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What is rms value?

Discussion in 'Electronic Basics' started by Matthew Ling, May 6, 2004.

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  1. Matthew Ling

    Matthew Ling Guest

    Hi, I am confused over the term rms.
    From an AC waveform, the average should be zero.
    how is rms value derived from the AC waveform?
    I know that in order to obtain rms, it is just dividing
    the max value by square root of 2, BUT WHY??!
    Any help is appreciated.
  2. the average value of an AC current waveform is NOT the same as the RMS value
    RMS is the effective value of the AC current...

    so... it's quite simple... say we have two circuit:
    1) an AC source connected with a resistor R in series
    2) an equivelant DC source connected with a resistor R in series

    the key point is
    the average power dissipated in R in the AC circuit should be the same as in
    an equivelant DC circuit...
    key point
    hint: <P> = <R*i^2> and i=Io*sin(wt+PS) in AC

    after hving done some math you will get something like this:
    in AC:
    <P> = (R*Io^2)/2 <-- still in AC
    so Io^2=<P>*2/R <--- AC

    in DC: ( current does not alternate in DC so the power is a constant )
    I^2 = P/R
    because Pac = Pdc
    you divide the first equation by the seconde euqation, and u get
    (Io/I)^2 = 2
    I = Io / 2^0.5
    here, I is the current in an equivelant DC circuit... it's also called the
    RMS value...
  3. RMS stands for the three mathematical operations (Root Mean Square)
    needed to calculate the equivalent DC voltage (or current) that would
    produce the same power in a resistor as the AC waveform. So, first,
    you square the waveform, then you take the average (mean) of that
    squared waveform over any time that includes an integer number of
    cycles (or long enough time to capture a representative average of all
    possible variation if the waveform is not cyclic) an then you take the
    square root of that mean.

    It is just a mathematical coincidence that if the wave is a sine the
    RMS result happens to be the peak value divided by square root of 2.
    For other wave shapes there are other results. For instance, for a
    symmetrical square wave (equal amplitude positive and negative half
    cycles) the RMS is the same as the peak.
  4. Bill Vajk

    Bill Vajk Guest

    Yes. The time averaged value of a symmetrical sine wave
    AC is zero.

    Power, however, depends on the absolute value of voltage
    and current.
    John has written a succinct explanation to you.
    For extra points work out why instead of dividing by
    the square root of two you can get the same result by
    multiplying by 1/2 the square root of two (aproximately
    ..707--close enough for most work.)
  5. JeffM

    JeffM Guest

    Have you ever been paid for teaching by an institution of (higher?) learning?
  6. Only by the U. S. Army.
  7. Bob Penoyer

    Bob Penoyer Guest

    It's not simple to look at an AC waveform (like a sine wave, for
    example) and "see" how it compares to DC. It's clear, though, that DC
    will heat a resistor by delivering power to it. It's also clear that
    AC, too, will heat a resistor. POWER is what heats a resistor. So, if
    a given AC waveform heats a resistor exactly as much as a particular
    DC voltage, then we can say that their "heating effect" is the same.
    An AC voltage's RMS value is often referred to as its heating effect.

    When a DC voltage heats a resistor to some temperature, it does it by
    delivering power to the resistor continuously. When an AC voltage
    heats the same resistor to the same temperature, it does it by
    delivering the same AVERAGE POWER to the resistor. (This is true
    because power is what heats the resistor. An AC waveform's
    "instanteous" power--the power delivered at a particular instant of
    time--varies with time because its voltage varies with time. But the
    resistor's temperature tends to vary slowly, much slower than the
    variation of the AC waveform. So it's AVERAGE POWER from the AC
    waveform that we care about.)

    As John Popelish explained, "RMS stands for the three mathematical
    operations (Root Mean Square) needed to calculate the" AC voltage that
    is equivalent to a DC voltage. R, M, and S are the mathematical
    operations in reverse. First, square the voltage. This makes sense
    because voltage squared is used to calculate power.

    Next, take the mean (average) of the squared voltage. That is,
    calculate the average power.

    Finally, take the square root. This makes sense because the square
    root of power can be used to calculate voltage.

    So, by operating on an AC waveform, first by squaring, then averaging,
    then by square-rooting, you convert the AC voltage to an equivalent DC

    Here's an example:

    A sine wave has a peak value of 10 volts. What is its RMS (or
    equivalent DC) value?

    V = 10 * sin(2 * pi * f * t)

    V^2 = 100 [sin(2 * pi * f * t)]^2

    but [sin(x)]^2 = 0.5 [1 - cos(2x)] (Check a trig reference if you're
    not familiar with trigonometry.)

    so V^2 = 50 - 50 * cos(4 * pi * f * t)

    Ave(V^2) = 50 - 50 * Ave[cos(4 * pi * f * t)]

    Since the average of the cosine function over a full period is zero,
    Ave(V^2) = 50 - 0 = 50

    SquareRoot[Ave(V^2)] = SquareRoot[50] = 7.07, approximately.

