Thevenin equivalent diamonda shaped

[DottZakapa]

_【Mod Note: thread moved from technical forum, so homework template is missing_】

Hi every one! I am encountering difficulties on finding the Req. in the sense that i can't really see which is in parallel end which is not, especially on the diamond shape. I haven't really understood how to treat such cases.
I short circuit the voltage source,
the Z=(25-j30)Ω is in parallel with?
Could please anyone help me figure out how to approach such configurations once and for all?
Thank you
See attachment

 

[Averagesupernova]

25 - j30 would be in parallel with a wire. Wouldn't you agree?

 

[phinds]

Could please anyone help me figure out how to approach such configurations once and for all?

FORGET about the shape of the circuit as drawn. It's totally irrelevant. Just look at every node. Find elements that both end in the same 2 nodes. They are parallel.

 

[DottZakapa]

25 - j30 would be in parallel with a wire. Wouldn't you agree?

I agree, the parallel would give zero, then i'll have ( -j20 || j10 ) + ( j50 || j50 ) am I right?

 

[DottZakapa]

FORGET about the shape of the circuit as drawn. It's totally irrelevant. Just look at every node. Find elements that both end in the same 2 nodes. They are parallel.

When I see nodes in between like A B i get lost

 

[Averagesupernova]

I agree, the parallel would give zero, then i'll have ( -j20 || j10 ) + ( j50 || j50 ) am I right?

That is the way I see it.

 

[Baluncore]

I short circuit the voltage source,

DO NOT short circuit the voltage source.
Label the point on the RHS between 25 and -j30 with a “c”.

The j100 V source is then connected to three separate current paths, that are in parallel with each other. Each of those parallel paths has a mid-point, now labelled a, b or c.
Each path has two components in series. Like resistors, series impedances add, so the three impedances, (named after their midponts), become;
Za = 0 +j (50+50)
Zb = 0 +j (10–20)
Zc = 25 +j (–30)
Note: The current that flows in each of those parallel paths will be ( 0 –j 100 ) / Z.

To solve parallel impedances, convert them to admittances by taking their complex reciprocals, then add those admittances to get the total admittance. Convert that admittance with another reciprocal back to impedance, giving you the Zeq of all three parallel paths.
1/Zeq = 1/Za +1/Zb +1/Zc

 

[NascentOxygen]

DO NOT short circuit the voltage source.
Label the point on the RHS between 25 and -j30 with a “c”.

The task is to determine the Thévenin equivalent between points a and b.

 

[NascentOxygen]

I agree, the parallel would give zero, then i'll have ( -j20 || j10 ) + ( j50 || j50 ) am I right?

That's half the task solved. Besides impedance you also need to determine the Thévenin voltage.