- #1
Apteronotus
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I have a quick question regarding how the complex impedance of an AC circuit is derived.
The voltage and current in an AC circuit are given by the equations
The impedance is given by
Do they actually mean that
if so...
What am I failing to see?
The voltage and current in an AC circuit are given by the equations
[tex]
V=V_{m}cos(\omega t) = V_{m}Re(e^{i\omega t})
[/tex]
[tex]
I=I_{m}cos(\omega t-\phi) = I_{m}Re(e^{i(\omega t-\phi)})
[/tex]
V=V_{m}cos(\omega t) = V_{m}Re(e^{i\omega t})
[/tex]
[tex]
I=I_{m}cos(\omega t-\phi) = I_{m}Re(e^{i(\omega t-\phi)})
[/tex]
The impedance is given by
[tex]
Z = \frac{V}{I}} = \frac{V_{m}Re(e^{i\omega t})}{I_{m}Re(e^{i(\omega t - \phi)})}
[/tex]
Some online resources state that Z = \frac{V}{I}} = \frac{V_{m}Re(e^{i\omega t})}{I_{m}Re(e^{i(\omega t - \phi)})}
[/tex]
[tex]
Z=\frac{V_{m}}{I_{m}}e^{i\phi}
[/tex]
Z=\frac{V_{m}}{I_{m}}e^{i\phi}
[/tex]
Do they actually mean that
[tex]
Z=\frac{V_{m}}{I_{m}}Re(e^{i\phi})
[/tex]?
Z=\frac{V_{m}}{I_{m}}Re(e^{i\phi})
[/tex]?
if so...
[tex]
\frac{Re(e^{i\omega t})}{Re(e^{i(\omega t - \phi)})}\neq Re(\frac{e^{i\omega t}}{e^{i(\omega t - \phi)}})
[/tex]
\frac{Re(e^{i\omega t})}{Re(e^{i(\omega t - \phi)})}\neq Re(\frac{e^{i\omega t}}{e^{i(\omega t - \phi)}})
[/tex]
What am I failing to see?
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