Question 2.65: In the circuit of Fig. 1, use Millman’s theorem to find curr......

The given circuit can be redrawn as shown in Fig. 2. In the circuit of Fig. 2 each voltage source has a series resistance which can be considered as internal resistance of the source. Hence, 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 sources

$R_{eq}$=Equivalent internal resistance.

Now, by Millman’s theorem,

$\mathrm{R_{eq}}={\frac{1}{\frac{1}{\mathrm{R_{1}}}+{\frac{1}{\mathrm{R_{2}}}}+{\frac{1}{\mathrm{R_{3}}}}}}={\frac{1}{\frac{1}{\frac{1}{8} +{\frac{1}{2}}+{\frac{1}{10}}}}}={\frac{1}{0.725}}\,=\,1.3793\,\Omega$

${E}_{\mathrm{eq}}\ =\left({\frac{{E}_{1}}{{R}_{1}}}+{\frac{{E}_{2}}{{R}_{2}}}+{\frac{{E}_{3}}{{R}_{3}}}\right){R}_{\mathrm{eq}}$

$=\left({\frac{20}{8}}+{\frac{10}{2}}+{\frac{5}{10}}\right)\times1.3793\,=\,11.0344\,V$

The circuit of Fig. 2 can be redrawn as shown in Fig. 3. Let, I be the current through 4 Ω resistance. With reference to Fig. 3, by Ohm’s law we can write,

$I=\;{\frac{\mathrm{E_{eq}}}{\mathrm{R_{eq}}+4}}\;=\;{\frac{11.0344}{1.3793+4}}\;=\;2.0513\,A$

RESULT

Current through 4 Ω resistance = 2.0513 A