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Parallel LRC circuit analysis help

Discussion in 'Electronics Homework Help' started by spartan09, Apr 16, 2014.

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  1. spartan09

    spartan09

    1
    0
    Apr 16, 2014
    Hello I have a parallel RLC circuit and I am struggling to find the following parameters:

    1) Total Circuit current
    2) phase angle relative to the supply
    3) Total Circuit Impedance using j operation (complex notation)
    4) Power dissipated in the resistor

    Link to Circuit diagram:
    http://i695.photobucket.com/albums/vv313/patel5557600/kjadsbcksjdb_zps4027ce59.png
    II have also attached a simple circuit diagram in this thread

    My workings so far:

    V=12V peak= 12/√2=8.49 V_RMS
    f=10KHz
    R=10Ω
    L=150μH
    C=330nF

    1 - Circuit Current

    I_R=V/R=8.49/10=0.849A
    I_L=V/X_L =8.49/9.42=0.901A
    I_C=V/X_L =8.49/48.23=0.176A

    I= √(〖I_R〗^2+(I_L - I_C )^2 )
    I= √(〖0.849〗^2+(0.901-0.176)^2 )
    I= 1.116A

    2 - Phase Angle

    〖G=〗⁡〖1/R〗
    〖G=〗⁡〖1/10〗
    〖G=〗⁡〖0.1 s〗
    〖θ=tan〗^(-1)⁡〖G/Y〗
    〖θ=tan〗^(-1)⁡〖0.1/0.132〗

    θ=⁡〖37.15〗

    3 - Impedance Using J operator

    1/Z=√((1/R)^2+(1/X_L -1/X_C )^2 )
    1/Z=√((1/j10)^2+(1/(-j9.42)-1/j48.23)^2 )
    1/Z=√(((1∠0)/(10∠90))^2+((1∠0)/(9.42∠-9... )
    Y=1/Z=√((0.1∠-90)^2+((0.106∠-90)-(0.020... )
    Y=1/Z=√((0.1∠-90)^2+((0.106∠-90)-(0.020... )

    4 - Power Dissipated

    P=VIcosϑ
    P=(8.49)(1.116)cos⁡(37.15)
    P=7.56W



    Can someone please confirm these results? If I have gone wrong, could someone please correct me. Any help would be much appreciated.

    Thank you
     

    Attached Files:

  2. duke37

    duke37

    5,364
    772
    Jan 9, 2011
    This is partly beyond me but 4. is wrong.
    There is 8.49V rms across the 10R resistor. The other compnents do not come into it since they do not dissipate power.

    Also in 1. Ic is given as a function of Xl. I presume this is a typo.
     
  3. Arouse1973

    Arouse1973 Adam

    5,177
    1,093
    Dec 18, 2013
    Look very good. Only thing I can see that might be wrong is you state R as being j10 as being an imaginary number. Shouldn't it be just 1/10. The average power in the coil and capacitor if ideal components will be zero. The phase angles are +90 and -90 for ideal inductor and capacitor that being the voltage and current. The resistor does not have a phase.
    Adam
     
  4. Laplace

    Laplace

    1,252
    185
    Apr 4, 2010
    My calculation gives the same total current as above.
     

    Attached Files:

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