the mistake for yourself.

Taking extra care to be meticulous may seem like a waste of time, but is actually a time saver in the long run. You need to learn that sooner than later. In the real world, you don't get partial credit because your mistake was "only" flipping a sign. If it doesn't work, it doesn't work, and it's on you.

There is no place for sloppiness in engineering. Grow up.

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1 comment thread

@#36396 I dont know how to use LaTeX and I used one of the online editors I found.I tried to find an ... (5 comments)

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