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Inductance Question simple.

  1. Oct 6, 2008 #1
    1. The problem statement, all variables and given/known data

    An AC generator produces a voltage of 230<45º volts. It is connected across an impedance of 17.4<-33º ohms.

    The supply frequency = 50Hz

    I got my assignment back and this question wrong, maybe you can guide me, please because I’m not sure entirety. Help is appreciated.

    Question: Calculate the value of inductance that will make the circuit become series resonant at 50Hz?

    2. Relevant equations

    1/2π fc = 2 π fl

    3. The attempt at a solution

    I= 17.4 sin33 = 9.47
    C= 1/2 π (50x9.47) = 3.36x10^-6
    Would it be: 2 π fl > 2 π(50) = 314 ?
  2. jcsd
  3. Oct 6, 2008 #2


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    Hmm, check your math. If

    [tex]X_c=\frac{1}{j\omega C}[/tex]

    then C = 3.36e-4. When I plug C into

    [tex]L=\frac{1}{\omega^2 C}[/tex]

    I get L=0.030.
  4. Oct 6, 2008 #3
    Thanks for the responce, but I dont undestand why you say check my math, what wrong with it. You've confused me now.


    Also where did those two other equations come from cause my lecturer has not used them, there not the ones i was using?

    1/2π fc = 2 π fl

  5. Oct 6, 2008 #4


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    The values are correct but the answer is wrong. Try it again.

    so in your notation,

    [tex]l=\frac{1}{(2\pi f)^2 c}[/tex]

    I get l=0.030.
  6. Oct 7, 2008 #5

    I re-calculated it and got this for C:

    C= 1/2 π (50x9.47) = 743.7


    Using : [tex]l=\frac{1}{(2\pi f)^2 c}[/tex]

    1/(2πx50)^2 = 1.013x10^-5

    (1.013x10^-5) x C (743.7) = 7.5x10^-3

    Which is utterly wrong.

    I broke it down into chucks thinking tit would help, either way i get the same answer above.
  7. Oct 7, 2008 #6


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    No, that's why I wrote down the equations before. The correct equation in your notation is
    [tex]C=\frac{1}{2\pi f I}.[/tex]

    Take a look in your book! Physics Forums is not a replacement for studying.
    Last edited: Oct 7, 2008
  8. Oct 8, 2008 #7
    Im sorry my limited amount of knowleadge, i find it hard to understand even on a good day (It takes me a couple of times, before it all sinks in, esp, when the lecturer is not even good, nor text books or websites make it clear).

    [tex]C=\frac{1}{2\pi f I}.[/tex] => [tex]C=\frac{1}{2\pi 50 9.47}.[/tex] = 3.361*10^-4


    [tex]l=\frac{1}{(2\pi f)^2 c}[/tex] => 1 / (2pi x 50)^2 x 3.361x10^-4) = 0.030155 = 30155uH

    So basically,that will make the circuit become series resonant at 50Hz, is that it marcus.
    Last edited: Oct 8, 2008
  9. Oct 8, 2008 #8


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    Yes, that's it!
    Watch out for the number of digits in your answer, though. Since the problem specified three significant digits (17.4 ohms), your answers should also have 3 significant digits: 0.030 H and 3.36e-4 F.
  10. Oct 8, 2008 #9
    Thanks for the help Marcus, it was most kind.
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