RLC Circuit

Discussion in 'Homework Help' started by Evilsanta31, Dec 30, 2014.

  1. Evilsanta31

    Thread Starter New Member

    Dec 29, 2014
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    Not sure what Im doing wrong here but think im needing some assistance.

    An RLC series circuit resonates at a frequency of 10Khz with a resistor value of 4 ohms and capacitor value of 400nF. If it is connected across a 10V AC supply calculate the following:

    A. Value of the Inductor

    I used the equation L=√3R/2πF so L=√ 3x4/2xPIx(10,000/1000) and I get 0.11mH

    B. Supply current at resonance

    I=V/R 10/4= 2.5A

    C. Voltage across the inductor and capacitor at resonance
    I know these values should be equal as they cross at the resonance point but when I use the equations

    VL=IXL = I(2piFL) = 2.5x2pix(10,000x0.00011) = 17.28V
    Vc=IXc = I/2piFC = 2.5/2pix(10,000x400x1e-9) =0.0157V

    I dont get the same answers for both so I know somethings went wrong, so it must be the value of the inductor where Ive went wrong.
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Let's tackle (A) before we proceed further.

    What does it mean for a series RLC circuit to be in resonance? What happens at resonance?

    Where does the factor of 3 come from?

    Doesn't it seem odd that you don't use the value of the capacitor at all?

    What is and what is not under the square root operator? Use parens to make your expression unambiguous.

    It appears that you are doing ((√3)R)/(2πF), but I had to figure that out by trail and error. Make your expressions clear -- engineering is not about guessing.

    Why is F replaced with 10,000/1000?
     
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  3. Evilsanta31

    Thread Starter New Member

    Dec 29, 2014
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    RLC circuit to be in resonance means that Inductance and Capacitance reactances are equal in magnitude but 180 degrees apart and so cancel each other out.

    The factor of 3 I got from a text book I was reading, Electrical and Electronic principles by John Bird, seems Im using a 3 phase equation??
    I used the capacitor in the Vc=IXC equation should I be using it before then?

    I will make my expressions more clear, but seems my calculator skills have a lot to be desired.
    I replaced F with 10,000/1000 as thats the way it was in the John Bird book possibly changing the frequency from KHz to Hz.

    My tutor gives us handouts in class and reads them to us rather than actually demonstrating the equations on the board, if you ask for help he gives it one on one rather than standing at the front and telling everyone, so were all at different learning levels. Im self teaching at the minute and seem to be having an epic fail at this.
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    Okay, so set the sum of the inductive and capacitive impedances to zero (which is the same as setting their magnitudes equal to each other but having their phases 180 degrees apart). What do you get?

    Are you familiar with impedance and using complex numbers, or just reactances?

    Diving 10,000 by 1000 is converting 10,000 Hz to 10 kHz.

    Because of the way that you did things, it coincidentally has the affect of converting henries to millihenries, but it is purely by chance that that is what you wanted to do.

    Always, always, ALWAYS track your units!
     
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  5. Evilsanta31

    Thread Starter New Member

    Dec 29, 2014
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    Im not really familiar with impedance and complex numbers, Im learning to wind motors at present and I dont do this day in day out. I struggle using the calculator thats how things are with myself.

    VL=IXL = I(2piFL) so this should be I(2piF) ?
    Vc=IXc = I/2piFC and this I/2piF

    Thanks for the help, Im sure its frustrating trying to help someone with probably the basics in engineering and I will keep a track on my units.
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    Not a problem -- we were all there once.

    What should be I(2piF)? That would have to be VL/L. Is that what you want?

    So set the magnitudes of the two reactances equal to each other and then solve for L.

    <br />
X_L \; = \; X_C<br />
2\pi f L = \frac{1}{2 \pi f C}<br />
     
  7. Evilsanta31

    Thread Starter New Member

    Dec 29, 2014
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    Arghhhh I dont know what I want anymore.

    So if I find the circuit current and the inductive reactance and as VL=VC I can find VL=I x XL
     
  8. Evilsanta31

    Thread Starter New Member

    Dec 29, 2014
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    L=1/4π² x F²C
     
  9. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi,
    Why are you trying to compound everything into a single formula.?

    Look at 'WB's formula's for XC and XL, post# 6.
    You are given all the parameters to solve for Xc, do that first to find Xc, then use that value to calculate the Inductance.

    E
     
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  10. Evilsanta31

    Thread Starter New Member

    Dec 29, 2014
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    So I should be doing the Xc equation first as I have the known total for the capacitor.

    so 1/2xPi x (10,000x400x1e-9) which gives me 40V
     
  11. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi,
    What is the Xc impedance value for the 400nF capacitor at 10KHz.?

    E
     
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  12. Evilsanta31

    Thread Starter New Member

    Dec 29, 2014
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    39.79ohms
     
  13. Evilsanta31

    Thread Starter New Member

    Dec 29, 2014
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    this isnt sinking in at all, Ive been at this one question for hours and just cant grasp what im doing. Ive been at this assignment almost non stop for the last three days trying to get it finished. Its not due for another three weeks but dont want to be struggling at the last minute.
     
  14. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    OK,
    So knowing that at resonance, that XL = XC, transpose the formula Xl= 2 * pi * F * L , to find the inductance of value of L.

    What do you calc the inductance to be.?
    E
     
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  15. Evilsanta31

    Thread Starter New Member

    Dec 29, 2014
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    so I would be doing L= XC/2*pi*f

    And I would get 0.0006H or 6mH

    Please say its right.
     
  16. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi,
    Its close, I make it ~0.627milli Henry
    Are you sure that 0.0006 = 6mH....... milli means its 1000 times smaller,

    Now that you know L= 0.627mH and that both XL and XC = 39.39R at 2.5Amps, calc the Vc and Vl values..

    E
     
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  17. Evilsanta31

    Thread Starter New Member

    Dec 29, 2014
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    oops forgot about the extra zero so its 0.6mH, thanks for your help and patience, feel a lot better now its moving on. Ill do the rest of the calculations and post up here what I get and hopefully should be done. But ill be reading this over and over until it sinks in.

    Thanks to both of you.
     
  18. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi,
    Always break the problem down into bite sized chunks, don't try to create a single formula.
    In that way its easier to find a mistake,
    E
     
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  19. WBahn

    Moderator

    Mar 31, 2012
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    Yes, provided I try to figure out what you meant to write and not what you wrote.

    Remember the order of precedence -- exponentiation then multiplication/division and then addition/subtraction, but within those groups it is done left to right. This means that what you wrote:

    L=1/4π² x F²C

    exponentiation first:

    L=1/4(π²) x (F²)C

    then multiplication/division done left to right:

    L=(((1/4)(π²)) x (F²))C

    What you meant to write was:

    L=1/(4π² x F²C)
     
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  20. Evilsanta31

    Thread Starter New Member

    Dec 29, 2014
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    so that would mean VL=Vc which is I x XL = 2.5 x 39.79 = 98.475V

    Q-factor then would be Vc/Vs 98.475 / 10 = 9.8475V
     
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