| Line 35: | Line 35: | ||
Once we know b we can use the critically damped equation to solve for C.<br /> | Once we know b we can use the critically damped equation to solve for C.<br /> | ||
<math>\begin{align} | <math>\begin{align} | ||
| − | 8^2 - \frac{4}{2C} = 0 | + | 8^2 - \frac{4}{2C} = 0\\ |
| − | 64 = \frac{2}{C} | + | 64 = \frac{2}{C}\\ |
| − | C = \frac{1}{32} | + | C = \frac{1}{32}\\ |
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 14:51, 2 May 2015
Critically Damped Practice
Practice question for ECE201: "Linear circuit analysis I"
By: Chinar Dhamija
Topic: Critically Damped Second Order Equation
Question
Find the value for C that will make the zero input response critically damped with roots at -4.
Answer
For a response to be critically damped we know that:
$ b^2 - 4c = 0 $
The next step would be to simplify the circuit as shown in the image below. Once simplified it becomes a parallel RLC circuit where we know:
$ b = \frac{1}{RC} $ and $ c = \frac{1}{LC} $
Since the root was given to be -4 we can find b.
$ \frac{-b}{2} = s $ so we get: $ \frac{-b}{2} = -4 $ therefore b = 8.
Once we know b we can use the critically damped equation to solve for C.
$ \begin{align} 8^2 - \frac{4}{2C} = 0\\ 64 = \frac{2}{C}\\ C = \frac{1}{32}\\ \end{align} $
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