Thévenin’s theorem – wikipedia 76 gas station credit card login

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• The equivalent resistance R th is the resistance that the circuit between terminals A and B would have if all ideal voltage sources in the circuit were replaced by a short circuit and all ideal current sources were replaced by an open circuit.

• If terminals A and B are connected to one another, the current flowing from A to B will be V th/R th. This means that R th could alternatively be calculated as V th divided by the short-circuit current between A and B when they are connected together.

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. [1] [2] [3] [4] [5] [6]

Thévenin’s theorem and its dual, Norton’s theorem, are widely used to make circuit analysis simpler and to study a circuit’s initial-condition and steady-state response. [7] [8] 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. [6] [9]

The Thévenin-equivalent voltage V Th is the open-circuit 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 V out and the other terminal to be at the ground point.

The Thévenin-equivalent resistance R Th is the resistance measured across points A and B "looking back" into the circuit. The resistance is measured after replacing all voltage- and current-sources with their internal resistances. That means an ideal voltage source is replaced with a short circuit, and an ideal current source is replaced 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.

The replacements of voltage and current sources do what the sources would do if their values were set to zero. A zero valued voltage source would create a potential difference of zero volts between its terminals, regardless of the current that passes through it; its replacement, a short circuit, does the same thing. A zero valued current source passes zero current, regardless of the voltage across it; its replacement, an open circuit, does the same thing. Example [ edit ]

In the example, calculating the equivalent voltage: V T h = R 2 + R 3 ( R 2 + R 3 ) + R 4 ⋅ V 1 {\displaystyle V_{\mathrm {Th} }={R_{2}+R_{3} \over (R_{2}+R_{3})+R_{4}}\cdot V_{\mathrm {1} }} = 1 k Ω + 1 k Ω ( 1 k Ω + 1 k Ω ) + 2 k Ω ⋅ 15 V {\displaystyle ={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 2 ⋅ 15 V = 7.5 V {\displaystyle ={1 \over 2}\cdot 15\,\mathrm {V} =7.5\,\mathrm {V} }

(notice that R 1 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 R 1 and therefore no voltage drop along this part)

Calculating equivalent resistance ( R x ‖ R y {\displaystyle R_{x}\|R_{y}} is the total resistance of two parallel resistors): R T h = R 1 + [ ( R 2 + R 3 ) ‖ R 4 ] {\displaystyle R_{\mathrm {Th} }=R_{1}+\left[\left(R_{2}+R_{3}\right)\|R_{4}\right]} = 1 k Ω + [ ( 1 k Ω + 1 k Ω ) ‖ 2 k Ω ] {\displaystyle =1\,\mathrm {k} \Omega +\left[\left(1\,\mathrm {k} \Omega +1\,\mathrm {k} \Omega \right)\|2\,\mathrm {k} \Omega \right]} = 1 k Ω + ( 1 ( 1 k Ω + 1 k Ω ) + 1 ( 2 k Ω ) ) − 1 = 2 k Ω . {\displaystyle =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 [ edit ]

A Norton equivalent circuit is related to the Thévenin equivalent by R T h = R N o {\displaystyle R_{\mathrm {Th} }=R_{\mathrm {No} }\!} V T h = I N o R N o {\displaystyle V_{\mathrm {Th} }=I_{\mathrm {No} }R_{\mathrm {No} }\!} I N o = V T h / R T h {\displaystyle I_{\mathrm {No} }=V_{\mathrm {Th} }/R_{\mathrm {Th} }\!} Practical limitations [ edit ]

• 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 implemented.

The proof involves two steps. The first step is to use superposition theorem to construct a solution. Then, uniqueness theorem is employed to show that the obtained solution is unique. It is noted that the second step is usually implied in literature.

By using superposition of specific configurations, it can be shown that for any linear "black box" circuit which contains voltage sources and resistors, its voltage is a linear function of the corresponding current as follows V = V E q − Z E q I . {\displaystyle V=V_{\mathrm {Eq} }-Z_{\mathrm {Eq} }I.}

Here, the first term reflects the linear summation of contributions from each voltage source, while the second term measures the contributions from all the resistors. The above expression is obtained by using the fact that the voltage of the black box for a given current I {\displaystyle 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 {\displaystyle I} (Alternatively, one can use an ideal current source of current I {\displaystyle I} ). Moreover, it is straightforward to show that V E q {\displaystyle V_{\mathrm {Eq} }} and Z E q {\displaystyle Z_{\mathrm {Eq} }} are the single voltage source and the single series resistor in question.

As a matter of fact, the above relation between V {\displaystyle V} and I {\displaystyle I} is established by superposition of some particular configurations. Now, the uniqueness theorem guarantees that the result is general. To be specific, there is one and only one value of V {\displaystyle V} once the value of I {\displaystyle I} is given. In other words, the above relation holds true independent of what the "black box" is plugged to. See also [ edit ]