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Capacitance required to continue in a power lapse

  1. Feb 1, 2005 #1
    A capacitor is often used in electronics to keep energy flowing even if there is a momentary loss of power from the electric company. What capacitance would be required to a 150-watt televisiion (plugged into a standard 120 volt ac outlet) to protect it from a .1 second lapse in power?

    I have done fine with every other question in this section but this one doesnt seem to be going as well. Im just curious how to begin. I know..

    C=E0 x A/d (E0=permitivity of free space, 8.85x10^-12)
    The time = .1 s

    I just dont see how it all goes together
     
  2. jcsd
  3. Feb 2, 2005 #2

    Curious3141

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    The energy released from a discharging capacitor that starts out with voltage V between its plates and continues to full discharge is given by : [tex]E = \frac{1}{2}CV^2[/tex]. This is the total energy released by the discharging capacitor. The energy needed to tide the TV over the lapse time is given by [tex]E = Pt[/tex] where P is the power and t is the time. Equate the two and figure out C.
     
  4. Feb 2, 2005 #3
    Thanks that makes a lot of sense
     
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