Millman's Theorem

Millman’s theorem provides a shortcut for finding the common voltage across any number of parallel branches with different voltage sources. A typical example is shown in fig. 10-16 For all the branches, the ends at point Y are connected to chassis ground. Furthermore, the opposite ends of all the branches are also connected to the point X. The voltage V/x y, therefore, is the common voltage across all the branches.











Finding the value of V/x y gives the net effect of the all source in determining the voltage at X with respect to chassis ground. To calculate this voltage
 v/x y=V1/R1 + V2/R2 + V3/R3 ……etc
1/R1 + 1/R2 +1/R3
This formula is derived from converting the voltage source to current sources and combining the results. The numerator with V/R terms is the sum of the parallel current sources. The denominator with 1/R terms is the sum of the parallel conductances. The net V/x y then is the form of I/G or 1*R, which is in units of voltage.
CALCULATING V/XY
For the values in Fig.10-16,
Vxy =32/4 + 0/2 – 8/4
1/4 + 1/2 + 1/4
=8 + 0 -2
1
Vxy=6 V
Note the branches 3, V3 is considered negative because it world make point X negative. However, all the resistances are positive. The positive answer for Vxy means that point X is positive with respect to Y.
In branch 2, V2 is zero because this branch has no voltage source however, R2 is still used in the denominator.
This method can be used for any number of branches, but they must all be in parallel, without any series resistance between the branches. In a branch with several resistance, they can be combined as one Rt. When a branch has more than one voltage source, they can be combined algebraically for one Vt.