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**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:

Not only elements involving derivatives. Elements involving integrals too. In the case of electrical elements, these involve capacitors and inductors.

If you apply the Laplace transform to an integro-differential equation, the derivatives become s and the integrals, 1/s, where [tex]s = \sigma + j\omega[/tex] is the Laplace variable.

The real term [tex] \sigma[/tex] corresponds to the transient response, while the imaginary term [tex]j\omega[/tex] corresponds to the steady state response.

If you are interested only in the steady state, you replace s by [tex]j\omega[/tex] .

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

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