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Finding RLC from Freq and Impediance (AC Ckt)

  1. Apr 22, 2008 #1
    [SOLVED] Finding RLC from Freq and Impediance (AC Ckt)

    1. The problem statement, all variables and given/known data

    1. Determine the values of R, L and C in an AC circuit to satisfy the following criteria:
    (i) the impedance at resonance is 5 k Ohms
    (ii) the resonant frequency is 60 Hz (USA standard)
    (iii) the voltage lags the current in the circuit by 2.8 degrees at a frequency of 50
    Hz (the European standard).
    (iv) Draw the phasor diagram for the impedance at 50 Hz.

    2. Find the impedance and phase angle for this circuit at f = 1 kHz. Then draw the impedance diagram at this frequency.

    2. Relevant equations

    R: Xr=R
    L: Xl=wL

    3. The attempt at a solution

    Well i know that Frequency = w/2(pi) so i can plug in 60hz=w/2(pi) to get a angular frequency=376.99 rad/s and since w=2(pi)/T i can get a time of .0166seconds...

    but i get stuck after that point, cause i can't find equations that relate any more variables without the others.
    Last edited by a moderator: Apr 23, 2008
  2. jcsd
  3. Apr 22, 2008 #2
    after speaking with a friend, i realize that at resonance the impedance is the resistance, so R=5000ohms, but that still doesn't give me much in solving for L or C...
  4. Apr 22, 2008 #3


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    Hi i4k20c,

    Looking at parts ii and iii: What is the expression for the resonance frequency, and the expression for the phase angle between the current and voltage in the AC circuit? Both of those quantities depend on L and C.
  5. Apr 23, 2008 #4
    Can someone please help me with this same problem!? I have no clue how to do it!
  6. Apr 23, 2008 #5


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    Hi hinuqueen88,

    In post #2 it was shown how the resistance was found. (At resonance the inductive reactance [itex]X_L[/itex] equals the capacitive reactance [itex]X_C[/itex]. Do you see how when that is true the impedance equals resistance?)

    The resonant frequency is given; what is the formula for resonant frequency? The phase angle (at a non-resonant frequency) is given; what is the formula for the phase angle? Once you have those written down, you should be able to get L and C.

    (In the criteria given in the original problem, (i) is referring to the impedance if the RLC part is attached to a USA standard voltage source with f=60 Hz; (iii) is referring to the phase angle if it is instead attached to a European standard voltage source with f=50 Hz.)
    Last edited: Apr 23, 2008
  7. Apr 23, 2008 #6
    [tex]\theta[/tex]=(tan^-1)[(Xl-Xc)/R] however using what you said in the post below, if Xl and Xc are equal at resonance than it would equal 0, and 0/Resoancne * tan-1 would equal 0; however it is not, the angle is 2.8 as stated by the problem...

    which is where i get confused! :confused:
  8. Apr 23, 2008 #7


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    When the frequency is 60 Hz, it is at resonance and Xl=Xc.

    When the frequency is at 50 Hz, Xl is not equal to Xc, and the angle is 2.8 degrees.

    So Xl at 50 Hz is not equal to Xl at 60 Hz (and the same for Xc). Xl and Xc depend on what frequency you are using so they change when you change the frequency.
  9. Apr 23, 2008 #8
    sorry for the multitude of questions, i want to say i truly appreciate your help.

    the equation for resonant frequency is fr=[tex]\omega[/tex]/2(Pi) and the equation for thetha==(tan^-1)[(Xl-Xc)/R]

    so would i be plugging in 2.8=(tan^-1)[(Xl-Xc)/5000ohms]?

    and i know that Xl=[tex]\omega[/tex]L and Xc=1/[tex]\omega[/tex]C

    so either way i feel like i am left with two unknowns in one equation.
  10. Apr 23, 2008 #9


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    Hi ik420c,

    The relationship [itex] \omega = 2 \pi f[/itex] is true for any frequency. But the resonant frequency is

    \omega_r = \frac{1}{\sqrt{LC} }

    (To see why this is true, remember that at resonance [itex]X_L = X_C[/itex]. But [itex]X_L=\omega L[/itex] and [itex]X_C = 1/(\omega C)[/itex]. Setting these equal gives:

    \omega L = \frac{1}{\omega C} \Longrightarrow \omega^2 = \frac{1}{LC}\Longrightarrow \omega = \frac{1}{\sqrt{LC}}

    That should give you two equations in two unknowns. What do you get?
  11. Apr 23, 2008 #10
    hmmm. okay, from that i get something like this.

    [tex]\omega[/tex]L=1[tex]/\omega[/tex]L => [tex]\omega[/tex]=1[tex]/\sqrt{LC}[/tex] = 60/2(pi) so LC=(2pi[tex]/60[/tex])^2 and rewriting it in terms of L=(2pi[tex]/60[/tex])^2 * (1[tex]/C[/tex]

    We also know that
    [tex]\vartheta[/tex]=(tan^-1) [2(pi)(50)]-([1/(2(pi)(50))]C)[tex]/R[/tex]

    so plugging in our unknowns i should get

    (5000[tex]\Omega[/tex])tan(2.8)= [100(pi)L]-[1/(100(pi)(C))][tex]/R[/tex]

    and plugging in subsitution for L, i get...
    (5000[tex]\Omega[/tex])tan(2.8)= [100(pi)(((2)(pi)/60)^2)(1/C)]-[1/(100(pi)(C))][tex]/R[/tex]

    and when solving for that i get a quadratic that looks like 0.0314C^2+244C-3.44=0 giving me a C=729.66... however when i plug this in to get a L=1.50x10^-5 and plug both my L and C into the theta equation to check my work and see if i get 2.8, i don't =(
  12. Apr 24, 2008 #11


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    I don't think that should give you a quadratic (by the way the R on the right side should not be there since you have already moved it to the left). Your equation with numerical values is:

    5000 \tan(2.8) = 100 \pi \frac{4 \pi^2}{60^2}\frac{1}{C} -\frac{1}{100\pi} \frac{1}{C}

    You can factor out a 1/C from both terms on the right side:

    5000 \tan(2.8) = \left(100 \pi \frac{4 \pi^2}{60^2} -\frac{1}{100\pi}\right) \frac{1}{C}

    What do you get for C and L?
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