Question 2.65: In the circuit of Fig. 1, use Millman’s theorem to find curr......
In the circuit of Fig. 1, use Millman’s theorem to find current through the 4 Ω resistance.
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
