- #1

rafehi

- 49

- 1

## Homework Statement

See picture:

http://img88.imageshack.us/img88/8319/87671967.jpg

The answer is given as phasors

I = 0.3536[tex]\angle[/tex]45

Z = 2.828[tex]\angle[/tex]-45

## Homework Equations

## The Attempt at a Solution

Z

_{L}= jwL = 2j Ohms

Z

_{C}= -j/(wC) = -4j Ohms

Can't do it by zeroing the sources as there's a dependent source (if it's possible, we haven't been taught it).

So then we'll have to find both the open source voltage V

_{OC}and the short circuit current I

_{SC}, correct?

Starting with V

_{OC}with a node b/w the inductor and resistor (which is equal to V

_{OC}because there is no voltage drop through the resistor due to no current going through it):

(groud at bottom node)

V

_{OC}: [tex]\frac{Voc}{-4j}[/tex] + [tex]\frac{Voc - 2}{2}[/tex] = 1.5I

_{L}

I

_{L}= [tex]\frac{Voc - 2}{2}[/tex]

Subbing in and arranging gives:

[tex]\frac{Voc}{-4j}[/tex] = [tex]\frac{Voc - 2}{4}[/tex]

Solving gives:

V

_{oc}= -1.414[tex]\angle[/tex]45,

however I'm fairly sure it's wrong given the answers ( IZ != V).I'm not sure how to find I

_{SC}because of the dependent current source. Have tried and can't get the correct answer - not sure if I'm going about it the right way.

i

_{1}= top loop current

i

_{2}= left loop

i

_{3}= right loop

Taking positive to be CW,

i

_{1}= 1.5i

_{L}

i

_{L}= i

_{1}- i

_{2}

Therfore,

i

_{L}= 2i

_{2}

Loop two (left):

-2i

_{2}- 4j (i

_{2}- i

_{3}) = 2[tex]\angle[/tex]0

Loop three (right):

2i

_{3}- 4j (i

_{3}- i

_{2}) = 0

Solving the two linear equations gives

i

_{3}= i

_{N}= 2j

which isn't the correct answer.Any help would be greatly appreciated...

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