# Thevenin equivalent diamonda shaped

• DottZakapa
In summary: That voltage is given by:Z = (1+jZa) + (1+jZb) + (1+jZc)When I see nodes in between like A B i get lostTry drawing a circuit using only points that are connected by lines. Then you will be able to see how nodes in between affect the answer.
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

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

DottZakapa said:
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.

Averagesupernova said:
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?

phinds said:
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

DottZakapa said:
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.

DottZakapa said:
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

Last edited:
Baluncore said:
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.

Averagesupernova
DottZakapa said:
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.

Averagesupernova

## 1. What is the Thevenin equivalent diamond shaped circuit?

The Thevenin equivalent diamond shaped circuit is a simplified electrical circuit that represents a more complex circuit. It consists of a voltage source in series with a resistor, with the voltage source and resistor connected by two branches of resistors in parallel. The Thevenin equivalent circuit has the same voltage and current characteristics as the original circuit, making it easier to analyze and troubleshoot.

## 2. How do you calculate the Thevenin equivalent resistance for a diamond shaped circuit?

The Thevenin equivalent resistance for a diamond shaped circuit can be calculated by first finding the resistance of the two parallel branches, and then adding that value to the resistance in series with the voltage source. This can be expressed as RTH = R1 || R2 + R3, where R1 and R2 are the resistances in the parallel branches and R3 is the resistance in series with the voltage source.

## 3. What is the purpose of using a Thevenin equivalent diamond shaped circuit?

The Thevenin equivalent diamond shaped circuit is used to simplify complex circuits and make them easier to analyze. It also allows for easier troubleshooting and can help in predicting the behavior of the original circuit.

## 4. Can a diamond shaped circuit have multiple Thevenin equivalent circuits?

No, a diamond shaped circuit will only have one Thevenin equivalent circuit. This is because the Thevenin equivalent circuit is a simplified representation of the original circuit, and there can only be one simplified version.

## 5. What are the limitations of using Thevenin equivalent diamond shaped circuits?

One limitation of using Thevenin equivalent diamond shaped circuits is that they can only be used for linear circuits. They also may not accurately represent the behavior of the original circuit in certain situations, such as when there are non-linear components or when the circuit operates at high frequencies.

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