# Thevenin equivalent by source transformation

• Castello
In summary, the conversation discusses the application of Thevenin's Theorem in finding the Thevenin equivalent of a circuit. The method of using source transformation to convert the circuit into a series voltage source and resistor is not always applicable, especially for circuits with dependent sources. Instead, Thevenin's Theorem suggests finding the open circuit voltage and short circuit current, which can be used to determine the equivalent voltage source and series resistance. This method is applicable for all circuits and ensures that the U-I characteristics of the original circuit and its Thevenin equivalent are the same.
Castello

## Homework Statement

Find the Thevenin equivalent at terminals a-b

V1 = 70V
V2 (dependent) = 4Vo
R1 = 10Kohm
R2 = 20Kohm

## Homework Equations

##V=RI##
##1/req = 1/r1 + 1/r2 + ... +1/rn##

## The Attempt at a Solution

Found the current i = 1mA by KVL using Vo = 10ki then did source transformation for both tension sources: 7mA and 2mA, parallel with 10kohm and 20Kohm respectively. Then, calculated the equivalent resistor (20/3)kohm and subtracted the 7mA by the 2mA current sources. Transforming again, It gives me Vth = 33,3V and Rth = 6,6kohm but the answer is 60V and 2.9kohm.

I've tried many times this exercise with source transformation (as some others), had different results, and sometimes looks like that the source transformation doesn't work so generally that i thought (or I'm applying it in a wrong way).

#### Attachments

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Castello said:

## Homework Statement

Find the Thevenin equivalent at terminals a-b

V1 = 70V
V2 (dependent) = 4Vo
R1 = 10Kohm
R2 = 20Kohm

View attachment 236913

## Homework Equations

##V=RI##
##1/req = 1/r1 + 1/r2 + ... +1/rn##

## The Attempt at a Solution

Found the current i = 1mA by KVL using Vo = 10ki then did source transformation for both tension sources: 7mA and 2mA, parallel with 10kohm and 20Kohm respectively. Then, calculated the equivalent resistor (20/3)kohm and subtracted the 7mA by the 2mA current sources. Transforming again, It gives me Vth = 33,3V and Rth = 6,6kohm but the answer is 60V and 2.9kohm.

I've tried many times this exercise with source transformation (as some others), had different results, and sometimes looks like that the source transformation doesn't work so generally that i thought (or I'm applying it in a wrong way).
Determine the Thevenin equivalent emf and resistance from the open circuit voltage Uo and short-circuit current Is between a and b: the emf is equal to Uo and Rth=Uo/Is.

Castello
ehild said:
Determine the Thevenin equivalent emf and resistance from the open circuit voltage Uo and short-circuit current Is between a and b: the emf is equal to Uo and Rth=Uo/Is.

Sorry, i didn't get how to do this. I mean, what i previously understood from source transformation with thevenin theorem is that if i could transform a circuit just in a tension source in series with a resistor, than it would be the Vth and the Rth. Isn't it? This kind of resolution worked in many exercises, and some others not.

Castello said:
Sorry, i didn't get how to do this. I mean, what i previously understood from source transformation with thevenin theorem is that if i could transform a circuit just in a tension source in series with a resistor, than it would be the Vth and the Rth. Isn't it? This kind of resolution worked in many exercises, and some others not.
Yes, according to Thevenin theorem, a circuit between two points can be replaced by a single voltage source and series resistor. The question is, how to do this. In general, the Thevenin source voltage is equal to the voltage between a and b if nothing is connected between these points - the open-circuit voltage (Uo).
To get the Thevenin-equivalent resistance, one needs the short--circuit current (Is) between a and b, that is, the current that would flow between a and b if those points were connected by a single wire with zero resistance. The Thevenin-equivalent resistance is Rth = Uo/Is. This method is general, applicable for all circuits. For circuits with independent sources only, there are other methods to get Rth, but these methods might fail for circuits with dependent sources. Your circuit contains a dependent source.

Castello
This circuit does not appear to be amenable to source transformation (current source to voltage source or vice versa) as a way to reduce the circuit. If you convert the independent voltage source and its series 10 k resistance to an equivalent current source then you would "lose" the sense resistor for Vo. Converting the dependent source and its series resistor to a current source leaves you no further ahead either. So you will need to use another strategy to solve this circuit.

Last edited:
Castello
When connecting different loading resistors to the output of a linear circuit, the relation between the output voltage and current is represented by a straight line, like in the figure. The idea of Thevenin's Theorem is to replace the circuit with a single voltage source and series resistance producing the same U-I characteristics.As a straight line is determined by two points, it is enough if the open circuit voltages (I=0) and short circuit currents (U=0) are the same for the circuit and its Thevenin-equivalent.

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Castello
Ok, now i see that it's not as general as i was thinking. Thank's! @ehild @gneill

## 1. What is Thevenin equivalent by source transformation?

Thevenin equivalent by source transformation is a method used to simplify a complex circuit into a single voltage source and a single resistor. This simplification allows for easier analysis of the circuit.

## 2. How is the Thevenin equivalent voltage calculated?

The Thevenin equivalent voltage is calculated by open-circuiting the load resistance and finding the voltage across the load terminals. This voltage is equal to the Thevenin equivalent voltage.

## 3. What is the purpose of finding the Thevenin equivalent by source transformation?

The purpose of finding the Thevenin equivalent by source transformation is to simplify a complex circuit into a more manageable form for analysis. This allows for easier calculation of circuit parameters such as current and power.

## 4. Can Thevenin equivalent by source transformation be used for any circuit?

Yes, Thevenin equivalent by source transformation can be used for any linear circuit with independent sources. It is particularly useful for circuits with multiple sources and resistors in parallel or series.

## 5. Is Thevenin equivalent by source transformation the only method for circuit simplification?

No, there are other methods for circuit simplification such as Norton equivalent and superposition theorem. Thevenin equivalent by source transformation is just one of the many techniques used by scientists and engineers to analyze and simplify circuits.

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