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Thevenin's Theorem Thevenin's theorem for electrical networks states that any combination of voltage sources and resistors with two terminals is electrically equivalent to a single voltage source V and a single series resistor R. The theorem can also be applied to general impedances, not just resistors. The theorem was first discovered by German scientist Hermann von Helmholtz in 1853, but was then rediscovered in 1883 by French telegraph engineer Lon Charles Thvenin (1857-1926). To calculate the equivalent circuit: - Remove the load circuit.
- Calculate the voltage, V, at the output from the original sources.
- Now replace voltage sources with shorts and current sources with open circuits.
- Replace the load circuit with an imaginary ohm meter and measure the total resistance, R,looking back into the circuit, with the sources removed.
- The equivalent circuit is a voltage source with voltage V in series with a resistance R in series with the load.
The Thevenin-equivalent voltage is the voltage at the output terminals of the original circuit. When calculating a Thevenin-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. In the example, -
V_\mathrm{AB} = {R_2 + R_3 \over (R_2 + R_3) + R_4} \cdot V_\mathrm{1} -
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= {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} -
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= {1 \over 2} \cdot 15 \mathrm{V} = 7.5 \mathrm{V} The Thevenin-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. In the example, -
R_\mathrm{AB} = \left ( \left ( 1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega \right ) \| 2\,\mathrm{k}\Omega \right ) + 1\,\mathrm{k}\Omega = 2\,\mathrm{k}\Omega See also External link
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