half-wave rectifier with filter capacitor

Discussion in 'Homework Help' started by notoriusjt2, Oct 3, 2010.

  1. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    [​IMG]

    book states

    [​IMG]

    thats basically 6540 micro farads, but that answer was deemed incorrect when I submitted my assignment. so my question is where did i go wrong?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    782
    Your value for R of 288Ω is incorrect because you based this on the RMS source voltage - rather than the estimated DC load voltage.

    The approximate DC load voltage will be Vm-Vr/2, where Vr is the peak-to-peak ripple.
     
  3. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    so the value for Rload would be

    Rload = ((Vm-Vr)/2)^2/50
    = ((169.7-1.5)/2)^2/50
    =141.46ohms

    is this correct?
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Not quite - I wrote Vm-Vr/2 not (Vm-Vr)/2

    So Vdc would be 120*√2-0.75=168.96V
     
  5. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    Rload=(168.96^2)/50
    Rload=570.95ohms

    so

    =(120(1.414))/(50)(570.95)(1.5)
    =169.7/42821.25
    =.00396297
    =3962microF

    thats really not even close to any answers on there
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    This was my approach ....
     
  7. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    thank you for your help
     
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