Understanding the Relationship Between i*cos and sin in Circuit Analysis

[hutchphd]

You have not supplied any exact example of the process so I no longer know what you are talking about. This is why exact referrals to published works is usually required by PF. You need to be more definitive about what "it" is.......

 

[jaydnul]

You have not supplied any exact example of the process so I no longer know what you are talking about. This is why exact referrals to published works is usually required by PF. You need to be more definitive about what "it" is.......

That's a fair point. I guess in electronics you can derive a transfer equation when you input a sinusoid into the circuit, but quantum mechanics isn't exactly similar in that way

 

[DaveE]

In circuit analysis, everything seems to work out when you set i*cos = sin. But thats not a legitimate equation, so why does that work? Is there a proof that this is a real equation?

Just a bunch of dogmatic explanations so far like just listen to us, thats not how you are supposed to do it

What do you want from us?

Sometimes education is dogmatic, sometimes that's the best way to get people on the right track. You have several people that know a lot about this subject trying to help. Do you want to listen or are you seeking irrational affirmation?

 

[hutchphd]

Ask a particular question please. One with well-defined terms. If you wish to be bemused, please do it privately.
Comparison similar to electronic I/O in QM is often called scattering theory. The arithmetic is similar.

 

[vanhees71]

In AC analysis, it's simplifying things tremendously, when you use the fact that you deal with linea differential equations with real coefficients for real functions ##U(t)## and ##I(t)## ("voltage" and currents). Further usual household voltage is harmonic.

Take for example the series of a coil and a resistande. The corresponding equation reads
$$L \dot{I} + R I=U_0 \cos(\omega t).$$
The idea is that to find the solution for this equation it's much simpler to introduce complex quantities ##i(t)## and ##u(t)## and define the physical real quantities as the real part of these complex quantities, i.e., ##U=\mathrm{Re} u##, ##I=\mathrm{Re} i##. The external AC voltage is given by ##U_0 \exp(\mathrm{i} \omega t)## (using the engineers' sign convention in the exponential, because then ##\mathrm{Re} [U_0 \exp(\mathrm{i} \omega t)=U_0 \cos (\omega t).##
Then we can solve the equation for the complex quantities,
$$L \dot{i} + R i=U_0 \exp(\mathrm{i} \omega t).$$
Obviously the real part of the solution for ##i## then gives you ##I##, because ##L##, ##R##, and ##U_0## are all real.

To find the "stationary state" you can simply make the ansatz
$$i(t)=i_0 \exp(\mathrm{i} \omega t),$$
and plugging this into the equation you only have to solve an algebraic equation for the complex amplitude ##i_0##:
$$(\mathrm{i} \omega L + R) i_0 = U_0 \; \Rightarrow \; i_0 = \frac{U_0}{R+\mathrm{i} \omega L} = \frac{R-\mathrm{i} \omega L}{R^2+(\omega L)^2}.$$
The physical current thus is
$$I(t)=\mathrm{Re} i(t)=\frac{U_0}{|R^2+(\omega L)^2} [R \cos(\omega t) + \omega L \sin(\omega t)].$$
This becomes a bit more convenient if you write the complex "inverse resistance" (impedance) in "polar form"
$$\frac{R-\mathrm{i} \omega L}{R^2+(\omega L)^2} = \frac{1}{\sqrt{R^2+(\omega L)^2}} \exp(\mathrm{i} \phi), \quad \varphi=-\arccos \left (\frac{R}{\sqrt{R^2+(\omega L)^2}} \right).$$
Then you get
$$I(t)=\frac{U}{\sqrt{R^2+(\omega L)^2}} \cos(\omega t+ \phi).$$
This gives you immediately the amplitude of the current,
$$I_0=\frac{U}{\sqrt{R^2+(\omega L)^2}},$$
and the phase shift ##\phi##, given above. It's negative, which means that the current's phase is always behind the external voltages phase, as expected for an inductance.

With this complex treatment you can calculate any AC circuit's stationary state as if it were a DC circuit, using Kirchhoff's two rules but use for resistors, capacitors, and inductances the corresponding complex impedances,
$$Z_R=R, \quad Z_L=\mathrm{i} \omega L, \quad Z_C=-\frac{\mathrm{i}}{\omega C}.$$
At the end of the cacluation you get all real physical currents and voltages by taking the real part of the corresponding complex quantities.