# Thevenin Equal Resistance - Strange Circuit

• staggarhs
In summary: There's some problems with your figure. Just for example, R2 should connect between A & B, after Vs2 is shorted.

## Homework Statement

I want to find Rτ (Thevenin Equal Resistance) if the load is R4 in the attached image's circuit.
As easy it may seem, I get confused because it asks for the equal resistance between A and B.

## The Attempt at a Solution

We remove the voltage sources and s/c them. Then we have: R1 is parallel to (R2+R3) -> R1//(R2+R3).
R5 parallel to the rest. So we have in total: Rτ= [R1//(R2+R3)] // R5

#### Attachments

• circuit.jpg
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Welcome to PF

I see no current sources in the figure, those are voltage sources. So, what do we do with voltage sources?

Also, note R2 & R3 are not actually in series. Because node A has a path leading to R4, they (R2 & R3) do not have an equal current going through them.

Sorry my mistake. I meant Voltage sources. I need to find the Thevenin Equivalent Resistance if i take R4 as the load. So I replace voltage sources with short-circuit and also remove the load R4. From this point could you help me figure out the connections of the resistors ?

Thanks again...

Your approach looks ok. But the calculations are wrong. Check again how to calculate the paralleled resistances. Have you gotten ET already?

Kruum said:
Your approach looks ok. But the calculations are wrong. Check again how to calculate the paralleled resistances. Have you gotten ET already?

I haven't done any calculations yet, I just indicated how I think the resistors are connected. No I just want to find RT

staggarhs said:
I haven't done any calculations yet, I just indicated how I think the resistors are connected. No I just want to find RT

Oh, okay. I thought those //'s were division marks.

staggarhs said:
Sorry my mistake. I meant Voltage sources. I need to find the Thevenin Equivalent Resistance if i take R4 as the load. So I replace voltage sources with short-circuit and also remove the load R4. From this point could you help me figure out the connections of the resistors ?

Thanks again...

I would redraw the figure (with voltage sources shorted), with "A" at the very top and "B" at the bottom. Redraw/move the resistors around until you see some that are in parallel or series, and combine them.

Redbelly98 said:
I would redraw the figure (with voltage sources shorted), with "A" at the very top and "B" at the bottom. Redraw/move the resistors around until you see some that are in parallel or series, and combine them.

I came up with this one, but the equivalent is different than the first one i posted ( Rτ= [R1//(R2+R3)] // R5 )

In this case it is: RΤ=[(R1+R2)//R3 + R5]

Please give me any hint as I am very confused.
Thanks a lot

#### Attachments

• circuit2.jpg
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staggarhs said:
Please give me any hint as I am very confused.
Thanks a lot

I'm not sure if this is any help to you, but what can be said about any open circuit (i.e. unconnected wires)?

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staggarhs said:
I came up with this one, but the equivalent is different than the first one i posted ( Rτ= [R1//(R2+R3)] // R5 )

In this case it is: RΤ=[(R1+R2)//R3 + R5]

Please give me any hint as I am very confused.
Thanks a lot

There's some problems with your figure. Just for example, R2 should connect between A & B, after Vs2 is shorted.

## What is Thevenin Equal Resistance?

Thevenin Equal Resistance is a concept in electrical circuit theory which states that any linear circuit can be represented by an equivalent circuit with a single voltage source and a single resistance in series. This equivalent circuit is known as the Thevenin equivalent circuit and the resistance in it is called the Thevenin resistance.

## How is Thevenin Equal Resistance calculated?

The Thevenin resistance is calculated by removing all voltage sources in the circuit and short-circuiting all current sources. Then, the resistance between the two terminals where the voltage source was removed is measured. This resistance is the Thevenin resistance and it is equal to the total resistance in the original circuit seen from those two terminals.

## What is the purpose of Thevenin Equal Resistance?

Thevenin Equal Resistance is used to simplify complex circuits into simpler and more manageable circuits. It allows us to analyze a circuit by replacing it with a simpler one, which makes calculations easier. The Thevenin equivalent circuit is also useful in designing and analyzing more complex circuits.

## What are the limitations of Thevenin Equal Resistance?

Thevenin Equal Resistance is only applicable to linear circuits, which means that the relationship between voltage and current must be linear. Nonlinear elements such as diodes and transistors cannot be represented using Thevenin equivalent circuit. Additionally, Thevenin Equal Resistance is only accurate for DC circuits and may not be accurate for AC circuits.

## How is Thevenin Equal Resistance different from Norton Equal Resistance?

Thevenin Equal Resistance and Norton Equal Resistance are two equivalent circuit representations of a complex circuit. While Thevenin equivalent circuit uses a single voltage source and a single resistance in series, Norton equivalent circuit uses a single current source and a single resistance in parallel. Both representations are valid and can be used interchangeably, depending on the type of analysis being performed.