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Phase angle between peak source voltage and peak voltage across RLC?

  1. Apr 15, 2014 #1
    1. The problem statement, all variables and given/known data

    An LRC series circuit with R = 200Ω , L = 31mH , and C = 1.8μF is powered by an ac voltage source of peak voltage V0 = 540V and frequency f = 550Hz. What are the peak voltages across the inductor, capacitor, and resistor and their phase angles relative to the source voltage?

    2. Relevant equations

    Phase shift = [itex]tan^{-1}[/itex]([itex]\frac{XL - XC}{R})[/itex]

    I already found the peak voltages across the components.

    3. The attempt at a solution

    I can't figure out how to find the phase angles between the source voltage and the peak voltage across the circuit components. Is the phase angle between the voltages found using this equation?
     
  2. jcsd
  3. Apr 15, 2014 #2

    BiGyElLoWhAt

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    I don't wanna say no, but not necessarily, the phase shift function is different for all circuits. Basically, you have a real part and an imaginary part (this comes from your complex analysis that i'm assuming you've done already).

    You'll end up with something like this:
    [itex]
    V_{out} = V_{in}(f(R,L,C) + g(R,L,C)i)[/itex]
    and this corresponds to the phase shift,

    What you get is:
    [itex]\phi=tan^{-1}(\frac{g(R,L,C)}{f(R,L,C)})[/itex]

    See where this comes from? just plot it and I think it becomes rather obvious. The imaginary term is the y-axis and the real part is the x-axis.
     
  4. Apr 15, 2014 #3

    gneill

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    Staff: Mentor

    Suggestion: Start by finding the impedance values for each of the components (complex representation). Then write the total impedance as their sum, Z = R + ZL + ZC. Find the current phasor for the circuit from the given source voltage and total impedance, I = E/Z, where E is the given 540V. Can you then find the individual voltage drops in their complex forms?
     
  5. Apr 15, 2014 #4

    BiGyElLoWhAt

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    your v_out function (g and f) include all your
    [itex][STRIKE]Z[/STRIKE]_{c} & X_{L} and R[/itex] terms. probably should've mentioned that previously.
     
    Last edited: Apr 15, 2014
  6. Apr 15, 2014 #5
    Well I have [itex]\frac{V_{out}}{V_{in}}[/itex] = [itex]\frac{XC}{\sqrt{R^{2}+(XL-XC)^{2}}}[/itex]

    How can I apply this to finding those functions that are part of the phase shift formula?
     
  7. Apr 15, 2014 #6
    Is the individual impedance of the capacitor equal to XC? And for the inductor XL and resistor R?

    I found the RMS voltage using I = Vsource/Z. After that I found the voltage across each component by multiplying the individual impedances by the RMS voltage.
     
  8. Apr 15, 2014 #7

    gneill

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    Impedance is a complex value. XL and XC are called reactances, which is the magnitude of the impedance for those components. To make it more clear, the impedances of the three components are:

    ZR = R + 0j = R
    ZL = 0 + jωL = jXL
    ZC = 1/(jωC) = -jXC

    Since you're looking for peak values according to the question statement, and since the source voltage is given as a peak value, there's no reason to involve RMS conversions. And I think you meant that you found the current, not the voltage, using Vsource/Z. Other than that terminology issue your method sounds okay. Can you show your results?
     
  9. Apr 15, 2014 #8

    BiGyElLoWhAt

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    If you have that, then you have:
    [itex]V_{out}[/itex] = [itex]V_{in}\frac{XC}{\sqrt{R^{2}+(XL-XC)^{2}}}[/itex]

    And i'm confused... is it XC or [itex]X_{c}[/itex]

    But basically your goal at this point is to separate the real part (the R in the denominator) from the imaginary part the XC's and XL's and do what I said previously.
     
  10. Apr 15, 2014 #9

    BiGyElLoWhAt

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    To clarify (again) where the v_in is is irrelevent. I just threw it on the RHS to relate it to my previous post.
     
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