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# Thévenin equivalent

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### Thévenin equivalent

Thévenin's theorem holds, to illustrate in DC circuit theory terms, that (see image):

• Any linear electrical network with voltage and current sources and only resistances can be replaced at terminals A-B by an equivalent voltage source Vth in series connection with an equivalent resistance Rth.
• This equivalent voltage Vth is the voltage obtained at terminals A-B of the network with terminals A-B open circuited.
• This equivalent resistance Rth is the resistance obtained at terminals A-B of the network with all its independent current sources open circuited and all its independent voltage sources short circuited.

For AC systems, the theorem can be applied to reactive impedances as well as resistances.

The theorem was independently derived in 1853 by the German scientist Hermann von Helmholtz and in 1883 by Léon Charles Thévenin (1857–1926), an electrical engineer with France's national Postes et Télégraphes telecommunications organization.

Thévenin's theorem and its dual, Norton's theorem, are widely used for circuit analysis simplification and to study circuit's initial-condition and steady-state response. Thévenin's theorem can be used to convert any circuit's sources and impedances to a Thévenin equivalent; use of the theorem may in some cases be more convenient than use of Kirchhoff's circuit laws.

## Calculating the Thévenin equivalent

To calculate the equivalent circuit, the resistance and voltage are needed, so two equations are required. These two equations are usually obtained by using the following steps, but any conditions placed on the terminals of the circuit should also work:

1. Calculate the output voltage, VAB, when in open circuit condition (no load resistor—meaning infinite resistance). This is VTh.
2. Calculate the output current, IAB, when the output terminals are short circuited (load resistance is 0). RTh equals VTh divided by this IAB.

The equivalent circuit is a voltage source with voltage VTh in series with a resistance RTh.

Step 2 could also be thought of as:

2a. Replace the independent voltage sources with short circuits, and independent current sources with open circuits.
2b. Calculate the resistance between terminals A and B. This is RTh.

The Thévenin-equivalent voltage is the voltage at the output terminals of the original circuit. When calculating a Thévenin-equivalent voltage, the voltage divider principle is often useful, by declaring one terminal to be Vout and the other terminal to be at the ground point.

The Thévenin-equivalent resistance is the resistance measured across points A and B "looking back" into the circuit. It is important to first replace all voltage- and current-sources with their internal resistances. For an ideal voltage source, this means replace the voltage source with a short circuit. For an ideal current source, this means replace the current source with an open circuit. Resistance can then be calculated across the terminals using the formulae for series and parallel circuits. This method is valid only for circuits with independent sources. If there are dependent sources in the circuit, another method must be used such as connecting a test source across A and B and calculating the voltage across or current through the test source.

### Example

In the example, calculating the equivalent voltage:



V_\mathrm{Th} = {R_2 + R_3 \over (R_2 + R_3) + R_4} \cdot V_\mathrm{1}



= {1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega \over (1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega) + 2\,\mathrm{k}\Omega} \cdot 15 \,\mathrm{V}



= {1 \over 2} \cdot 15 \, \mathrm{V} = 7.5 \,\mathrm{V} (notice that R1 is not taken into consideration, as above calculations are done in an open circuit condition between A and B, therefore no current flows through this part, which means there is no current through R1 and therefore no voltage drop along this part)

Calculating equivalent resistance:



R_\mathrm{Th} = R_1 + \left[ \left( R_2 + R_3 \right) \| R_4 \right]



= 1\,\mathrm{k}\Omega + \left[ \left( 1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega \right) \| 2\,\mathrm{k}\Omega \right]



= 1\,\mathrm{k}\Omega + \left({1 \over ( 1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega )} + {1\over (2\,\mathrm{k}\Omega ) }\right)^{-1} = 2\,\mathrm{k}\Omega.

## Conversion to a Norton equivalent

Main article: Norton's theorem

A Norton equivalent circuit is related to the Thévenin equivalent by the following:

$R_\mathrm\left\{Th\right\} = R_\mathrm\left\{No\right\} \!$
$V_\mathrm\left\{Th\right\} = I_\mathrm\left\{No\right\} R_\mathrm\left\{No\right\} \!$
$I_\mathrm\left\{No\right\} = V_\mathrm\left\{Th\right\} / R_\mathrm\left\{Th\right\}. \!$

## Practical limitations

• Many, if not most circuits are only linear over a certain range of values, thus the Thévenin equivalent is valid only within this linear range and may not be valid outside the range.
• The Thévenin equivalent has an equivalent I–V characteristic only from the point of view of the load.
• The power dissipation of the Thévenin equivalent is not necessarily identical to the power dissipation of the real system. However, the power dissipated by an external resistor between the two output terminals is the same regardless of how the internal circuit is represented.

## A proof of the theorem

The proof involves two steps. First use superposition theorem to construct a solution, and then use uniqueness theorem to show the solution is unique. The second step is usually implied. Firstly, using the superposition theorem, in general for any linear "black box" circuit which contains voltage sources and resistors, one can always write down its voltage as a linear function of the corresponding current as follows

$V = V_\mathrm\left\{Eq\right\}-Z_\mathrm\left\{Eq\right\}I \!$

where the first term reflects the linear summation of contributions from each voltage source, while the second term measures the contribution from all the resistors. The above argument is due to the fact that the voltage of the black box for a given current $I$ is identical to the linear superposition of the solutions of the following problems: (1) to leave the black box open circuited but activate individual voltage source one at a time and, (2) to short circuit all the voltage sources but feed the circuit with a certain ideal voltage source so that the resulting current exactly reads $I$ (or an ideal current source of current $I$). Once the above expression is established, it is straightforward to show that $V_\mathrm\left\{Eq\right\}$ and $Z_\mathrm\left\{Eq\right\}$ are the single voltage source and the single series resistor in question.

## See also

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