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Finding resistance and capacitance

  1. Jul 20, 2013 #1
    I am stuck, can someone help me, please? I am not sure if I started the right way, it seems tedious :(

    The magnitude of V1 is 2 times greater than V2, and its phase leads π/2 ahead. The angular frequency ω of the power supply voltage is 5rad/s.
    Find the values of the resistance R and Capacitance C.
     

    Attached Files:

  2. jcsd
  3. Jul 20, 2013 #2
    OK, I think I did a mistake on the impedance across the capacitor. It's a parrallel set, right?
    So is this formula correct? Z2=R//(1/jwc)

    So here's what I've done so far. What should I do next?
     

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    Last edited: Jul 20, 2013
  4. Jul 20, 2013 #3
    Or perhaps there is another easier way to do it?
     
  5. Jul 21, 2013 #4

    gneill

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    Staff: Mentor

    Perhaps there is. Rather than hammering away at the phase angles right away, remember that when you divide two complex numbers the resulting angle is the difference between the numerator's angle and the denominator's angle. So, if you divide V1/V2 you want the phase angle of the result to be ##\pi/2##.

    Why not suppose that the source driving the circuit is a 1A current source. Then from your work,

    ##V1 = 1 + 2j##

    ##V2 = \frac{R}{1 + 5RCj}##

    Can you carry on?

    (HINT: The tricky bit lies in realizing the implications of the angle being ##\pi/2##)
     
  6. Jul 21, 2013 #5
    Ok, so I have to find arg(V1/V2), right? I got arg(V1/V2) = arctan((5CR+2)/(1-10R))

    Am I doing this right? I wanna make sure before proceeding.
     
  7. Jul 21, 2013 #6

    gneill

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    That's the idea. Check the denominator though, I think you've dropped a 'C'.

    In order for the arctan of this expression to be ##\pi/2##, what condition must hold?
     
  8. Jul 21, 2013 #7
    Errr......I'm guessing that the real part of the complex number aka the x-coordinate to be 0?
    So, that means 1-10CR=0, and CR= 0.1 !!!!I think that must be correct.
     
    Last edited: Jul 21, 2013
  9. Jul 21, 2013 #8

    gneill

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    :approve:

    Yes, in the limit as the denominator goes to zero, the slope goes to infinity. That's the same as saying that the angle goes to vertical --- 90° or ##\pi/2## radians.

    I suggest that for now you write it as ##C = \frac{1}{10 R}##. (You'll see why a bit later).

    Can you now use the information about the ratio of magnitudes of V1 and V2 to proceed?
     
  10. Jul 21, 2013 #9
    Yes, actually I was solving it while waiting for your reply. The answer for C is 1/25 and R is 5/2. Are my answers correct?
    I can't show you my working yet because it's too messy LOL.
     
  11. Jul 21, 2013 #10

    gneill

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    They're off by a factor of two. No doubt this will be discovered as you "tidy up" :smile:
    Ha! :rofl:
     
  12. Jul 21, 2013 #11
    Yeah, I had a careless mistake there. Here it is, tell me if I did it right. You're a genius, man, thank you so much!!!You explain better than my own professor does, haha.
     

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  13. Jul 21, 2013 #12

    gneill

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    Staff: Mentor

    Looks good!
    That's very kind of you to say. Thanks!
     
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