Question 2.66: In the circuit of Fig. 1, determine V0 using Millman’s theor......

In the given circuit the parallel branches with 8 Ω and 4 Ω resistances can be assumed to have a zero value voltage source as shown in Fig. 2. In the circuit of Fig. 2, each source has a series resistance, which can be considered as internal resistance of the source. Therefore, the parallel connected voltage sources with internal resistance can be converted into a single equivalent source using Millman’s theorem.

Let, E_{eq}= Equivalent emf of parallel connected source

R_{eq}= Equivalent internal resistance.

Now, by Millman’s theorem,

\mathrm{R_{eq}}={\frac{1}{{\frac{1}{\mathrm{R_{1}}}}+{\frac{1}{\mathrm{R_{2}}}}+{\frac{1}{\mathrm{R_{3}}}}}}={\frac{1}{{\frac{1}{4}}+{\frac{1}{8}}+{\frac{1}{4}}}}={\frac{1}{0.625}}\,=\,1.6\,\Omega

{E}_{\mathrm{eq}}\;=\;\left({\frac{{E}_{1}}{{R}_{1}}}+{\frac{{E}_{2}}{{R}_{2}}}+{\frac{{E}_{3}}{{R}_{3}}}\right){R}_{\mathrm{eq}}\;=\;\left({\frac{10}{4}}+{\frac{0}{8}}+{\frac{0}{4}}\right)\times1.6\;=\;4\,V

The circuit of Fig. 2, can be redrawn as shown in Fig. 3. With reference to Fig. 3, by voltage division rule, we can write,

\begin{array}{r}{\operatorname{V}_{0}\,=\,\operatorname{E}_{\mathrm{eq}}\times{\frac{5}{5+\left(1.6+1.4\right)}}\,=\,4\times{\frac{5}{8}}\,=\,2.5\,V}\end{array}