    So, a sine wave with a peak value of 10 has an RMS value of 7.07. This
    means, of course, that a sine wave AC voltage with a peak value of 10
    volts will heat a given resistor exactly as much as a DC voltage of
    7.07 volts. For a sine wave, then, to find the RMS value, just divide
    the peak value by the square root of 2: 10 / sqrt(2) = 7.07.
  8. Bob Myers

    Bob Myers Guest

    John, I am SO tempted at this point, but I'm going to let
    this one pass's just too much of an easy target...:)

    Bob M.
  9. Ratch

    Ratch Guest

    ENERGY is what heats a resistor. Power determines the energy dissipation
    and ultimately the temperature rise of the resistor. Ratch
  10. POWER is the rate of energy dissipation...
  11. Bob Penoyer

    Bob Penoyer Guest

    Well, let's see. If I dump 1000 joules into a 1-watt resistor but I
    take all of February to do it, the resistor won't get very hot. But,
    if I do it over a period of 100 seconds, the resistor will get damn
    hot. Rate! Rate is important. It's power that heats the resistor.
    Hence, terms like 1-WATT resistor.
  12. i wonder if you did your lil dumping-joul experiment in VACUMM AND there is
    no heat RADIATING from that resistor!!!

    joul is the unit of energy, also the unti of heat!!!
    which means... no matter how many jouls you dump into a object...
    as long as there's no heat lost... the temp of that object will go up by a
    certain degrees...

    for the case you were talking about... it's becuase a partial heat that
    CURRENT generated is LOST in air and radiation... that's why you dont feel
    it really hot if you take a whole month to dump that 1000J into the
    for only 100s... you feel it much hotter becuase not much heat is running

    as the current thru a R goes up, the montion of free electrons becomes more
    active in order to keep that current.( remember the definition of current I?
    it's I = Q/t... )therefore, the molecules are moving in a faster rate in
    high current than in low current... thus, the collision among molecures
    occurs in a higher rate which generate more HEAT...
    so it's CURRENT that heats up the resistor... not POWER...

    furthermore, power is defined as POWER = ENERGY / T
    doesn't this look similar to VELOCITY = DISPLACEMENT / T ??

    :) have a nice day~~~
  13. Rich Grise

    Rich Grise Guest

    Ya wanna think this one through again?
  14. could you list any material that is not radiating energy when they are not
    in thermaldynamic equilibrium??
  15. Bob Penoyer

    Bob Penoyer Guest

    Heat radiates just fine from a resistor in a vacuum. If heat couldn't
    radiate through a vacuum, the sun wouldn't do us much good.
    A resistor won't be insulated under normal circumstances. So heat will
    be lost by conduction, convection, and/or radiation.
    Nonsense. As I said in my earlier post, it is rate that effects the
    heating of the resistor. If you dump energy into the resistor faster
    (i.e., apply more power), the resistor will get hotter.
    It's pretty tough to pump current through a resistor without
    encountering P = (I^2)R. Now, if I apply the same current to a 10-ohm
    resistor as I do to a 20-ohm resistor, the 20-ohm resistor will get
    hotter--even though I use the same current in both cases. It's power
    that heats the resistor, not current. That's why resistors are rated
    in watts, not amps.
    Only because they are both RATES. Other than that, I don't see a
    You, too.
  16. in vacuum and radiation are different statements... they are not a bit
    alrite... one simple Q
    do u measure heat in Jouls or in Watts?
  17. Ratch

    Ratch Guest

    Heat is the lowest form of energy. Heat/energy is measured in joules for
    International System units. Heat/energy rate of gain/loss is measured in
    watts for International System units. Ratch
  18. Stop it, you are making my head hurt...

    The amount of power radiated from a resistor is related to its
    temperature. The higher the temp, in relation to the environment, the
    faster the object will radiate energy.

    If you add energy by passing current through it, it will heat up until
    the rate of radiation = the rate of addition, ie, it will reach a
    thermal equilibrium.

    If it is unable to radiate energy for some reason (maybe the
    surroundings are kept at the same increasing temperature) the resistor
    will continue to heat up, because energy is getting added by the
    current flow.

    Thus, you are both right, it's energy that heats a resistor, but in
    the real world, its power that determines the final temperature,
    because, given a particular resistor configuration, the resistor will
    radiate at a particular rate dependent on temperature. Thus, the
    temperature will satisfy a formula something like

    I^2*R = P(T)

    where P(T) is the function of radiated power for the object in its
    environment, given a temperature T.

    Bob Monsen
  19. Thank you...
    Bob, i think you should read this...
    heat , which is a form of a energy, is measured in Jouls...
  20. Bob Penoyer

    Bob Penoyer Guest

    It was your proposition. You must have thought it made sense.

    Here's a question: Two resistors of equal power-handing capability are
    connected in series. One resistor is twice the value of the other. A
    current is passed through the resistors so that they are warm to the
    touch. Which of the following choices is correct?

    A. The larger resistor gets hotter.
    B. The smaller resistor gets hotter.
    C. The resistors are equally warm because it is current that
    determines the amount of heating.
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