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Chapter 1: Q5E (page 1)

In Exercise 5, write a matrix equation that determines the loop currents. [M] If MATLAB or another matrix program is available, solve the system for the loop currents.

Short Answer

Expert verified

The matrix equation is

\[\begin{array}{c}Ri = v\\\left[ {\begin{array}{*{20}{c}}{11}&{ - 5}&0&0\\{ - 5}&{10}&{ - 1}&0\\0&{ - 1}&9&{ - 2}\\0&0&{ - 2}&{10}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{I_1}}\\{I{ & _2}}\\{{I_3}}\\{{I_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{50}\\{ - 40}\\{30}\\{ - 30}\end{array}} \right]\end{array}\].

And its solution is \[i = \left[ {\begin{array}{*{20}{c}}{3.68}\\{ - 1.9}\\{2.57}\\{ - 2.49}\end{array}} \right]\].

Step by step solution

01

Write the equation for each loop

In loop 1, apply Ohm鈥檚 law as

\[{I_1} + 5{I_1} + 4{I_1} + {I_1} = 11{I_1}\].

In loop 2, apply the voltage law as

\[11{I_1} - 5{I_2} = 50\].

In loop 2, apply the current law as

\[ - 5{I_1} + 10{I_2} - {I_3} = - 40\].

In loop 3, apply the current law as

\[ - {I_2} + 9{I_3} - 2{I_4} = 30\].

In loop 4, apply the current law as

\[ - 2{I_3} + 10{I_4} = - 30\].

02

Write the matrix equation

The matrix equation of the above system is

\[\left[ {\begin{array}{*{20}{c}}{11}&{ - 5}&0&0\\{ - 5}&{10}&{ - 1}&0\\0&{ - 1}&9&{ - 2}\\0&0&{ - 2}&{10}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{I_1}}\\{I{ & _2}}\\{{I_3}}\\{{I_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{50}\\{ - 40}\\{30}\\{ - 30}\end{array}} \right]\].

That is \[Ri = v\].

03

Solve the system

Use the following MATLAB command to obtain the reduced row echelon form of R:

\[\begin{array}{l} \gg {\rm{R}} = \left[ {11\,\, - 5\,\,\,0\,\,0;\,\, - 5\,\,10\,\, - 1\,\,0;\,\,\,0\,\, - 1\,\,9\,\, - 2;\,\,0\,\,0\,\, - 2\,\,10} \right];\\ \gg {\rm{rref}}\left( {\rm{R}} \right)\end{array}\]

\[R = \left[ {\begin{array}{*{20}{c}}{11}&{ - 5}&0&0&{50}\\{ - 5}&{10}&{ - 1}&0&{ - 40}\\0&{ - 1}&9&{ - 2}&{30}\\0&0&{ - 2}&{10}&{ - 30}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0&{\frac{{265}}{{72}}}\\0&1&0&0&{ - \frac{{137}}{{72}}}\\0&0&1&0&{\frac{{185}}{{72}}}\\0&0&0&1&{ - \frac{{179}}{{72}}}\end{array}} \right]\]

Hence the loop currents are \[{I_1} = \frac{{265}}{{72}} \approx 3.68,{I_2} = - \frac{{137}}{{72}} \approx - 1.9,{I_3} = \frac{{185}}{{72}} \approx 2.57,\] and \[{I_4} = - \frac{{179}}{{72}} \approx - 2.49\]. That is,

\[i = \left[ {\begin{array}{*{20}{c}}{3.68}\\{ - 1.9}\\{2.57}\\{ - 2.49}\end{array}} \right]\].

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