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[[Category:ECE201]]  
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[[Category:ECE]]  
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[[Category:ECE201Spring2015Peleato]]
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[[Category:circuits]]
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[[Category:linear circuits]]
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[[Category:problem solving]]
  
=Chinar_Dhamija_Critically_Damped_Problem_ECE201S15=
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=Critically Damped Practice=
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<center><font size= 4>
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'''Practice question for [[ECE201]]: "Linear circuit analysis I" '''
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</font size>
  
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By: Chinar Dhamija
  
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Topic: Critically Damped Second Order Equation
  
Put your content here . . .
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</center>
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----
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==Question==
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Find the value for C that will make the zero input response critically damped with roots at -4.
  
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[[File:ECE201P6.png|500px|center]]
  
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----
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----
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===Answer ===
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For a response to be critically damped we know that:<br />
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<math>b^2 - 4c = 0\\</math>
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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:
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<math> b = \frac{1}{RC}    and    c = \frac{1}{LC}\\</math>
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[[File:ECE201P6_1.png|500px|center]]
  
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Since the root was given to be -4 we can find b.<br />
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<math> \frac{-b}{2} = s\\ so we get: \frac{-b}{2} = -4\\ therefore b = 8</math><br />
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Once we know b we can use the critically damped equation to solve for C.
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<math>\begin{align}
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8^2 - \frac{4}{2C} = 0//
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64 = \frac{2}{C}//
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C = \frac{1}{32}//
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\end{align}
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</math>
  
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----
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==Questions and comments==
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If you have any questions, comments, etc. please post them below
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*Comment 1
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**Answer to Comment 1
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*Comment 2
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**Answer to Comment 2
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----
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[[2015 Spring ECE 201 Peleato|Back to 2015 Spring ECE 201 Peleato]]
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[[ECE201|Back to ECE201]]

Revision as of 14:44, 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.

ECE201P6.png


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}\\ $

ECE201P6 1.png

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} $


Questions and comments

If you have any questions, comments, etc. please post them below

  • Comment 1
    • Answer to Comment 1
  • Comment 2
    • Answer to Comment 2

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