What is rms power?

  1. 1. The problem statement, all variables and given/known data
    What is the meaning of rms power? Is it useful?


    2. Relevant equations
    Prms=Pmax / √2


    3. The attempt at a solution
    I' m not very sure about the equation I write above. Is that correct?

    I am learning about AC. I find that the average current is the same as Irms and average voltage is the same as Vrms, but average power is not the same as Prms.

    Average power is the value of electrical energy received by a load every second. And what about Prms?

    Thanks

    EDIT: Oh my god, I posted in the wrong place. Please move my post to the correct place. I am sorry
     
  2. jcsd
  3. LCKurtz

    LCKurtz 8,386
    Homework Helper
    Gold Member

    For a sinusoidal voltage and current

    Vrms is defined to be the constant voltage that supplies the same power over a single period to a resistance R as the sunusoidal voltage:
    [tex] P = \frac{V_{rms}^2}{R}[/tex]
    and you have the same type of definition for the current:
    [tex]P = I_{rms}^2R[/tex]
    If you multiply these two equations by each other you get:
    [tex]P^2=\frac{V_{rms}^2}{R}\cdot I_{rms}^2R = V_{rms}^2I_{rms}^2[/tex]
    [tex]P = \sqrt{V_{rms}^2I_{rms}^2}=V_{rms}I_{rms}[/tex]
    Does that answer your question?
     
  4. BruceW

    BruceW 3,597
    Homework Helper

    I think you got a bit confused here. The average current is not the same as Irms and the average voltage is not the same as Vrms. Usually, in AC electricity the average voltage is zero, since half the time it is negative and half the time it is positive, the average voltage is zero.
    Vrms gives an idea of a typical value you might get for the absolute value of the voltage if you measured it. (Absolute meaning unsigned). Vrms=Vmax/√2 For a sinusoidal wave. So for a sinusoidal wave, a typical value for the unsigned voltage is roughly 0.7 times the max voltage.
    The power depends on the voltage, so it changes all the time, but it is always positive. The average power is equal to the max power divided by 2. This is also equal to the power when the voltage is equal to Vrms.
    In other words, if we replaced the voltage at all times with Vrms, the average power would remain unchanged.
    I don't know what you mean by Prms. Do you mean the power given by a voltage equal to Vrms? In this case it is just the average power. Or do you mean the squared variance of the power output?
     
  5. HallsofIvy

    HallsofIvy 40,673
    Staff Emeritus
    Science Advisor

    "RMS", "root-mean-square" is (almost) exactly what it says. Given a list of numbers, Find the mean (arithmetic average) of the squares of the numbers, then take the square root of that. It is one of many different kinds of "averaging".
     
  6. BruceW

    BruceW 3,597
    Homework Helper

    That's not quite right. Its defined as:
    [tex]\sqrt{\bar{x^2} - {\bar{x}}^2}[/tex]
    (Where [itex]\bar{z}[/itex] says 'take the mean of this set of numbers denoted by z').
    (As long as the list of numbers is much greater than one).
    So I guess you're right as long as the mean of the numbers is equal to zero.
     
  7. I think that the term RMS is required only for quantities which can take both positive and negative values. Since power is always positive, what is the use of RMS power?
     
  8. BruceW

    BruceW 3,597
    Homework Helper

    Not necessarily. The RMS gives an idea of how much something varies. So the RMS of the power would tell us how much the power varies by. But I don't see why that would be important in circuits. I would have thought only the average power was important...
     
  9. I see the thread hasn't been moved to physics so I assume it's okay to continue the discussion here :biggrin:

    Unless I'm mistaken, I think your answer doesn't answer my question. You explain how to get the formula of average power in terms of Vrms and Irms but my question is asking about Prms not Paverage. What is the actual meaning of Prms in electrical circumstances? Is Prms useful? Do we use Prms to understand or calculate something?

    Ok, now I understand average current is not the same as Irms. No, I don't think Prms is the power delivered by Vrms, it should be average power in my opinion. My teacher told me that average and rms power are not the same. That's the reason I'm looking for explanation about it. I want to know what Prms is and whether it is useful or not.

    Maybe I didn't phrase the question good enough. I know how to find the rms value using the method you explain. I just want to know the meaning of Prms in electrical circumstances. Based on what you said, rms is one of many different kinds of averaging, it means that we have two values of 'average' power, which are Paverage and Prms. What is the difference of the two average power? We can use average power to know the amount of energy delivered per unit time to a load. What do we use Prms for?


    I have the same opinion that average power is more useful than average power but my teacher said that Prms is more convenient and better than average power. I don't understand why. I google a bit and find a web that claims rms power is meaningless without further explanation. I'm still confused what rms power is and whether it is useful or not

    thanks
     
  10. ehild

    ehild 12,363
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    Gold Member
    2014 Award

    RMS means root-mean-square.

    The AC current is I=I0cos(wt). If it flows through a resistor, the instantaneous power is P(t) = RI2(t). When using an AC device you are interested in the average power. If the power of a heater is said to be 2000 W, it means the average power.

    The average power of a time dependent current flowing through a resistor R is

    [tex]P_{av}=1/T \int_0^T{R I^2(t) dt}=R(1/T \int_0^T{ I^2(t) dt})[/tex]

    that is, the average power is R times the mean of the squared current.
    We call the square root of the integral "root-mean-square current", Irms.

    [tex]I_{rms}=\sqrt{1/T \int_0^T{I^2(t) dt}}[/tex]

    For the AC current, [tex]P_{av}=R/T\int_0^T{(I_0cos(\omega t))^2 dt}=RI_0^2/T \int_0^T{\frac{(1+cos(2 \omega t))}{2} dt}\rightarrow P_{av}=R\frac{I_0^2}{2}=RI_{rms}^2[/tex].

    So the average power is the same as that of a DC current Irms=I0/√2

    The average power is [tex]P_{av}=RI_{rms}^2=\frac{V_{rms}^2}{R}[/tex]

    There is no use of rms power: The average power is connected to rms current and voltage.

    ehild
     
  11. Thanks for your explanation :)
     
  12. nsaspook

    nsaspook 1,133
    Science Advisor

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