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RLC Parallel circuit

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi
    Can a kind person please check my solution? This is the question: Demonstrate the total current in an RLC parallel circuit using the following circuit. I cant draw it but the inductive reactance is 20ohms, capacitive reactance is 5ohms and resistor of 10ohms. All are in parallel. The voltage is 5v in series with 20ohms inductive reactance.


    2. Relevant equations



    3. The attempt at a solution
    This is my attempted solution:
    Resistive Branch: I = E/R = 5/10 = 0.5A
    Capacitive Branch : E/Xc = 5/5 = 1A
    Inductive Branch = E/XL =5/20 = 0.25A
    Resultant current, Ix = 1 - 0.25 = 0.75A

    Total current = (I^2R + I^2X)^1/2
    = (0.75^2 + 0.5^2)^1/2
     
  2. jcsd
  3. Mar 28, 2012 #2

    gneill

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    Staff: Mentor

    Hello aurao2003.

    Is this more or less what your circuit looks like?

    attachment.php?attachmentid=45639&stc=1&d=1332944855.gif

    I suspect that the question wants to know the value of the current I as produced by the voltage source.

    Note that since the inductor is in series with the voltage source there will be a potential drop across that inductor before the current reaches the paralleled R & C.

    Do you know how to work with impedances (complex numbers)?
     

    Attached Files:

  4. Mar 28, 2012 #3
    Yes. That's what my circuit looks like. I have done complex numbers in my maths class.

    Is my solution on track? Please let me know.
     
  5. Mar 28, 2012 #4

    gneill

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    Staff: Mentor

    Alas, your solution is not "on track" because you have neglected the potential drop that must occur across the inductor -- the total current passes through the inductor before it ever reaches the capacitor or resistor.

    Since you have experience with complex numbers, treat the components if they were plain resistors but with values as shown in the diagram I supplied. What would be the total "resistance" as seen by the voltage source? This "resistance" is the circuit's impedance.

    You can find the total current if you have the voltage supplied and the impedance. So first find that impedance.
     
  6. Mar 28, 2012 #5
    Thanks for your help. I mislead you about the diagram. The inductor, capacitor and resistor are all in parallel with the supply voltage at the bottom of the circuit. Does this clarify matters? I apologize for not presenting a diagram.
     
  7. Mar 29, 2012 #6

    gulfcoastfella

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    Gold Member

    You didn't provide a frequency for the voltage source. Is the voltage DC or AC?
     
  8. Mar 29, 2012 #7

    gneill

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    Staff: Mentor

    Okay, no problem. So the "new" arrangement looks like this:

    attachment.php?attachmentid=45673&stc=1&d=1333022658.gif

    In this case your original method will work fine. This is because all three passive components are being driven by the same voltage, so you can determine the individual branch currents quite easily. Your answer looks fine.

    Note that the complex impedance method would work too (in fact it always works).
     

    Attached Files:

    Last edited: Mar 29, 2012
  9. Mar 29, 2012 #8
    Thanks a million. You are amazing! I might some other. Hope it's okay to let you know.

    Cheers!
     
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