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RLC circuit question

  1. May 26, 2014 #1
    1. The problem statement, all variables and given/known data

    In an RLC series circuit, the rms potential difference provided by the source is V = 150 V, and the frequency is f = 220/pi Hz. Given that L = 0.7 H, C = 70 µF, and V_R = 80 V, find:

    a) I (rms);
    I = A

    b) R;
    R = Ω

    c) V_L (rms);
    V_L = V

    d) V_C (rms).
    V_C = V

    2. Relevant equations

    X_L=WL_i , X_C= 1/WC ; w= 2(pi)(f) ,

    3. The attempt at a solution

    A part from the equations listed above I don't really know what I need. I am kinda at lost with RLC circuits. my teacher taught it really quickly and it really confuses me. I know
    I_rms = I (peak) / sqrt(2) but I'm not even sure how to get the peak current.

    Any help to lead me in the right direction would be appreciated!
     
  2. jcsd
  3. May 26, 2014 #2

    gneill

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    Staff: Mentor

    What about the basic circuit laws and methods? You know, like KVL, Ohm's Law, etc?
     
  4. May 26, 2014 #3
    Well I do know them, ( I don't know if you mean Kirchoff law by KVL?). I haven't been taught to apply them in RLC circuits. I know it sounds dumb because they never change ( i.e:voltage will be the same between 2 parallel conductor and so on). I am not very good with electricity, I try my best but it takes time to make sense in my head.

    A part from that, how would I be able to apply these law for my current if the latter alternates?
     
  5. May 26, 2014 #4

    gneill

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    Staff: Mentor

    Have you been taught about impedance (complex valued "resistance") yet, or only reactance for inductors and capacitors?
     
  6. May 26, 2014 #5
    I do know the formula for impedance, as in z= sqrt(R^2 (X_C+X_L)^2) but not the meaning of it or when to use it.
     
  7. May 26, 2014 #6

    gneill

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    Okay, that's actually a formula for the magnitude of the impedance of a series RLC connection, which will work for you just fine. Impedance is actually a complex-valued analog to resistance, but like any complex value it has a magnitude.

    You need to be a bit careful about the signs you assign to the two reactances; in a series circuit capacitive reactance and inductive reactance work against each other when they are summed.

    You can use that impedance magnitude like you would a total resistance to determine things like current magnitude. Hint: you might try using the voltage divider rule to find an expression for the voltage on the resistor (which you've been given a value for).
     
  8. May 26, 2014 #7
    Would it be weird if I said I haven't heard of the voltage divider rule? Plus, now if I understand, can I take the impedance magnetude ( in ohms) and use V = IR?? Or is that completely wrong??
     
  9. May 26, 2014 #8

    gneill

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    Staff: Mentor

    Suppose you have a battery and two resistors in series (Say E, R1, and R2). Derive an expression for the voltage across R2. That will show you the form of the voltage
    divider rule. It's very handy indeed and something well worth becoming familiar with.

    Yes, Ohm's law applies for the magnitudes of the voltage, impedance, and current.
     
  10. May 26, 2014 #9
    ooooohh VDR would be R_2/R_T * V_T = V_R2!! And can this rule be applied to find X_L and X_C?
    Also, I have few more questions however... you said above

    "You need to be a bit careful about the signs you assign to the two reactances; in a series circuit capacitive reactance and inductive reactance work against each other when they are summed. "

    Does that mean in my impedance formula I'd have X_C-X_L?
     
  11. May 26, 2014 #10

    gneill

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    It can. Although first you need to determine R (the resistor value). Your impedance magnitude expression serves as the "R_T" of the voltage divider expression in this case.

    Yes. Or ##X_L - X_C##. Either way, since the result is squared so the sign of the difference doesn't matter in this case.
     
  12. May 26, 2014 #11
    And going back to the initial question, for an average i, I can use the average voltage but I don't have my resistance, even using the voltage divider rule I can't find it because I still have 2 unknown ( Z and R_3). I am stuck there because I tried to use V_rms = I_rms * Z
     
  13. May 26, 2014 #12

    gneill

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    You're given the voltage across the resistor. You're given the supply voltage. You have an expression for the total magnitude of the impedance which includes the R in it. Use the voltage divider rule to write an equation and solve for R; it doesn't involve I as a variable at all.
     
  14. May 26, 2014 #13
    Well I get 327 ohms for my R_3 by doing this:

    R3 = (VR3/VT) √(R32+(XC+XL)2)

    Which gave me R3 and to find my Irms = Vrms / Z which I got using my impedance formula, I got Irms = .33 A. It is still wrong though, I don't know if someone can tell me if its a calculation mistake or I got the wrong step?
     
  15. May 26, 2014 #14

    gneill

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    I'm not sure how you arrived at that formula for ##R_3##, and again, beware of the sub-expression for the total reactance... the reactances don't just add. How did you derive the expression?

    The results you're getting for the resistance and current don't look right to me.
     
  16. May 26, 2014 #15
    They aren't right ( according to my assignment).

    I derived the formula for R3 from the voltage divider rule ( the one I derived earlier) where (R2/ RT) * VT=VR2

    I just replaced RT with Z because my impedance is the total resistivity of the circuit ( unless I misunderstood). So because I have XL and XC in my circuit, respectively inductance reactance and capacitive reactance (I think those are the terms?), then I can find my Z but because I am lacking R3 I thought I could use the voltage divider rule to find my R3 as I had its respective voltage
     
  17. May 27, 2014 #16

    gneill

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    The term "resistivity" has a specific definition and doesn't fit here. Z is the magnitude of the impedance. No, you've understood correctly. But something has gone wrong with your implementation of the plan :smile:

    You don't have Z because your R3 is part of it and it is unknown. What you do have is the voltage across is though.

    Suppose XL and XC are the reactances of the inductor and capacitor respectively, R is the unknown resistor, VR is the given voltage across VR, and E is the source voltage driving the circuit. Then the magnitude of the impedance of the series circuit is
    $$Z = \sqrt{R^2 + (XL - XC)^2}$$
    Note that R appears in that expression. Note also the use of the DIFFERENCE between the reactances; inductive and capacitive reactances CANCEL each other.

    Use the above expression for Z as the total impedance for your voltage divider.
     
  18. May 27, 2014 #17
    So I got it right but I used the ratio Vr / R3 = Vt/Z after realizing that it is proportional. I don't know if that's what you meant by voltage divider rule but it kinda made me think of this. Thank you!!
     
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