# Converting to a circuit's Thevenin equivalent?

• Engineering
• Jayalk97
In summary, The user struggled with finding the Thevenin equivalent and attempted to find the Norton equivalent before converting it. They encountered difficulties and asked for help, realizing they needed to find the voltage from a to b instead of the voltage difference. They were grateful for the assistance.
Jayalk97

## Homework Equations

Finding the Thevenin equivalent?

## The Attempt at a Solution

I found the thevenin resistance to be 185/47 ohms, which came out to be correct.

I've been struggling with this question all day. I ended up attempting to find the norton equivalent and then converting it, since the question asks for both anyways. I started with a source conversion on the 3A and 5ohm current source and resistor, turning it into a 15V voltage source and a 5 ohm resistor in series (going into the b terminal). I shorted the a-b terminal and used KVL to find the current going through as 3A, typed that into my webwork page and it said I was wrong. Is there anything I'm missing here?

Are you given values for ##I## and ##R## ?

Oh yeah sorry, I = 2.5A and R = 4.5ohms

gneill said:
Are you given values for ##I## and ##R## ?
On that note actually, these problems look fairly straightforward and I'm afraid that I misunderstand something fundamental about circuits...

Jayalk97 said:
Oh yeah sorry, I = 2.5A and R = 4.5ohms
Okay.

Hint: Work from left to right towards the output, converting and combining sources and resistances as you go.

gneill said:
Okay.

Hint: Work from left to right towards the output, converting and combining sources and resistances as you go.
If I have a voltage source>resistor>voltage source>resistor in series could I just rearrange and combine them? If so I converted the leftmost current source and resistor to a voltage source in series with it, combined the 2 resistors and voltage sources to get 25.25V and 18.5 ohms. I converted those to a current source in parallel with a resistor to get 101/74A and 18.5ohms. Adding those to the current source in parallel with the resistor on the right I get 121/74A (pointing down) with 185/47ohms, giving me a voltage difference of 605/94 but the program says that is incorrect...

gneill said:
Okay.

Hint: Work from left to right towards the output, converting and combining sources and resistances as you go.
Oh my god it explicitly wanted the voltage from a to b, and that is -605/94. I've been trying to figure this out for four hours... Thank you for the help haha.

Jayalk97 said:
Oh my god it explicitly wanted the voltage from a to b, and that is -605/94. I've been trying to figure this out for four hours... Thank you for the help haha.

## 1. What is a Thevenin equivalent circuit?

A Thevenin equivalent circuit is a simplified representation of a complex circuit that consists of a single voltage source and a single resistor. It is used to analyze and solve circuit problems in a more efficient manner.

## 2. Why is it useful to convert a circuit to its Thevenin equivalent?

Converting a circuit to its Thevenin equivalent allows for easier analysis and calculation of circuit variables, such as current and voltage. It also helps in simplifying complex circuits and reducing the number of components needed.

## 3. How do you determine the Thevenin equivalent voltage and resistance?

The Thevenin equivalent voltage (Vth) can be found by open-circuiting the circuit and measuring the voltage across the open terminals. The Thevenin equivalent resistance (Rth) can be found by short-circuiting the circuit and calculating the total resistance between the open terminals.

## 4. Can a circuit have multiple Thevenin equivalents?

No, a circuit can only have one Thevenin equivalent. However, different equivalent circuits can be found by considering different load conditions.

## 5. How does converting to a Thevenin equivalent affect circuit performance?

Converting to a Thevenin equivalent does not affect the performance of the circuit, as the equivalent circuit represents the same behavior as the original circuit. It only simplifies the analysis process and makes it easier to calculate circuit variables.

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