Question 2.67: In the circuit of Fig. 1, apply Millman’s theorem to find Th......

Let us remove \overline{Z}_L and redraw the circuit of Fig. 1 as shown in Fig. 2. In the circuit of Fig. 2, each voltage source has a series impedance which can be considered as internal impedance of the source.

Therefore, the parallel connected voltage sources with internal impedance can be converted into a single equivalent source using Millman’s theorem.

Let, \overline{E}_{\mathrm{eq}}\ = Equivalent emf of parallel connected sources.

\overline{Z}_{\mathrm{eq}}\ = Equivalent internal impedance.

Now by Millman’s theorem,

\overline{{{Z}}}\mathrm{eq}\,=\,\frac{1}{\frac{1}{\overline{{{Z}}}_{1}}+\frac{1}{\overline{{{Z}}}_{2}}}\,=\,\frac{1}{\frac{1}{4+\mathrm{j}4}+\frac{1}{2-\mathrm{j}6}}

=\left[(4+\mathrm{j}4)^{-1}+(2-\mathrm{j}6)^{-1}\right]^{-1}=\,5.6-\mathrm{j}0.8\,\Omega

\overline{E}_{\mathrm{eq}}\;=\;\left(\frac{\overline{E}_{1}}{\overline{Z}_{1}}+\frac{\overline{E}_{2}}{\overline{Z}_{2}}\right)\overline{{{Z}}}_{\mathrm{eq}}\;=\;\left(\frac{20\angle0^{\circ}}{4+j4}+\frac{12\angle\ 90^{\circ}}{2-j6}\right)\times\left(5.6-j0.8\right)

Now, the parallel connected sources in Fig. 2 can be represented as shown in Fig. 3 by Millman’s theorem.

∴ \overline{{{\mathrm{V}}}}_{\mathrm{th}}\,=\,2.4\,{-}\,\mathrm{j}\,1\,1.2\,V\,=\,1.45432∠-77.9^{\circ}\,V

The Thevenin’s equivalent of the given circuit at terminals A-B is shown in Fig. 4. Let us connect the load impedance \overline{Z}_L  at terminals A-B of Thevenin’s equivalent as shown in Fig. 5. Now, by maximum power transfer theorem, for maximum power transfer to\overline{Z}_L , the value of \overline{Z}_L should be conjugate of \overline{Z}_{th} .