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- Voltages phasors in AC circuit - Physics Stack Exchange
The following involves phasor notation, which is in the frequency domain You need to add ac voltages vectorially for RC and LC circuits because the voltage and currents in inductors and capacitors are 90 degrees out of phase with each other, whereas the voltage and current are in phase for a resistor
- Kirchhoffs laws in phasor domain - Physics Stack Exchange
My understanding is that this phasor method to find the amplitude and phase offset in AC analysis is effectively a shortcut to solve the differential equation that describes the circuit formed from Kirchhoff's laws because it is known that the steady-state solution of the differential equation oscillates at the driven frequency of the source
- homework and exercises - Finding frequency from phasor in EM wave . . .
The general form of a phasor discribing a plane wave (spherical, cylindrical or Cartesian) travelling in the direction $\hat n$ is $$ \vec E_p=\vec E_0(x,y,z) e^{-jk\hat n\cdot \vec r} $$ where $\vec E_0(x,y,z)\cdot\hat n=0$ The amplitude need not be constant: in the case of a spherical wave it would go like $1 r$ for $\hat n=\hat r$ for instance
- Phasor Notation Convention - Which Projection to Consider?
A phasor is a rotating vector If, at time t it makes angle $(\omega t + \phi)$ with the x axis, its projection (for a phasor of unit magnitude) on to x axis is, by definition, $\cos(\omega t + \phi)$, and its projection on to y axis is $\sin(\omega t + \phi)$ Which of these we choose to represent a sinusoidal displacement, voltage or whatever
- Phasor form of Maxwells Equations - Physics Stack Exchange
I'm interested in the transformation from the standard Maxwell's equations to their phasor equivalents From the literature, this means injecting: \begin{equation} E = Re(\boldsymbol{E}e^{j\omega t}) \end{equation}
- Computing Poynting Vector from phasors - Physics Stack Exchange
As the computation of the Poynting Vector is not a linear operation, taking the vector product of the phasors would not result in a phasor whose real part equals $\mathbf{N}$ However, is it valid to say that the following (with $*$ being the complex conjugate operator) $$\widetilde{\mathbf{N}} =\widetilde{\mathbf{E}}\times\widetilde{\mathbf{H
- Why do phasors add like vectors? - Physics Stack Exchange
A phasor is actually a complex number, which is isomorphic to $\mathbb{R}^2$, and that's a well known vector space: the usual arrows Translation: a phasor is a complex number, and you add complex numbers in the same way you do with vectors
- Description of phasor form of Maxwells equations in linear media
The phasor representation accounts for this phase shift by the complex form of $\epsilon$ and $\mu$ $\epsilon$ is the complex permittivity of the medium, and $\mu$ is the complex permeability of the medium I don't understand this part:
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