Question

Question with circuits

Answer 1

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.

Answer 2

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?

Answer 3

Wait, i think i posted the wrong picture there.

Answer 4

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.

Answer 5

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.

Answer 6

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?

Answer 7

@radar

Answer 8

@ybarrap

Answer 9

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)

Answer 10

hmm, your Isc is calculated wrong, i believe.

Answer 11

You don't see it like this?