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Curry

  • one year ago

Question with circuits

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  1. Curry
    • one year ago
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  2. Curry
    • one year ago
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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.

  3. sidsiddhartha
    • one year ago
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    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?

  4. Curry
    • one year ago
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    Wait, i think i posted the wrong picture there.

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  5. Curry
    • one year ago
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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.

  6. Curry
    • one year ago
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    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.

  7. Curry
    • one year ago
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    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?

  8. Curry
    • one year ago
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    @radar

  9. Curry
    • one year ago
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    @ybarrap

  10. ybarrap
    • one year ago
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    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)

  11. Curry
    • one year ago
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    hmm, your Isc is calculated wrong, i believe.

  12. ybarrap
    • one year ago
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    |dw:1433729454567:dw| You don't see it like this?

  13. Curry
    • one year ago
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    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.

  14. ybarrap
    • one year ago
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    KVL reveals that Vc = 0: |dw:1433804815187:dw| So $$ i_{sc}= -\cfrac{V_s}{R} $$

  15. ybarrap
    • one year ago
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    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 \(i_{sc}\).

  16. Curry
    • one year ago
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    Yes, i understand much better now, thank you!

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