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Why voltage in a circuit is in a sinusoid form?

  1. Jul 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Could someone explain to me why voltage in a circuit is in a sinusoid form?


    2. Relevant equations
    (voltage)(sinusoid) [tex]\Rightarrow[/tex] [tex]V_Scos(\omega t)[/tex]


    3. The attempt at a solution
    Shouldn't all sinusoidal voltages in a circuit be in the form [tex]V_Scos(\omega t + \theta)[/tex], or is [tex]V_Scos(\omega t)[/tex] correct? I mean it's trivial, but the former notation is better because we can assign [tex]\theta[/tex] = 0 which is the same as the latter.

    thanks,

    JL
     
  2. jcsd
  3. Jul 5, 2009 #2
    Re: Voltage

    It isn't always in that form. You'll sometimes see voltage written in phasor form, which includes the phase. But what typically matters is the phase relative to other voltages in a circuit so for simplicity you can arbitrarily assign one of the voltages to have zero phase shift.

    At least, that's my take on it; I'm not an EE.
     
  4. Jul 5, 2009 #3

    rl.bhat

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    Homework Helper

    Re: Voltage

    When you start observing a variable voltage which changes just like a sine function, if the voltage is zero at t=0, then we say that the phase of the voltage is zero. If the voltage is not zero, then we say there is a phase θ. That is the difference between the two expressions.
     
  5. Jul 5, 2009 #4
    Re: Voltage

    Ok got it. So when a voltage is in that form, it's in phasor notation- which is known to have it's parameters constant. Now I just have to search the web to find out why a voltage is multiplied with a sinusoid.

    Thanks,

    Jeff
     
  6. Jul 6, 2009 #5

    turin

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    Re: Voltage

    I would comment that any single voltage or current does not "have a phase". Phases are always relative to something. Probably the most commonly encountered phase in circuit analysis is the impedance phase, which is the phase of the voltage across some element relative to the current through that element, at a given frequency (sinusoidal voltage and current).

    If you are wondering why sinusoid, as opposed to any other arbitrary functional form:

    Sinusoids are used because they are the basis for Fourier (frequency) analysis. The lowest order approximation for a circuit is to assume that it is "linear", or at least has linear modes of operation. This means that, for example, you can connect two different voltage sources, and then add the results as if you had connected each one individually. There's some kind of mathematical theorem or something that claims that you can construct (pretty much) arbitrary functions by adding up sinusoids in the appropriate way. (See Fourier Series/Transform.)
     
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