RMS Vs Peak Values Part 2 - RMS 0.707 Proved

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  • เผยแพร่เมื่อ 25 ธ.ค. 2024

ความคิดเห็น • 39

  • @orras263
    @orras263 3 ปีที่แล้ว +1

    ขอบคุณมากๆค่ะ เข้าใจขึ้นเยอะเลย👍👍

  • @wi_zeus6798
    @wi_zeus6798 7 ปีที่แล้ว +4

    Good video.
    When I think of integrals I like to think of what you called "all instantaneous values" as simply the area between the function and the x-axis. It just seems to make more sense in my eyes.

    • @gugunkhongsai13
      @gugunkhongsai13 ปีที่แล้ว

      Yea but area under curves just comes from the fact that it happens to be equal to the sum of all instantaneous values.

  • @MikeMBMr
    @MikeMBMr 6 ปีที่แล้ว +4

    If you just want the average or 'mean" of a sine wave from 0 to pi there is no reason to square the sine (the value is always positive from 0 to pi versus from 0 to 2pi) . The average value of a sine wave of amplitude 1 from 0 to pi is: 1/pi times the integral of sine from 0 to pi. This yields 2 divide by pi or approximately 0.637. This differs from the RMS value by about 10%. This is because is for most wave forms you cannot square it over a period where the value changes significantly, find the average (of the squared wave form) and take the square root of the result and expect to get the average. The only periodic wave form for which the RMS value is exactly the same as the average is a pure square wave.

    • @Jpres
      @Jpres 5 ปีที่แล้ว +1

      This confused me so much previously. I saw so many people explaining that the significance of the RMS value was the average value of the sin wave when it really wasn't.

    • @uniflover5921
      @uniflover5921 4 ปีที่แล้ว

      so if you want the rms of a sine wave from 0 to 2pi this is the time should you use the squared?

    • @uniflover5921
      @uniflover5921 4 ปีที่แล้ว

      can you pls elaborate the part "This is because is for most wave forms you cannot square it over a period where the value changes significantly" thank you!

    • @johntryl8009
      @johntryl8009 ปีที่แล้ว

      @@Jpres omg same

    • @freeeeeeeeak
      @freeeeeeeeak ปีที่แล้ว

      We need to use square because we are operating with Power value (Power = U^2 / R = I^2 * R). Not pure voltage or current.

  • @ParaQueTudoIsso
    @ParaQueTudoIsso 6 ปีที่แล้ว +3

    Thank you very much this was perfect. The only think that i'm curious to know it's why to use the root in all integration? Does have any explanation or it's just the rule?

  • @AG-cx1ug
    @AG-cx1ug 3 ปีที่แล้ว

    Could we have also gone only till Pi/2 to calculate the RMS? Also why do we square sin^2(theta) @ 5:15? Thank you

  • @AG-cx1ug
    @AG-cx1ug 3 ปีที่แล้ว

    If the peak wasn't 1, and it was derived like this, then would it be (P*sintheta)^2 in the first step? And then the integral, average, square root, would that result in P coming out just as a factor = 0.707*P? Thank you

  • @muhannadalhazmi8341
    @muhannadalhazmi8341 5 ปีที่แล้ว

    i don't get why you multiply the final answers with sqrt ?

  • @mariamsoultan5891
    @mariamsoultan5891 4 ปีที่แล้ว

    Why do we multiply 0.707 by the amplitude value?

    • @itsamemario6588
      @itsamemario6588 3 ปีที่แล้ว

      Oof. 9 months later haha. So, we multiply by 0.707 for a few reasons, but the way I think about it is that the RMS voltage is A) the universal standard for specifying the AC voltage used in electrical systems (or at least residential and low / medium voltage industrial voltages), all the way from your phone charger to 480VAC hydraulic pumps. B) the RMS is the DC equivalent voltage which is useful for power calculations. If you have a motor that runs at 120VAC 1.0A, you can multiply 120 x 1 and say "alright, this thing uses 120W of power to run." If we used peak voltage, you couldn't do that! You see, the power consumption isn't constant in an AC system, right? In a purely resistive load, it's also sinusoidal. Then we're getting into apparent vs. reactive loads and that's a whole other topic of discussion... But the point is, if you look at the power consumed by a simple resistor over hours or days or weeks, it will be the same if it has 120V DC or 170V AC across it. Again, that's because for a lot of the time during the sine wave's period the voltage across the resistor is actually less than 120V. RMS is just sort of a magic tool for getting the V=IR voltage from sinusoidal sources. It just works haha

