Refer to Fig. 3.38. Use mesh analysis to obtain an expression for the voltages ν1 , ν2 , ν3 .

Question

Refer to Fig. 3.38. Use mesh analysis to obtain an expression for the voltages [latex] ν_1, ν_2, ν_3 [/latex]

Accepted Answer

There are four meshes, as shown in Fig. 3.38 , but two of the mesh currents ( [latex] i_0[/latex] and [latex] 3 i_0[/latex] ) are known. Thus we need only the two equations for the unknown mesh currents [latex] i_1, i_2[/latex] .

[latex] \begin{array}{r} -R_1 i_0+\left(R_1+R_2+R_3 ight) i_1-R_3 i_2=0 \\ -3 R_4 i_0-R_3 i_1+\left(R_3+R_4+R_5 ight) i_2=0 \end{array} [/latex]

To simplify notation, let

[latex] R_A=R_1+R_2+R_3, \quad R_B=R_3+R_4+R_5 . [/latex]

Then

[latex] R_A i_1-R_3 i_2=R_1 i_0, \quad R_3 i_1+R_B i_2=3 R_4 i_0 [/latex]

which yield

[latex] i_1=\frac{R_1 R_B+3 R_3 R_4}{R_A R_B-R_3^2} i_0, i_2=\frac{R_1 R_3+3 R_A R_4}{R_A R_B-R_3^2} i_0, [/latex]

and thus

[latex] \begin{aligned} &ν_1=\left(i_0-i_1 ight) R_1=\frac{R_A R_B-R_3^2-R_B R_1-3 R_4 R_3}{R_A R_B-R_3{ }^2} R_1 i_0, \\ &ν_2=\left(i_1-i_2 ight) R_3=\frac{R_B R_1+3 R_3 R_4-R_1 R_3-3 R_A R_4}{R_A R_B-R_3^2} R_3 i_0, \\ &ν_3=i_2 R_5=\frac{R_1 R_3+3 R_A R_4}{R_A R_B-R_3{ }^2} R_5 i_0 . \end{aligned} [/latex]

Question 3.15: Refer to Fig. 3.38. Use mesh analysis to obtain an expressio...

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