Camara Example 31.6, am I crazy?

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Dec 26, 2016
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I'm looking for some verification here.

In Camara on page 31-2, Section 2, First-order analysis, the guidelines for solving this kind of problem uses a Thevenin equivalent circuit to isolate the inductor or capacitor - by themselves without the resistor R2.  I get this, and I understand how examples 31.1-5 operate on this principle.  These examples use a DC source.

Then, for example 31.6 on page 31-5, (which has an AC source, but is a similar layout to Ex. 31.1, shouldn't matter, right?) the problem isolates the resistor, R2 WITH the inductor L and is solved together as the combined load.  OK, I get that, and what has me confused is that Camara solves for the Thevenin resistance using R1 and R2 because their answer is Zrth 5 ohms (top right column of page 31-5), and I can't figure out how they come up with Vth = 60V at 30 deg.  If the further solution uses the combined impedance of R2 with L as the load (where Zth = Zrth+Zl), shouldn't the Thevenin resistance exclude R2? 

Another way to say this is: once points are chosen to solve for the Thevenin equivalent circuit in this method, they should not move, and what I read is, when you consider the circuit diagram, first they these points are to the right of R2 (to get a Thevenin resistance of 5 ohms) and then they are to the left or R2 (to get the impedance that includes R2 and L).

I'd really appreciate if someone would take a look and help me check my sanity - or interpretation - or understanding.



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