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Curry
 one year ago
Question with circuits
Curry
 one year ago
Question with circuits

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Curry
 one year ago
Best ResponseYou've already chosen the best response.0the 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.

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.1ok 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?

Curry
 one year ago
Best ResponseYou've already chosen the best response.0Wait, i think i posted the wrong picture there.

Curry
 one year ago
Best ResponseYou've already chosen the best response.0So 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.

Curry
 one year ago
Best ResponseYou've already chosen the best response.0But 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.

Curry
 one year ago
Best ResponseYou've already chosen the best response.0Find 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?

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0I get the following equations for \(V_{oc}\). \(V_c\) is voltage at capacitor. $$ \cfrac{V_sV_c}{600}+\cfrac{V_{oc}}{300}=\cfrac{V_c}{(1/j\omega C)}\\ V_s+(V_sV_c)2V_cV_{oc}=0 $$ Two equations, two unknowns, solvable. Next, \(i_{sc}\): $$ i_{sc}+\cfrac{V_sV_c}{600}=\cfrac{V_c}{(1/j\omega C)} $$ $$ R_{th}=\cfrac{V_{oc}}{i_{sc}} $$ (Note my \(V_c\) is the same as "V" in your drawing)

Curry
 one year ago
Best ResponseYou've already chosen the best response.0hmm, your Isc is calculated wrong, i believe.

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433729454567:dw You don't see it like this?

Curry
 one year ago
Best ResponseYou've already chosen the best response.0well 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.

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0KVL reveals that Vc = 0: dw:1433804815187:dw So $$ i_{sc}= \cfrac{V_s}{R} $$

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0Also 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 \(i_{sc}\).

Curry
 one year ago
Best ResponseYou've already chosen the best response.0Yes, i understand much better now, thank you!
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