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    Hey guys, I've come across this questions and although a shady method gives me the correct answer, I would appreciate it if anyone can provide a detailed working.

    "Mains-operated power supplies have large-value capacitors to help keep the output voltage constant. In a full-wave rectified supply, without a capacitor the output falls to zero every 10 micro seconds. If the output must be maintained at a value at least 90% of its maximum value when a load of 1.0 Kilo Ohms is connected to the output terminals, show that the minimum capacitance needed in the power supply circuit is about 100 micro Farads."

    Quick answers, even if brief will be wholly appreciated.
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    (Original post by Aahmsil)
    Hey guys, I've come across this questions and although a shady method gives me the correct answer, I would appreciate it if anyone can provide a detailed working.

    "Mains-operated power supplies have large-value capacitors to help keep the output voltage constant. In a full-wave rectified supply, without a capacitor the output falls to zero every 10 micro seconds. If the output must be maintained at a value at least 90% of its maximum value when a load of 1.0 Kilo Ohms is connected to the output terminals, show that the minimum capacitance needed in the power supply circuit is about 100 micro Farads."

    Quick answers, even if brief will be wholly appreciated.
    Let's see your "shady method". It may be ok.
    If not we can suggest a better one.
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    (Original post by Stonebridge)
    Let's see your "shady method". It may be ok.
    If not we can suggest a better one.
    Okay, I just simply equated this:

    e^(-t/RC) = 0.9

    And solved for capacitance, C. Is that alright? Is there any other way?
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    (Original post by Aahmsil)
    Okay, I just simply equated this:

    e^(-t/RC) = 0.9

    And solved for capacitance, C. Is that alright? Is there any other way?
    Yes. That's how to do it.
    V=Voe(-t/RC)
    And you need the value of C such that in a time t the voltage V can be no less than 90% of Vo
    So V/Vo = 0.9 = e(-t/RC)
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    (Original post by Stonebridge)
    Yes. That's how to do it.
    V=Voe(-t/RC)
    And you need the value of C such that in a time t the voltage V can be no less than 90% of Vo
    So V/Vo = 0.9 = e(-t/RC)
    Thank you very much for your time!
    Much appreciated.
 
 
 
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