## Curry one year ago Question with circuits

1. Curry

2. Curry

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

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

Wait, i think i posted the wrong picture there.

5. Curry

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

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

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

9. Curry

@ybarrap

10. ybarrap

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

hmm, your Isc is calculated wrong, i believe.

12. ybarrap

|dw:1433729454567:dw| You don't see it like this?

13. Curry

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

KVL reveals that Vc = 0: |dw:1433804815187:dw| So $$i_{sc}= -\cfrac{V_s}{R}$$

15. ybarrap

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

Yes, i understand much better now, thank you!