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B RLC circuit analysis

  1. May 4, 2016 #1

    Titan97

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    In an RLC series circuit let applied EMF be given ##V=V_0\sin\omega t##, $$Z=Z_C+Z_R+Z_L=R+i\left(\frac{1}{\omega C}-\omega L\right)$$
    $$|Z|=\sqrt{R^2+\left(\frac{1}{\omega C}-\omega L\right)^2}$$

    Then $$i(t)=\frac{V(t)}{Z}=\frac{V_0e^{i\omega t}}{R+i\left(\frac{1}{\omega C}-\omega L\right)}$$

    Its given in my book that
    $$i(t)=\frac{V_0(\sin\omega t+\phi)}{\sqrt{R^2+\left(\frac{1}{\omega C}-\omega L\right)^2}}$$

    Why are they considering a phase difference of ##\phi##?

    Also, why are they taking modulus of ##Z## and only the imaginary part of applied voltage?

    What is the difference between the first ##i(t)## and the second ##i(t)##?
     
  2. jcsd
  3. May 4, 2016 #2

    f95toli

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    Are you familiar with phasor notation?
     
  4. May 4, 2016 #3

    cnh1995

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    That phase difference is between voltage and current.
     
  5. May 4, 2016 #4

    Titan97

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    I get it. I can write $$R+i\left(\omega L-\frac{1}{\omega C}\right)=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}e^{i\phi}$$
    Hence,
    $$i(t)=\frac{V(t)}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}e^{i\phi}}$$
    $$i(t)=\frac{V(t)e^{-i\phi}}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}$$

    Now it's in phasor notation.
     
    Last edited: May 5, 2016
  6. May 4, 2016 #5

    SammyS

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    There's ##\ i\ ## and then there's ##\ i\ ## .

    You probably should write ##\ i(t)\ ## for the current
     
  7. May 5, 2016 #6

    Titan97

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