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anonymous

  • one year ago

pls help me in this. A 60 V peak full-wave rectified voltage is applied to a capacitor-input filter.? If f= 120Hz, RL = 10Kohms, and C= 10µF. what is the ripple voltage?

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  1. anonymous
    • one year ago
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    the answer should be 5.0 V

  2. anonymous
    • one year ago
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    i've tried to solve but i cant get the right answer

  3. KenLJW
    • one year ago
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    T = 1/f : TC = RL C ripple = (T/TC) Vp This gives you 5 V ripple, leave you to derive theory, please show

  4. anonymous
    • one year ago
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    \[V_r = \frac{V_m}{2fRC} \] substitution of the given values leaves me with 2.5 V. if it was half wave it would have been 5 V. I don't understand why it is 5.

  5. KenLJW
    • one year ago
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    I suspect the f in your equations is the frequency unretified, 60 HZ The 2 signifies fullwave.

  6. anonymous
    • one year ago
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    It is 5 v. RF=1/4*1.7328fcr and Vdc=Vrms/rf

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