rms voltage..

Discussion in 'General Electronics Chat' started by sndpgr, Dec 28, 2007.

  1. sndpgr

    Thread Starter Member

    Jun 22, 2006
    Quite an embarrassing question really.
    A DC through a resistor R will dissipate V^2/R watts.
    when a ac signal v(t) will be applied to same resistor R what will be the power dissipated,I know it will be V(rms)^2*R and I know the value of V(rms).but how do you arrive at that.
    kindly help!
  2. recca02

    Senior Member

    Apr 2, 2007
    see if this helps.
    also go thru this if you have the time.
    and then its AAC we are talking about.
  3. Audioguru


    Dec 20, 2007
    The average power from RMS AC at a certain voltage is exactly the same as power from DC at that same voltage.
    The proof is in a heater. Or a big lightbulb that is slow enough not to flicker with AC.
  4. beenthere

    Retired Moderator

    Apr 20, 2004
    The whole purpose of using RMS to measure AC is to make AC and DC power calculations exactly equivalent.
  5. thingmaker3

    Retired Moderator

    May 16, 2005
    I hate to be nit-picky, but there is a difference between "average power" and "RMS power." Just as there is a difference between "average voltage" and "RMS voltage"

    But, yes, RMS and DC will yield the same numbers for power, voltage, and current.
  6. Distort10n

    Active Member

    Dec 25, 2006
    To be strictly technical, "average power" implies RMS voltages and currents. "RMS Power" is a misnomer by many engineers:


    It is also fun to read the arguments on the Wikipedia "Talk" page about Audio Power:


    But you are correct, in regards to average or RMS voltage or current waveforms. Using a single cycle sine wave any any given frequency the RMS value is equal to 0.707 of the peak value, while the average is equal to 0.634 (I believe) of the peak value.
  7. sndpgr

    Thread Starter Member

    Jun 22, 2006
    thank's for the encouraging replies.
    This is what I was I was looking for-
    'The heating effect depends on I2R, and so an average of I2 is needed and not an average of I. ' from the website you gave .

    Thank you!again!
    the average square voltage is then V(t)^2 integrated on a time period and divided by time period.
    Am I on right direction.
    waiting comments!

    thank you! and again!