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Voltage and phase between two branches of ac circuit

  1. Feb 22, 2017 #1
    Mod Note: moved from technical forum, so homework template is missing

    In a parallel circuit, one branch contains two equal resistors of resistance ##R_1## connected in series. The other branch contains a resistor of resistance ##R_1## in series with an inductor of variable inductance ##L##.
    Show that the magnitude ##|V_{AB}|=\frac{V}{2}## and ##\phi=\tan^{-1}\big(-\frac{2\omega LR_1}{R_1^2-\omega^2L^2}\big)##


    https://gyazo.com/50cbf5ef913b0fc533ebcb0074b3b5a3 (Very sorry for the link, but the image uploader wasn't working for me)
    50cbf5ef913b0fc533ebcb0074b3b5a3.png {image inserted by moderator}


    I have worked out that total impedance =##|Z|=\frac{2R_1(R_1+\omega L)}{3R_1+\omega L}## Which I'm not sure is correct.

    I'm thinking that I have to work out what fraction of voltage ##R_1## gets in the ##B## branch, but I'm pretty confused.

    Edit: I have managed to prove ##|V_{AB}|=\frac{V}{2}## and am now stuck on the second part, it's only two marks so it shouldn't be difficult, I just don't know what impedance I'm meant to be taking

    Any help on this would be really appreciated :)
     
    Last edited by a moderator: Feb 22, 2017
  2. jcsd
  3. Feb 22, 2017 #2

    NascentOxygen

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    Hi Tom Moynihab. smiley_sign_welcome.gif

    You need to enclose your Latex inside double hash marks, not single dollars. You can use double dollars where you want it presented on a new line.
     
  4. Feb 22, 2017 #3
    oh sorry, which tex plugin is used?
     
  5. Feb 22, 2017 #4

    NascentOxygen

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    MathJax does its work at the server and it is sent to you as javascript, I think.
     
  6. Feb 22, 2017 #5
    oh hmm I'm not getting any tex, still all code. Also is this visible/what do i need to do to make it visible?
     
  7. Feb 22, 2017 #6

    NascentOxygen

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    If you are using the PF app it displays the raw MathJax code. You'll need to use a browser to see your LaTex creation in its full glory.
     
  8. Feb 22, 2017 #7
    I am currently using chrome though. Edit: fixed it, had to turn off my previous plugin
     
  9. Feb 22, 2017 #8
    Sorry my post currently visible?
     
  10. Feb 22, 2017 #9

    gneill

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    For the phase angle, you've probably done most of the work already when proving that ##|V_{AB}|=\frac{V}{2}##. Can you show that work up to the point where you found the magnitude?
     
  11. Feb 22, 2017 #10
    Well it might be a bit of a jump, but I said for A it was V/2 as the voltage simply splits 50/50. Then for B I said as the impedance between the resistor and the inductor is pi/2 out of phase, therefore voltage across the resistor varies between V and 0... and I just called it 0.
     
  12. Feb 22, 2017 #11

    gneill

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    I'm afraid that doesn't hold water as a derivation. You'll need to return to that part and do the math.

    The A branch is a simple voltage divider as you pointed out. But so is the B branch. The only difference is that one of the elements is an inductor rather than another resistor. You should be able write an expression for the potential at point B using the voltage divider formula.
     
  13. Feb 22, 2017 #12
    The part which confused me was that the voltage from A is already V/2, therefore B has to be 0, which means surely the circuit wouldn't work?
     
  14. Feb 22, 2017 #13

    cnh1995

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    That's not correct. Branch B contains an inductor. You need to consider the phase difference between voltages at A and B.
     
  15. Feb 22, 2017 #14

    gneill

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    Following on to what @cnh1995 said, if you do the algebra you'll find the answers to both the voltage magnitude and the phase angle questions.
     
  16. Feb 22, 2017 #15
    So as the impedance for a resistor is just R and the impedance for an inductor is [;j\omega L;] meaning they are [;\frac{\pi}{2};] out of phase with each other, meaning that when [;V_R=R*0;] and [;V_L=X_L*I_{max};] maybe? Am I at least on the right track? I am going to bed now as it is 3.30AM, but I will check tomorrow
     
  17. Feb 22, 2017 #16

    gneill

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    Do the math; it's just a bit of complex algebra. It'll automatically take care of the phase relationships without your having to make guesses. Certainly when the circuits get more complicated you won't be able to use handwavy arguments to analyze them.
     
  18. Feb 23, 2017 #17
    I think my problem is setting up the correct math, it's confusing to me what to do, could you help with that please?
     
  19. Feb 23, 2017 #18

    gneill

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    Look up: "voltage divider" (or "potential divider").
     
  20. Feb 23, 2017 #19

    cnh1995

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    Would you be comfortable with the phasor approach?
     
  21. Feb 23, 2017 #20
    Yeah I am, many thanks
     
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