Use of phasor representation in physics

Click For Summary
Phasor representation is essential in physics, particularly for analyzing AC circuits and linear systems, as it simplifies the treatment of waves with constant frequency. The transformation of Maxwell's equations into phasor form involves substituting the time derivative with jw, allowing for analysis in frequency space. This approach assumes electromagnetic fields are harmonic waves, facilitating easier calculations of media properties at specific frequencies. The principle of superposition enables the handling of multiple frequency components by solving Maxwell's equations for each and summing the results. Overall, phasors streamline the analysis of complex wave phenomena in electrical engineering.
Abel I Daniel
Messages
12
Reaction score
0
Why do we use phasor representation in physics..For example,why we need maxwells equation in phasor form as well??
 
Physics news on Phys.org
Phasors are good for waves of constant frequency ... hence electrical engineers use them for AC circuits, and they use them a lot.

So you will probably only use phasors while studying AC circuits; they are useful only for linear systems.
 
ok ,thank you for the reply...what made me ask this question is-i saw maxwells equations(electromagnetic) written in phasor form from dpoint form by just substituting d/dt with jw..So what my questin is ,what is the implication of removing that time factor from that equation??
 
They were working in frequency space ... same place you go with the Laplace transform, or its cousin the Fourier transform.

In this case they assumed that the electromagnetic field was a harmonic wave - and plugged this into Maxwell's equations - leaving you with the "Phasor form of Maxwell's equations".

Here is a lecture which includes the derivation: http://ivp.ee.cuhk.edu.hk/~ele3310/data/ELE3310_Tutorial_10.pdf
 
Another reason is that properties of media are "easy". For example, for electric field:
<br /> \mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P}<br />
but it is only for a single frequency that
<br /> \mathbf{P}(\omega) = \epsilon_0 \chi_e (\omega) \mathbf{E}(\omega).<br />
In the time domain we in general have a convolution,
<br /> \mathbf{P}(t) = \epsilon_0 \int^t d\tau \, \, \, \chi_e (\tau) \mathbf{E}(t-\tau).<br />

jason
 
Abel I Daniel said:
ok ,thank you for the reply...what made me ask this question is-i saw maxwells equations(electromagnetic) written in phasor form from dpoint form by just substituting d/dt with jw..So what my questin is ,what is the implication of removing that time factor from that equation??

You can assume a time dependence of exp(jwt) without losing generality due to the principle of superposition. For incident fields with multiple frequency components, you can solve Maxwell's equations for each frequency component, then sum the solutions at the end as required.

Claude.
 
Thread 'What is the pressure of trapped air inside this tube?'
As you can see from the picture, i have an uneven U-shaped tube, sealed at the short end. I fill the tube with water and i seal it. So the short side is filled with water and the long side ends up containg water and trapped air. Now the tube is sealed on both sides and i turn it in such a way that the traped air moves at the short side. Are my claims about pressure in senarios A & B correct? What is the pressure for all points in senario C? (My question is basically coming from watching...

Similar threads

  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 18 ·
Replies
18
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
23
Views
3K