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I do understand that the concept of complex impedance is the developed form of original impedance, where only the size of the impedance could be represented. By allowing the use of imaginary unit, we can represent not only the size of the impedance but also the direction of the impedance.

If complex impedance is (a+bi), then the size of the impedance is √(a²+b²). Here, the real part 'a' and the imaginary part 'b' are both considered.

However, the thing that I don't understand is that v = Ve^(iwt) = Vcos(wt) + iVsin(wt) is equivalent to v = Vcos(wt). In this case, only the real part of v, Vcos(wt), is considered.

Why does complex impedance considers both real part and complex part but complex voltage only considers the real part?

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Why does complex impedance considers both real part and complex part but complex voltage only considers the real part?

It doesn't.

Simply put, in the phasor domain (impedance is the ratio of the voltage phasor and current phasor), the voltage (current) phasors are not real but rather, complex constants that represent the amplitude and phase of a time domain sinusoidal voltage (current).

It is only 'at the end', after the circuit is solved in the phasor domain, do we add back the time dependence and take the real part to get the physical time domain voltage.

For example, if you find that the phasor voltage (a complex number) across an impedance $Z$ is

$$V_Z = 12\angle 37^\circ \mathrm{V}$$

then, to get the physical time dependent voltage, add the time dependence back and take the real part

$$v_Z(t) = \Re\{ V_Z\,e^{j\omega t} \} = 12\cos(\omega t + 37^\circ)\,\mathrm{V} $$

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When you first learned about circuits, you probably considered a battery (constant voltage) connected to a network of resistors. You learned about the rules for combining resistors in parallel and resistors in series and for simple circuits that allowed you to calculate all the (constant) currents through every part of the circuit and the (constant) voltages at every "interesting" point in the circuit. If the circuit was a bit more involved, you might need Kirchhoff's two laws to calculate currents and voltages. And maybe you learnt about parallel and series combinations of capacitors (which, for constant voltage and after the transient effects of charging them up, are just open circuits).

When you go to varying voltages and circuit elements like capacitors and inductors (in addition to resistors), you generally have to solve differential equations, rather than algebraic equations. Unless the voltage is a cosine: for a linear system, then every voltage and current is similarly a cosine (except that it may lag behind the input voltage by a phase). Then you can take advantage of the fact that you can represent the cosine as the real part of an exponential with an imaginary exponent using Euler's formula:

$$ V(t) = V_0 \cos \omega t = Re(V_0 e^{i\omega t})$$

You can then calculate things as if there is no time variation, using parallel and in-series rules and Kirchhoff's laws, and then at the very end, multiply by the complex exponential and take the real part. But there is one gotcha: you don't deal in bare resistances, capacitances and inductances any longer, you deal in impedances. The impedance of a circuit element is related to the "bare" quantity associated with the circuit element, but it is frequency-dependent and it is complex. Now you can pretend there is no time variation and go ahead and calculate using the usual algebraic rules, but with impedances. You come up with complex voltages and currents, but you know how the time variation can be accounted for: multiply by $e^{i\omega t}$ and take the real part. But note that if I take a complex voltage and multiply it by the exponential and then take the real part, there is going to be in general both a cosine and a sine term: in other words, the voltage at some point in the circuit is going to oscillate at the same frequency as the driving voltage, but it is generally not going to be in phase with the driving voltage.

Or you can set up the differential equations and solve them anew for every circuit you encounter. Then you don't introduce complex quantities at all, but I think you will agree that solving a circuit becomes much more cumbersome this way.

This is the idea but doing some actual calculations on simple circuits using both methods certainly helps. Here is a reference that you might enjoy: http://www.feynmanlectures.caltech.edu/II_22.html (but note the dependencies to a couple of lectures in the first volume - depending on your background, you might want to start with those).

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  • $\begingroup$ I comment on your comment therein : physics.stackexchange.com/questions/342123/… as follows : &Nick These are notes I prepared some years ago in order to dissolve my own confusion about coupling angular momenta. The notes were written in LaTeX and a large part of this "lot of work" was ready for copy-paste. I feel the need to share this experience in Physics SE with answers, not as a textbook. But the site has at least 40 users with name "Nick". $\endgroup$
    – Frobenius
    Jul 9 '17 at 19:11
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The idea is to avoid the manipulations of trig functions since various components introduce phase shifts in the voltage in addition to voltage drops. Since exponentials easily account for the phase and are easier to manipulate, one

  1. does all the initial calculations using phasors and complex numbers,
  2. converts to physical quantities at the very end by taking the real part of phasors for the appropriate quanties.

Thus the real part of a voltage might not be $\cos(\omega t)$ but rather $\cos(\omega t+\phi)$ once you account for the phases due to the various impedance in your system, i.e. the resulting voltage phasor might well be $Ve^{i(\omega t+\phi)}$ and so becomes $V\cos(\omega r +\phi)$ upon taking the real part.

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