  • @sridharchitta7321
    @sridharchitta7321 3 ปีที่แล้ว

    Current in a resistor is a start-stop motion of conduction band electrons due to their collision with the rocking lattice ions, and this causes a resistor with a sinusoidal voltage applied, to produce heat. The polarity reversals of an applied sinusoidal voltage (with the direction reversals of the applied electric field) do not affect electron collisions with the lattice ions. Electrons colliding with lattice ions from either direction will continue to produce heat and there is no cancellation of the heat developed!
    Mathematically, the average value of a sinusoid is zero, and so, the average value of the current will also be zero. The average values cannot therefore represent the heat developed in a resistor with a sinusoidal current.
    A resistor cannot develop heat due to a current in one half-cycle and then cool itself by a like amount of heat during the next half-cycle! It develops heat either way whether the current is positive or negative. The lattice ions vibrate from collisions irrespective of the directions in which the electrons collide with them. Therefore, since the average value is zero, it necessitates the use of the root-mean-square values of the voltage and current to compute the power, which is a statistical measure of the magnitude of a varying quantity and is the square root of the arithmetic mean of the square of the sinusoidal function.
    Electrostatics and circuits belong to one science not two. To learn the operation of circuits it is instructive to understand Current, the conduction process, resistors and Voltage at the fundamental level as in the following two videos:
    i. th-cam.com/video/REsWdd76qxc/w-d-xo.html and
    ii. th-cam.com/video/8BQM_xw2Rfo/w-d-xo.html
    It is not possible in this post to discuss in more detail average and rms values.
    The last frame References in video #1 lists textbook 4 which discusses in detail using a unified approach sinusoidal voltage, current, their average and root mean square values.

  • @electricfieldtv709
    @electricfieldtv709 6 ปีที่แล้ว +2

    can these formulas feed me to survive in real life? Hhh

  • @buffplums
    @buffplums 3 ปีที่แล้ว

    Helium?

  • @opaaaaaaaaaaa
    @opaaaaaaaaaaa 3 ปีที่แล้ว

    Awesome 👍

  • @niceguy100000
    @niceguy100000 5 ปีที่แล้ว +2

    Copy and paste this into WolframAlpha: square root of (integral of sin^2(x) from 0 to pi) divided by pi

  • @shemsnow3711
    @shemsnow3711 2 ปีที่แล้ว

    Why didn't you set the limits of integration to be 0 and 2 pi? You get the same result and I believe the definition of RMS requires you to set the limits to the start and end of a single period.

  • @harmeshsingh8086
    @harmeshsingh8086 5 ปีที่แล้ว

    Thanks for your help man!

  • @Someone_Unemployed
    @Someone_Unemployed 8 หลายเดือนก่อน

    Here in 2024 .......after 7years 😌

  • @Jdonovanford
    @Jdonovanford 7 ปีที่แล้ว

    Mmm, the integration is over theta. I would have thought it should be over pi.

  • @andrearomanelli6259
    @andrearomanelli6259 2 ปีที่แล้ว

    0 meter 0.707 bar mount Everest!33000 meters altitude!new Aquator 33643 km! End of world!Better Dooms day!neutronstar!10 bar 0 meter!

  • @danielsproelectricalservic5805
    @danielsproelectricalservic5805 6 ปีที่แล้ว

    What is 120v 40 ohms rms valvue

  • @TelekineticKhai
    @TelekineticKhai 6 ปีที่แล้ว

    SUPER
    PERMANENT
    INK

  • @Jdonovanford
    @Jdonovanford 7 ปีที่แล้ว

    Sorry for so many questions. I used Matlab to prove this. I created a sine wave 'd', and multiplied its maximum by the 1/sqrt(2) factor --> (max(d)*1/sqrt(2) but it doesn't equal rms(d) for some reason :/

    • @Jdonovanford
      @Jdonovanford 7 ปีที่แล้ว

      I can't say the same about your brain or genitals. No manifestation of neither.

    • @Jdonovanford
      @Jdonovanford 7 ปีที่แล้ว

      I'm here because I suck at Matlab, given that I started my Matlab for beginners course last week. Not so bad, in my opinion! :D As for you, imbecile, I got from you the reply I looked for. Thanks for your submission. Much appreciated.

    • @justadreamerforgood69
      @justadreamerforgood69 5 ปีที่แล้ว

      @@Jdonovanford
      Lmao he helped you out and you're being so rude 😂

  • @VINOBHAAJITH
    @VINOBHAAJITH 6 ปีที่แล้ว

    Nice

  • @LeatherCladVegan
    @LeatherCladVegan 7 ปีที่แล้ว

    Thank you :]

  • @alexmanbeck
    @alexmanbeck 7 ปีที่แล้ว

    thanks

  • @loser-nobody
    @loser-nobody 3 ปีที่แล้ว

    I love you

  • @lamarleachman1094
    @lamarleachman1094 5 ปีที่แล้ว

    173 volt in all

  • @user-be4yc2vr5c
    @user-be4yc2vr5c 4 ปีที่แล้ว

    This math is so far beyond my grasp. It's like ya'll just always making up random shiz and symbols and never giving in depth explanation ;( Even the math videos I try to watch to understand it lol