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The problems about power factor changed in RCL circuit, help~

  1. Oct 2, 2012 #1
    R = 100Ω, L=0.5H, supply voltage vs(t) = 12sin(377t) in Figure 1a

    XL = wL =377(0.5) = 188.5Ω
    vrms = 8.485∠0
    Z=213.38∠62.05 Ω
    I = 8.485∠0/ 213.38∠62.05 = 0.0398∠-62.05 A
    VR = 3.98∠-62.05 V
    VL = 7.502∠27.95 V
    PT = 0.03982(100) = 0.158W
    QT = QL = 0.299 VAR
    ST = 0.338∠62.05 VA
    Fp = 0.467


    a capacitor is added to the circuit in series in figure 1b, and Fp = 1
    S=0.158∠0
    Q=0
    PT=0.0346W
    I=0.0186A
    Z=455.7Ω
    C=14.1μF

    If Fp in figure 1b is equal to 0.8
    Q=0.02595VAR(ind)
    I=0.0051∠36.87 A
    Z=1662.82∠-36.87 Ω

    For the capacitance of here, I'm so interrogative.

    If I used the method of QT =QL - Qc
    Qc = 0.0049-0.02595 = -0.02105VAR
    Xc=809.3Ω
    But I used the method of Vs=Vc+VR+VL
    Vc = 8.485-0.0051(100)-0.0051(188.5)=7.01V
    Xc=1374.5Ω

    Why the answers are different when I used these two methods, and it's correct all of my above answers? If some are wrong, please feel free to indicated the mistake. The Figure 1a and 1b are uploaded the files.
    Thanks
     

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  2. jcsd
  3. Oct 2, 2012 #2

    gneill

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    Staff: Mentor

    Your calculations for the required capacitor in Figure (a) appear to be fine.

    For Figure (b), what are the given component values and conditions? Are the voltage source, resistance, and inductance the same as before?
     
  4. Oct 2, 2012 #3

    Yes, They are same as before, and then step and step add the capacitor in the circuit and change the power factor from 1 to 0.8.
    Thanks
     
  5. Oct 2, 2012 #4

    gneill

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    Staff: Mentor

    Just to be clear, the components start out as Vs, R, L, and C from part (a) where the power factor is 1, and you wish to add another capacitor to bring the power factor down to 0.8?
     
  6. Oct 2, 2012 #5
    The complete question is uploaded in the file.
    Thanks a lot.
     

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  7. Oct 2, 2012 #6

    gneill

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    Okay. Having the complete question helps.

    For part (c)(iv), finding the capacitance, if you know the power factor and the resistance (the real part of the impedance), then you should be able to find the corresponding reactance. After all,
    $$ pf = cos(\theta) = \frac{R}{\sqrt{R^2 + X^2}}$$
    X will be positive or negative depending upon whether the current is said to be lagging or leading. You can then find XC from X = XL + XC.

    I haven't been able to follow how you've approached parts (i) through (iii). I think if it were me I would have been tempted to do the parts in reverse order :smile:
     
  8. Oct 3, 2012 #7
    QT = QL-Qc
    Because all I are same in the in series circuit.
    XT=XL - Xc
    QT/I2=188.5-Xc
    Xc = 1186.19Ω

    another method,
    |Z|=1662.82
    |Z|2 = R2+Xc22
    1662.822 = 1002+Xc22
    Xc2=1659.8Ω
    QT=I2(1659.8)
    QT=0.0432VAR(cap)
    Qc=QL-QT
    Qc=0.0049+0.0432=0.0481VAR(cap)
    Xc=0.0481/0.00512=1849.3Ω


    There are also difference between the answers, why?
     
  9. Oct 3, 2012 #8

    gneill

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    Staff: Mentor

    Is Q a positive or negative value if the pf is leading?
    I have suspicions about your value for the impedance. Since you know the real part R is 100 Ohms, you should be able to find the magnitude of the impedance given the power factor. You should also be able to find the magnitude of the reactive part.
     
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