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Phasor Voltage and Voltage RMS value

  1. Sep 19, 2009 #1
    What's the difference between the two?

    RMS is divided by square root of 2?
     
  2. jcsd
  3. Sep 19, 2009 #2
    In short, it's used to easily calculate real power of AC circuits.

    Power is: P = IV

    if you use the rms values for current and voltage, then the power is the same as if I and V were DC instead of AC.
     
  4. Sep 19, 2009 #3

    Averagesupernova

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    Only for a sine wave.
     
  5. Sep 20, 2009 #4
    Phasor voltage represents the instantaneous value of the voltage. For example in time t the voltage may have a value of 120V and one second later it may have a voltage of 50V.
    The rms voltage is (let say) the ac voltage that is equivalent to the dc voltage. This is not very clear. Isn't it?
    So, let explain it in another way. A famous example that always is given for explanation of the rms concept is the associated heat generated by a resistor due to the current flowing through it.
    Let's apply an ac voltage to a resistor for one hour and measure the generated heat from the resistor. Afterward, we will aplly a dc voltage to the resistor that generate the same value of the heat and for the same time.
    It is said that the ac voltage in this case has an rms value that is equal to the applied dc voltage.
     
  6. Sep 23, 2009 #5
    Voltage phasor
    By using Euler's Formula, the voltage can be expressed as

    [tex]V = V_{0} \:cos(\omega t + \phi)= Re[V_{0}\:e^{j(\omega t + \phi)}]=Re[V_{0}\:e^{j\phi}e^{j \omega t}]=Re[\bar{V}e^{j \omega t}][/tex]

    where [tex]\bar{V}= V_{0}\:e^{j \phi}[/tex] is known as the phasor.

    In effect you have factored out the time-dependency, and you're left with a representation (the phasor) of voltage that only deals with the static quantities of amplitude and phase angle of the cosine function. Being able to represent voltage this way simplifies a lot of calculations, e.g. addition of two sinusoidally time-varying functions (of the same frequency) is reduced from a trigonometric problem to a simple algebraic problem. To sneak back into the time-domain, you just shamelessly tack on the [tex]e^{j \omega t}[/tex] to your result. Good stuff!

    RMS Value
    If you combine the fact that 1) when dealing with ac power, you're usually interested in the average value of power during each cycle of the waveform, with 2) power is proportional to the voltage squared, you get

    [tex]{V_{rms}}^2=\frac{V_{0}^{2}}{T}\int^{T}_{0}cos^{2}(\omega t) dt = \frac{V_{0}^{2}}{T}\int^{T}_{0}1+cos(2\:\omega t) dt = \frac{V_{0}^{2}}{2}[/tex]

    Or

    [tex]V_{rms}=\frac{V_{0}}{\sqrt{2}}[/tex]


    EDIT: For some reason I can't edit the tex code to correct the missing 1/2 factor. It's supposed to be

    [tex]\frac{V_{0}^{2}}{T}\int^{T}_{0}\frac{1}{2}+\frac{cos(2\:\omega t)}{2}\:dt [/tex]
     
    Last edited: Sep 24, 2009
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