Let us now consider the type of problem encountered in the introduction to mesh and nodal analysis in Chapter 8. Mesh analysis was applied to the network of Fig. 9.103 (Example 8.14). Let us now use Millman’s theorem to find the current through the 2 Ω resistor and compare the results.

Question

Accepted Answer

a. Let us first apply each step and, in the (b) solution, Eq. (9.10). Converting sources yields Fig. 9.104. Combining sources and parallel conductance branches (Fig. 9.105) yields

[latex]E_{eq}=\frac{\pm\frac{E_1}{R_1}\pm\frac{E_2}{R_2}\pm\frac{E_3}{R_3}\pm\cdot \cdot \cdot \pm\frac{E_N}{R_N} }{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\cdot \cdot \cdot +\frac{1}{R_N} } [/latex] (9.10)

[latex] I_T = I_1 + I_2 = 5 A +\frac{5}{3}A=\frac{15}{3}A+\frac{5}{3}A=\frac{20}{3}A\[0.5cm] G_T = G_1 + G_2 = 1 S +\frac{1}{6}S=\frac{6}{6}S+\frac{1}{6}S=\frac{7}{6}S[/latex]

Converting the current source to a voltage source (Fig. 9.106), we obtain

[latex]E_{eq}=\frac{I_T}{G_T}=\frac{\frac{20}{3}A }{\frac{7}{6}S }=\frac{(6)(20)}{(3)(7)}V=\frac{40}{7}V [/latex]

and [latex]R_{eq}=\frac{1}{G_T}=\frac{1}{\frac{7}{6}S }=\frac{6}{7}\Omega [/latex]

so that

[latex]I_{2\Omega}=\frac{E_{eq}}{R_{eq}+R_3}=\frac{\frac{40}{7}V }{\frac{6}{7}\Omega+2\Omega }= \frac{\frac{40}{7}V }{\frac{6}{7}\Omega+\frac{14}{7} \Omega }=\frac{40V}{20 Ω}=2 A [/latex]

which agrees with the result obtained in Example 8.14.

b. Let us now simply apply the proper equation, Eq. (9.10):

[latex]E_{eq}=\frac{+\frac{5V}{1\Omega}+\frac{10V}{6\Omega} }{\frac{1}{1\Omega}+\frac{1}{6\Omega} }= \frac{\frac{30V}{6\Omega}+\frac{10V}{6\Omega} }{\frac{6}{6\Omega}+\frac{1}{6\Omega} }=\frac{40}{7}V [/latex]

and

[latex]R_{eq}=\frac{1}{\frac{1}{1\Omega}+\frac{1}{6\Omega} }=\frac{1}{\frac{6}{6\Omega}+\frac{1}{6\Omega} }=\frac{1}{\frac{7}{6}S }=\frac{6}{7}\Omega [/latex]

which are the same values obtained above.

Question 9.20: Let us now consider the type of problem encountered in the i...

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