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anonymous

  • one year ago

A 120V rms potential at f=1000Hz is applied to a 30 mH inductor, a 40 microF capacitor and a resistor (R). Irms=400mA. What is the value of R? What is the power factor phase angle phi? The only equation I know that I could possibly use to find R is Vrms=IrmsR, when I plugged in my data, I got R=300 ohms. Using R=300 ohms, I rearranged the equation PF = cos phi = R/Z, where Z=sqrt(R^2+[XL-Xc]^2) to solve for R and I got phi=90 degrees. Which makes me suspicious. The data never specified that the circuit was in resonance, so was my original equation invalid?

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  1. IrishBoy123
    • one year ago
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    300\(\Omega\) is the total impedance Z where \(Z = \sqrt{R^2 + (X_L - X_C)^2)}\) the circuit will resonate when \(X_L = X_C \). i make that about 45 Hz but do it yourself and do not rely on my owlet packet calcs :p

  2. IrishBoy123
    • one year ago
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    @radar

  3. anonymous
    • one year ago
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    I'm not sure what you are saying. Is my equation wrong?

  4. IrishBoy123
    • one year ago
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    OK you said V = IR I am saying V = IZ Z is as described above make sense?

  5. IrishBoy123
    • one year ago
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    are you familiar with phasor diagrams? that is how you should "add" the equivalent "resistances" of the resistor inductor and capacitor

  6. anonymous
    • one year ago
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    So I solve for Z using Vrms and Irms and then solve for R by plugging Z, L, C, and f into the long equation? Where XL = 2 pi f L and Xc = 1/2 pi f C ?

  7. anonymous
    • one year ago
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    I know what a phasor diagram is but I don't know how to use it in this instance.

  8. IrishBoy123
    • one year ago
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    Imaginary parts: \(X_L = j \omega L\) and \(X_C = -j /\omega C\) Real part : R use \(w = 2 \pi f\) to get \(\omega\) from 1000Hz frequency |dw:1436290334762:dw| \(X_{net} = X_L - X_C = j \ ( \omega L - 1/ \omega C )\)

  9. anonymous
    • one year ago
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    I have never taken calculus. I am assuming that you are saying that Z is the hypotenuse of a triangle where R gives the X value and Xnet gives the Y? So I convert f to omega, say XL = omega L and Xc = 1/omega C, subtract Xc from XL to get Xnet and then use pythagorean theorem to get R?

  10. IrishBoy123
    • one year ago
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    sounds good. |dw:1436290978466:dw|

  11. IrishBoy123
    • one year ago
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    i would try to be more helpful but i got it in the ear recently for being too helpful so i must tread carefully :p

  12. anonymous
    • one year ago
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    Thank you, I am just trying to make sure I understand what you are saying. Is my original equation wrong because it is not in resonance?

  13. IrishBoy123
    • one year ago
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    to be in resonance \(X_L\) and \(X_C\) cancel each other out so \(f_{res} = \frac{1}{2 \pi \sqrt{LC}}\) you can calculate the resonant frequency. i think it's 145Hz but check for yourself

  14. IrishBoy123
    • one year ago
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    and yes you are right to say that, if it was resonating, you could just say V = IR because at that frequency R = Z.

  15. radar
    • one year ago
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    You do not state whether the 120 V was applied to those components connected in series or parallel|dw:1436304451073:dw| Would it make a difference?

  16. anonymous
    • one year ago
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    I am assuming that they are connected in series.

  17. radar
    • one year ago
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    I believe IrishBoy123 in his post has given you the direction you must ttavel. Calculate the reactance at 1,000 Hz if they are equal then consider resonance the mode, if not then the diagram that was provided by IrishBoy123 could be used using the net reactance.

  18. radar
    • one year ago
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    Then work under that assumption, what did you get for \[X _{L}\]\[X _{C}\]??

  19. radar
    • one year ago
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    The inductive reactance at the 1000 Hz source frequency would be around 188.5 Ohms

  20. anonymous
    • one year ago
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    This question was not meant to be overly difficult, I don't believe that my instructor would have meant for them to be calculated in parallel.

  21. radar
    • one year ago
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    For the capacitance reactance, I calculaten approximately 4 Ohms. The circuit is definitely not resonance. Series is fine. so the circuit is predominately inductive for a series circuit

  22. radar
    • one year ago
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    |dw:1436307500401:dw|

  23. IrishBoy123
    • one year ago
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    |dw:1436307866338:dw|

  24. radar
    • one year ago
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    |dw:1436311256873:dw|

  25. radar
    • one year ago
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    According to my calculator R=236.56 Ohm, @eabollich What did you get for R?

  26. radar
    • one year ago
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    PF =Real Power (PW)/Apparent Power (PA PF = (.4^2)(236.5)/(120)(.4) PF = 37.84/48 = 0.789

  27. radar
    • one year ago
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    This value is too much power factor for a lot of industry, many attempt to keep it higher than .8. Correction is usually "capacitor banks" compensation. This is because the biggest contributor is electric motors with their inherent inductive reactance.

  28. anonymous
    • one year ago
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    That's the resistance I got as well. Thank you both for your help! I figured out the second part of the question as well, it is asking for the phase angle, phi, which is the inverse cos of the PF.

  29. anonymous
    • one year ago
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    Are you saying that a lower number is more power factor?

  30. radar
    • one year ago
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    Power factor of 1 is perfect where the apparent power is all real power. Power factor of 0 is no real power and total reactive power. The reason PF of 1 is desirable is there is no reactive power, while reactive power is simply exchanged between source and load and it would not register on a watt hour meter, the additional currents flowing on the transmission lines, even though the resistance of the lines or low, there will be I^2R losses which the power utility does not like, and will encourage the user to clean up his act by keeping PF close to 1.

  31. anonymous
    • one year ago
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    Great! I think I understand it better now!

  32. radar
    • one year ago
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    Good luck with your studies. PF is also the Cosine of the angle between line voltage and line current. When they both are in phase (0 degrees) Cosine is 1. They will be in phase when the load is pure resistance and no reactance.

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