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Calculating half power frequency/cutoff frequency

  1. May 5, 2015 #1
    1. The problem statement, all variables and given/known data

    *(Problems attached as images)*

    For first problem:

    ##\cfrac{|V_2|}{|V_1|} = \cfrac{1}{\sqrt{4+(\omega RC)^2}}##

    For second problem:

    ##\cfrac{|V_2|}{|V_1|} = \cfrac{\sqrt{1+(\omega RC)^2}}{\sqrt{4 + (\omega RC)^2}}##

    I have to calculate half power frequency for both cases.

    2. Relevant equations

    At half power frequency, power absorbed by circuit = max power/2

    3. The attempt at a solution

    I cannot understand why the problem was solved differently in both cases. In first problem,##\cfrac{V_2}{V_1}## was set to ##\cfrac{1}{2\sqrt{2}}## and in the second problem it was set to ##\cfrac{1}{2}##. While in both problems, max value of ##\cfrac{V_2}{V_1}## occurs at ##\omega = 0## and is equal to ##\cfrac{1}{2}##.

    Can someone please clear this doubt?

    Thanks
     

    Attached Files:

  2. jcsd
  3. May 6, 2015 #2

    Svein

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    Science Advisor

    Power = Voltage * Current
     
  4. May 6, 2015 #3
    Thanks for the reply. But, I didn't get you. I am familiar with that formula. I understand that we use that formula to substitute ##\cfrac{|V_2|}{|V_1|} = \cfrac{1}{\sqrt{2}}## but how does it explain the ##\cfrac{1}{2\sqrt{2}}## part ?

    Thanks again.
     
  5. May 6, 2015 #4

    Svein

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    Science Advisor

    When the impedance is reduced, the current increases. Thus the power ratio is not equal to the voltage ratio.
     
  6. May 6, 2015 #5

    Svein

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    Science Advisor

    When the impedance is reduced, the current increases. Thus the power ratio is not equal to the voltage ratio.
     
  7. May 6, 2015 #6

    gneill

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    Staff: Mentor

    Check again the frequency and maximum output for the second circuit. How do capacitors behave with frequency?

    For the second solution shown, I think the magnitude of the voltage ratio should have been set to ##\frac{1}{\sqrt{2}}## rather than ##\frac{1}{2}##
     
  8. May 6, 2015 #7
    Thanks for the reply. I think I got it. The max output should be 1 when ##\omega \rightarrow \infty##?
     
  9. May 6, 2015 #8

    gneill

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    Staff: Mentor

    Yup. The capacitor starts to look like a short at high frequencies.
     
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