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The theorem of Millman is thus named in the honor of the American electronics specialist Jacob Millman.

In an electrical communication of branches in parallel, including/understanding each one a perfect generator of tension in series with a linear element, the terminal voltage of the branches is equal to the sum of the electromotive forces respectively multiplied by the Admittance of the branch, the whole divided by the sum of the admittances.

$V_m= \ frac {\ sum_ {k=1} ^N E_k.Y_k} {\ sum_ {k=1} ^N Y_k} = \ frac {\ sum_ {k=1} ^N \ frac {E_k} {Z_k}} {\ sum_ {k=1} ^N \ frac {1} {Z_k}}$

In the particular case of an electrical communication made up of resistances:

$V_m= \ frac {\ sum_ {k=1} ^N E_k.G_k} {\ sum_ {k=1} ^N G_k} = \ frac {\ sum_ {k=1} ^N \ frac {E_k} {R_k}} {\ sum_ {k=1} ^N \ frac {1} {R_k}}$

With G, the conductance.

Demonstration of the Theorem of Millman

One considers the diagram above.

Like the branches (Zk; Ek) are in parallel, one works with the Admittance S $Y_ {K} = \ frac {1} {Z_ {K}}$ and of the transformations Thévenin-Norton: $IN_ {K} =E_ {K} \ times Y_ {K}$ (convention generator)

For each branch (source of tension and impedance), one obtains, according to the law of ohm: $I_ {K} =Y_ {K} \ times (V_ {m} - E_ {K})$

Then, according to the law of the nodes, one a: $\ sum_ {k=1} ^N I_ {K} =0$

that is to say

$\ sum_ {k=1} ^N Y_ {K} \ times (V_ {m} - E_ {K}) =0$

while developing:

$\ sum_ {k=1} ^N Y_ {K} \ times V_ {m} = \ sum_ {k=1} ^N Y_ {K} \ times E_ {K}$

from where:

$V_ {m} = \ frac {\ sum_ {k=1} ^N E_ {K}. Y_ {K}} {\ sum_ {k=1} ^N Y_ {K}} = \ frac {\ sum_ {k=1} ^N \ frac {E_k} {Z_k}} {\ sum_ {k=1} ^N \ frac {1} {Z_k}}$

Practical case: This theorem is pleasant to use so moreover $V_ {m}$ is null (for example, the differential tension of a AOP in linear mode).

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Unua Libro

Demonstration of the Theorem of Millman

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