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Circuit elements (Real or Imaginary)

  1. Jul 20, 2009 #1
    Statement:
    I was wondering if someone could provide more background knowledge on why certain elements in a circuit (e.g. capacitors) are associated with imaginary components, whereas other elements (e.g. resistors) are only associated with the real components.

    In particular, the equation below is for capacitors (assuming a power source with [tex]Vcos(\omega t)[/tex]):
    [tex]
    C[\frac{d}{dt}(Vcos(\omega t))] = C[\frac{d}{dt}(V Re<e^{j(\omega t)}>] = C[j \omega V],[/tex] (#1). Note: [tex]Re<>, e^{j(\omega t)}[/tex] are assumed and can be taken off/ignored, by Phasor Relationship.

    Questions:
    Is it reasonable to assume if the power source was given be [tex]Vsin(\omega t)[/tex] then the resulting equation, similar to equation (#1), would be real instead of imaginary (thus not containing a "j"):
    [tex]
    C[\frac{d}{dt}(Vsin(\omega t))] = C[\frac{d}{dt}(V Re<e^{(\omega t)}>] = C[ \omega V],[/tex] (#2)
    If above (equation (#2)) is true, then elements in circuits that have derivatives (maybe integrals also)- capacitors- are not necessarily imaginary- and I suppose relative to the reference of the sinusoid.

    Or is the imaginary term "j" inherent to integrals and derivatives as Cel said:
    I am trying to find out why certain elements in circuits are imaginary and others are real, and from my previous post why elements like capacitors are imaginary:
    [tex]
    I = GV + C\frac{dV}{dt} = GV + j\omega CV
    [/tex]

    If someone could explain it to me carefully that would be great- I am not an engineer, so this is a little different to me.


    Thanks,


    JL
     
    Last edited: Jul 20, 2009
  2. jcsd
  3. Jul 22, 2009 #2

    Redbelly98

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    Since it is customary to represent the time-dependence of electrical signals by

    ejωt,​

    then differentiating or integrating involves multiplying by ±j.

    The current and voltage of a capacitor or inductor are related via differentiating one or the other. So when we express the ratio of voltage/current for these elements, we get a factor of ±j. Hence an imaginary number.
     
  4. Jul 22, 2009 #3

    cepheid

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    Hi Jeff,

    Are you familiar with the concept of impedance? Impedance, Z, is defined as the complex (phasor) voltage divided by the complex current. In this way, it is somewhat like resistance, although it takes the effects of the time-varying signals into account. Therefore, impedance can have both real (resistive) and imaginary (reactive) parts. I'm not sure how much sense it makes to say that a circuit component itself is real or imaginary. What can be said is that the impedance of a capacitor or inductor (which is a phasor) will always be imaginary (or entirely "reactive", to use the electrical engineering jargon). What this factor of "i" (or "j") on the impedance of a capacitor really means is that these elements will phase shift any signal sent into them by pi/2. For instance, if you put a sinusoidal current across an inductor, the corresponding voltage across it (which is proportional to the derivative of the current) will be phase shifted by +pi/2 (corresponding to a factor of +j in the impedance), corresponding to a time "advance" (as opposed to a time delay). The voltage is phase shifted compared to the current in such a way that whatever is happening with the current has already happened earlier with the voltage. We say that current "lags" voltage in an inductor, because the inductor is reacting to the rate of change of the current by producing an induced voltage. Therefore, when the rate of change of the current is highest (i.e. it building up to a peak), the voltage has already peaked.

    In contrast, if there is a certain sinusoidal current through a capacitor, the voltage across it will be phase shifted by -pi/2, (corresponding to a factor of -j in the impedance), which corresponds to a time delay. Whatever is happening with the current in a capacitor won't happen until some time later with the voltage. We say that current "leads" voltage in a capacitor. This makes sense, because the current through the capacitor is reacting to the rate of change of the voltage across it. At the point when the voltage is changing at the highest rate (i.e. building up to a peak), the current has already reached that peak.

    In both cases, the phase shift of the impedance follows logically from the differential equation that described what a capacitor or inductor does to time-varying signals (differentiates or integrates). Indeed, we can easily derive the impedances from these differential equations, and in so doing, we are going from the time domain to the frequency domain. Are you familiar with Fourier transforms? I talk some more about this here:

    https://www.physicsforums.com/showthread.php?t=325000

    I hope this helps.
     
    Last edited: Jul 22, 2009
  5. Jul 22, 2009 #4

    cepheid

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    By the way, this is not even correct:

    It's wrong because:

    [tex] \sin(\omega t) \neq Re[e^{\omega t}] [/tex]​

    I'm not sure where you got that from. The function eωt is real to begin with, and there is no way to turn it into a sine function! What it should have been was:

    [tex] \sin(\omega t) = Re[e^{j( \omega t - \pi/2)}] [/tex]​

    Therefore, if the voltage is given by a sine function, corresponding to a phasor voltage of:

    [tex] V = V_0 e^{j(-\pi/2)}e^{j\omega t} [/tex]​

    Then,

    [tex] C\frac{d}{dt}V = CV_0 e^{j(-\pi/2)}\frac{d}{dt}e^{j\omega t} = j \omega C V_0 e^{j(-\pi/2)}e^{j\omega t} [/tex]

    [tex]= j \omega C V[/tex]​

    The point is that it's still going to be jωC times the original phasor voltage, NO MATTER what that original phasor voltage is.
     
    Last edited: Jul 22, 2009
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