Millman’s Theorem for AC networks

Millman’s theorem for AC network states that if there as “n” number of voltage sources having magnitude V1, V2, V3,…..Vn hving internal magnitude Z1, Z2, Z3,…..Zn respectively, then these sources may be replaced by a single voltage source Vm having equivalent series internal impedance Zm given by the equations

 E = \dfrac{V_1Y_1 + V_2Y_2 + V_3Y_3 + V_nY_n}{Y_1 + Y_2 + Y_3}   …..(1)

And Z_m = \dfrac{1}{Y_1+Y_2+Y_3 + Y_n}     …..(2)

Where Y1, Y2,…..Yn are the admittances corresponding to impedance Z1, Z2, Z3 …..,Zn (figure 1)

millmans equivalent circuit

Let us see an example of Millman’s theorem

Example 1. Using Millman’s theorem find the current in the load ZL in figure 2.

example of millmans theorem

Solution, Since Millman’s theorem is applicable only to voltage source hence, the current source in the given circuit is converted to voltage source first as shown in figure 3.

reduction of example of millmans theorem

As per Millman’s theorem, the equivalent voltage source is given by.

 E = \dfrac{E_1Y_1 + E_2Y_2 + E_3Y_3}{Y_1 + Y_2 + Y_3}

 

\dfrac{1<0^0\times\dfrac{1}{1}+5<0^0\times\dfrac{1}{1}+25<0^0}{\dfrac{1}{1}\dfrac{1}{1}\dfrac{1}{5}}

 

=\dfrac{1<0^0+5<0^0+0.2\times25<0^0}{1+1+0.2}

 

=\dfrac{11}{2.2} = 5<0^0 V.

 

Therefore, I_L = \dfrac{E}{Z_L}=\dfrac{5<0^0}{2+j4} = 1.12<-63.43^0 A.