Question with circuits

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Question with circuits

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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the answer is 4/3 - 27.7(cos20t - 56). I got the seccond part of the answer, but i can't figure out the 4/3.
ok how did u proceed given \[\omega=20\\x_l=jwL=20*0.5j=10j\\x_c=\frac{ 1 }{ jw C }=\frac{ 1 }{ j.10*10^{-3}*20 }=-5j\] right?

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Wait, i think i posted the wrong picture there.
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So that was the real question in concern. Um, I calculated Voc correctly. And I was trying to calculate Isc. So as soon as I short a,b, the 300ohm resistor is shorted. Ok.
But why is the dependant voltage source turned off? the answer they calculated for Rt was 236<-23j. The only way you get that answer is through the following method.
Find Voc, which is 1.97<-j23. And then divide by Isc, which would be 5/600. The only way Isc can equal 5/600 is if 300ohm is shorted, the dependant voltage source is turned off, and the capacitor is also shorted. Can a short really get rid of a voltage source and a capacitor? I thought it can only eliminate resistors?
I get the following equations for \(V_{oc}\). \(V_c\) is voltage at capacitor. $$ \cfrac{V_s-V_c}{600}+\cfrac{V_{oc}}{300}=\cfrac{V_c}{(1/j\omega C)}\\ -V_s+(V_s-V_c)-2V_c-V_{oc}=0 $$ Two equations, two unknowns, solvable. Next, \(i_{sc}\): $$ i_{sc}+\cfrac{V_s-V_c}{600}=\cfrac{V_c}{(1/j\omega C)} $$ $$ R_{th}=\cfrac{V_{oc}}{i_{sc}} $$ (Note my \(V_c\) is the same as "V" in your drawing)
hmm, your Isc is calculated wrong, i believe.
|dw:1433729454567:dw| You don't see it like this?
well yes, but wouldn't just laving the equation in that form cause a hole as to what Vc is? Because then we have 2 unknowns. Isc and Vc.
KVL reveals that Vc = 0: |dw:1433804815187:dw| So $$ i_{sc}= -\cfrac{V_s}{R} $$
Also note that since \(V_c=0\) that \(i_c=c\frac{dV_c}{dt}=0\) (i..e. open circuit), which means all the current goes through \(i_{sc}\).
Yes, i understand much better now, thank you!

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