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Joules and watts

  1. Feb 10, 2015 #1
    How can a microwave I took apart have a .9 microfarad capacitor rated for 2100v? The microwave is rated at 1800 watts. What am I missing?
    1/2 CV*2=1.98 joules
     
  2. jcsd
  3. Feb 10, 2015 #2

    mfb

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    Staff: Mentor

    The capacitor won't be able to power it for a long time then (can you calculate for how long?), but that was never its purpose.
     
  4. Feb 10, 2015 #3
    I don't know why do you calculate capacitor energy. Do you know how a microwave power supply works?
     
  5. Feb 10, 2015 #4
    I don't know the formula for such a calculation. How can it draw more power then the capacitor is able to provide? The power supply isn't charging to higher voltages than it's rated for.
     
  6. Feb 10, 2015 #5
    I don't understand what you're asking me.

    A microwave power supply works by providing a high voltage ac current to a doubler circuit. It then feeds the rectified current into a magnetron. Im definitely not the one to be answer question of the sort though.
     
  7. Feb 10, 2015 #6

    Drakkith

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    Total energy stored in a capacitor is not the same thing as the power it is able to provide. I assume the purpose of this capacitor is to help smooth the output DC current/voltage? If so, then it's constantly being charged and discharged. The current and voltage during capacitor discharge should be nearly the same as they are at the peak of the AC signal (unless you have a poor quality filter circuit) so the output power should remain relatively constant with only a small ripple. Having too small of a capacitor for your output would greatly increase the ripple effect.

    I'd provide formulas for determining all this, but I don't happen to know them. Looking around on google did net me the following link though. Maybe it can help?
    http://scholarcommons.usf.edu/cgi/viewcontent.cgi?article=4856&context=ujmm

    I think the key point here is that the capacitor is only discharging for a short time before being charged again. If sized correctly, the drop in voltage and current between the start and end of discharge is small, so output DC power is relatively constant and equals the input power (ignoring losses).
     
  8. Feb 11, 2015 #7
    Then, if you know how it works, I don't know what is bothering you...
     
  9. Feb 11, 2015 #8
    I thought it was a power capacitor; that is what was bothering me. Drakkith already helped me, but thanks anyway.
     
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