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RLC AC Circuits

  1. Feb 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Information given:
    Resistance=105 ohms, , XL= 212 ohms, XC= 93 ohms
    V rms is 143 Volts.
    Frequency of the circuit is f. Not defined.

    Information Solved for:

    Impedance (Z) = 158.7

    I rms=.901 Amps

    Phase angle = 48.58

    The final question, which I cannot solve for some reason, is this:

    At the instant the voltage across the generator is maximum, what is the current I?

    I know that the current will be less than maximum as it lags in this case, but I don't know by what factor or how to compute it from here. Any suggestions?

    2. Relevant equations

    Vmax=ImaxZ=2^1/2*V rms
    Z= (XL - XC) / R

    3. The attempt at a solution
    I have tried everything here and it's been pathetic so. I tried using the angle difference from 45 degrees (3.58) as somehow why I will not be maximized as the current is out of phase with V. This was wrong. Any suggestions?
    Last edited: Feb 28, 2009
  2. jcsd
  3. Feb 28, 2009 #2


    User Avatar

    Staff: Mentor

    Does fixing your typos help you solve the problem?
  4. Feb 28, 2009 #3
    Not so far.
  5. Feb 28, 2009 #4


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    Staff: Mentor

    Let me re-phrase.

    Would your fixing your typos in your post help us to help you to learn?
  6. Feb 28, 2009 #5
    I don't even know where to start here, do you have a diagram you could post of the circuit for us?
  7. Feb 28, 2009 #6
    Does [tex]v(t)= \sqrt{2}V_{rms}sin( \omega t)[/tex] say anything to you? Is this given in the problem? Or has your teacher said this is the signal you can use in all of your calculations?
  8. Feb 28, 2009 #7
    Kruum: It was not given in the problem; it *is* relevant in the next problem, but I don't think in this one.

    As far as the circuit goes, It is a simple: Generator----R----L----C----(return to Generator) circuit.

    berkeman: aside from putting amps where I should have put volts, I don't see any other typos.
  9. Feb 28, 2009 #8
    You don't need the frequency or the time. All you need to know that sin(90) is 1. But since it's not given or you can't assume that is the signal, it's no use then, obviously.
  10. Feb 28, 2009 #9
    I don't see how I can use the reactances when I'm given neither the L or C values. I result in variables Lf or Cf either way, as the equations are XL=2pi*fL , and XC=(2pifC)^-1, respectively.
  11. Feb 28, 2009 #10
    Yeah, I was abit too hasty. I remembered the values were given, but it was only the reactance. But you really don't need that info as I wrote above all you need is the fact that sin(90) is 1. So [tex]i(t)= \sqrt{2}I_{rms}sin(90-48.58)[/tex] sould be the value for current. But if you can't assume that is the signal, then it's no use.
  12. Feb 28, 2009 #11
    It is indeed the signal. I tested this and it was correct. Thanks. However, I would like to know why this assumption is valid? It doesn't seem so apparent to me.
  13. Feb 28, 2009 #12
    Which assumption, that the signal for the voltage is what it is or the signal for the current?
  14. Feb 28, 2009 #13
    The assumption that it 2pi*ft = 90 (= 1 when computed in sin). I know the phase angle, and I know that v(t)=Vmax*sin(2pi*ft + phase angle), but that doesn't give me a logical pathway to i(t)=Imax*sin(2pi*ft - phase angle) nor describe why 2pi*ft = 90.

    In other words, my book does a poor, poor job of explaining this relationship.
  15. Feb 28, 2009 #14
    You probably have had some trigonometry and you know that [tex] \frac {\pi}{2}=90^{ \circ}[/tex] and [tex]sin( \frac {\pi}{2})=1[/tex]. So [tex]2 \pi f*t[/tex] must give us [tex] \frac {\pi}{2}[/tex] in order to get the maximum voltage, right?

    Now, the phase angle is quite a bit more complicated story. But basically it's due to the physical properties of the capasitor and the inductor. The inductor creates a magnetic field, which tries to oppose the rise of the current. And the capasitor, on the other hand, is gathering pontetial difference between it's plates. So the faster the potential of the capasitor rises, the faster the electrons has to go from plate to plate, thus stronger the current.

    Does that help?
    Last edited: Feb 28, 2009
  16. Feb 28, 2009 #15
    That helps quite a bit. Thanks for the in-depth explanation. That's what I was looking for.
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