Question

In Exercises 5-8, 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

\(\left[ {\begin{array}{*{20}{c}}{12.11}\\{8.44}\\{4.26}\\{1.90}\end{array}} \right]\)

Step-by-step solution

Find the resistance vector for loop 1

In loop 1, currents \({I_3}\) and \({I_4}\) are not flowing. Current \({I_1}\) has four \(RI\) voltage drops, and \({I_2}\) is negative as it flows in theopposite direction.

So, the resistance vector for loop 1 is

\({r_1} = \left[ {\begin{array}{*{20}{c}}4\\{ - 1}\\0\\0\end{array}} \right]\).

Find the resistance vector for loop 2

In loop 2, current \({I_4}\) is not flowing. Current \({I_1}\) has a negative voltage drop and \({I_2}\) has six \(RI\) drops. Voltage drop for \({I_3}\) is negative.

So, the resistance vector for loop 2 is

\({r_2} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\6\\{ - 2}\\0\end{array}} \right]\).

Find the resistance vector for loop 3

In loop 3, current \({I_1}\) is not flowing. Current \({I_2}\) has a negative voltage drop. \({I_3}\) has ten \(RI\) drops, and \({I_4}\) has a negative voltage drop.

So, the resistance vector for loop 3 is

\({r_3} = \left[ {\begin{array}{*{20}{c}}0\\{ - 2}\\{10}\\{ - 3}\end{array}} \right]\).

Find the resistance vector for loop 4

Currents \({I_1}\) and \({I_2}\) are not flowing in loop 4. Current \({I_3}\) has a negative voltage drop and \({I_4}\) has 12 \(RI\) drops in loop 4. So, the resistance vector for loop 4 is

\({r_4} = \left[ {\begin{array}{*{20}{c}}0\\0\\{ - 3}\\{12}\end{array}} \right]\).

Form the equivalent matrix

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{{r_1}}&{{r_2}}&{{r_3}}&{{r_4}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{I_1}}\\{{I_2}}\\{{I_3}}\\{{I_4}}\end{array}} \right] = \left[ v \right]\\\left[ {\begin{array}{*{20}{c}}4&{ - 1}&0&0\\{ - 1}&6&{ - 2}&0\\0&{ - 2}&{10}&{ - 3}\\0&0&{ - 3}&{12}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{I_1}}\\{{I_2}}\\{{I_3}}\\{{I_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{40}\\{30}\\{20}\\{10}\end{array}} \right]\end{array}\)

Convert the matrix into row-reduced echelon form

Consider the matrix \(A = \left[ {\begin{array}{*{20}{c}}4&{ - 1}&0&0&{ - 40}\\{ - 1}&6&{ - 2}&0&{ - 30}\\0&{ - 2}&{10}&{ - 3}&{ - 20}\\0&0&{ - 3}&{12}&{ - 10}\end{array}} \right]\).

Use the code in MATLAB to obtain the row-reducedechelon form, as shown below:

\[\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ {\begin{array}{*{20}{c}}4&{ - 1}&0&0&{ - 40}\end{array};{\rm{ }}\begin{array}{*{20}{c}}{ - 1}&6&{ - 2}&0&{ - 30}\end{array};{\rm{ }}\begin{array}{*{20}{c}}0&{ - 2}&{10}&{ - 3}&{ - 20}\end{array};{\rm{ }}\begin{array}{*{20}{c}}0&0&{ - 3}&{12}&{ - 10}\end{array}{\rm{ }}} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\]

\(\left[ {\begin{array}{*{20}{c}}4&{ - 1}&0&0&{ - 40}\\{ - 1}&6&{ - 2}&0&{ - 30}\\0&{ - 2}&{10}&{ - 3}&{ - 20}\\0&0&{ - 3}&{12}&{ - 10}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0&{ - 12.11}\\0&1&0&0&{ - 8.44}\\0&0&1&0&{ - 4.26}\\0&0&0&1&{ - 1.90}\end{array}} \right]\)

Find the general solution for loop currents using the echelon form

\(\left[ {\begin{array}{*{20}{c}}{{I_1}}\\{{I_2}}\\{{I_3}}\\{{I_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{12.11}\\{8.44}\\{4.26}\\{1.90}\end{array}} \right]\)

So, the loop currents in the given circuit are \(\left[ {\begin{array}{*{20}{c}}{12.11}\\{8.44}\\{4.26}\\{1.90}\end{array}} \right]\).