Norton equivalent circuit help

Thread Starter


Joined Oct 31, 2009

The answer is given as phasors
I = 0.3536 angle 45
Z = 2.828 angle -45


What I did:

Zl = jwL = 2j Ohms
Zc= -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 Voc and the short circuit current Isc, correct?

Starting with Voc with a node b/w the inductor and resistor (which is equal to Voc because there is no voltage drop through the resistor due to no current going through it):
(groud at bottom node)
Voc: \(\frac{Voc}{-4j}\) + \(\frac{Voc - 2}{2}\) = 1.5Il

Il = \(\frac{Voc - 2}{2}\)

Subbing in and arranging gives:

\(\frac{Voc}{-4j}\) = \(\frac{Voc - 2}{4}\)

Solving gives:
Voc = -1.414 angle 45,

however I'm fairly sure it's wrong given the answers ( IZ != V).

I'm not sure how to find I[SUB]SC[/SUB] 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.
i1 = top loop current
i2 = left loop
i3 = right loop

Taking positive to be CW,
i1 = 1.5iL
iL = i1 - i2

iL = 2i2

Loop two (left):
-2i2 - 4j (i2 - i3) = 2 angle 0

Loop three (right):
2i3 - 4j (i3 - i2) = 0

Solving the two linear equations gives
i3 = iN = 2j

which isn't the correct answer.

Any help would be greatly appreciated...

Thread Starter


Joined Oct 31, 2009
Attempting the question again this morning and I managed to get the right result for the current. Forgot to multiply iL by the impedance in loop 2...probably shouldn't be attempting questions when tired and sleepy after midnight.

Will have another go at the Voc and see if I can't get the right answer.

EDIT: Also got Voc. When finding the voltage across the inductor I divided by 2 instead of 2j. Spent probably a couple of hours on this question last night but managed it in 5-10 minutes this morning.
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