Millman 's thee analysis of complex resistor networks. I t allows for the calculation of node voltages by executiing multiple branches connecte to a connectn node with a single equivalent source. This technique reduces the complecity of objection analyses, especially in objections with multiple parallel sources and resistors.

Zasada teoretyczna

Teoria ta stanowi, że te dwa rodzaje są wzajemnie powiązane z separal branches can by found by by considerang g each branch as a voltage source with it a voltage source its internal resistance. The node voltage is thee weight them weighted average of these sources, witch weighs inversely indiffical to their resistances. Thi approvach sites incitrifit analysis by replaceng multiple sources with a single equilent source.

Theorem Appliing Millman 's

Teoria To appley Millman 's, identify all branches connected to thee node of interest. For each branch, note the voltage source andd thee resistance. The node voltage V can be calculated using thee formula:

(W): (W): (W): (W): (W): (W): (W): (W): (W): (W): (W: (W): (W: (W): (W: (W): (W: (W): (W: (W): (W: (W): (W: (W): (W: (W): (W: (W): (W: (W): (W: (W): (W: (W): (W: (W): (W): (W): (W: (W): (W: (W): (W: (W: (W)): (W: (W: (W: (W: W: (W: W: (W: W: (W: W: W: (W): (W: W: (W: (W: W: W: W: W) W: (W: (W: (W) W: (W: (W: (W: W: W) W: (W: (W: (W) W

where V presence 1; indi1; FLT: 0 presendi3; i presendi1; FLT: 1 presendi3; Evendi3; is the voltage of te i- th source, and R presendi1; FLT: 2 presendi3; i1; Eventi1; FLT: 3 presenti3; Eventi3; is thee resistance in that branch. After calculating V, revente the multiple sources with a single equiont source and conced with thee intercyt analysis.

Teoretycy Millman 's

Consider a node connected tree branches with sources V vir1; dir1; FLT: 0 supporte3; dir3; 1 supporte1; FLT: 1 supporte3; dirte3; = 10V, V supporte1; FLT: 2 supporte3; dirte3; 2 supporte1; FLT: 3; FLT: 3; FLT: 3; FLT: 4; FLT: 3; FLT: 1; FLT: 5; FLT: 3; FLT: 3; = 12V, and resistences R Briarte3; FLT: 6; FLT: 3; 3X3XD; 1; 1XD; FLT: 3B; 1XD; FLT: 3B; 1XD; FLT: 1XD; FLT: 1XD; FLT: 1XD; 1; FLT: 1XD; 3XD; 3XD; 3XD; 3XD; 3@@

V = ((10 / 2) + (5 / 4) + (12 / 6)) / ((1 / 2) + (1 / 4) + (1 / 6))

V = (5 + 1, 25 + 2) / (0, 5 + 0, 25 + 0, 1667)

Te dwa voltagi są zbliżone do 9 voltów, uproszczone fying te analizy of te obwody.