91Ó°ÊÓ

In loop 1, current \({I_3}\) is not flowing. Current \({I_1}\) has three \(RI\) voltage drops and \({I_2}\) is negative as it flows in the opposite direction. The voltage drop for \({I_4}\) is negative.

So, the resistance vector for loop 1 is

\({r_1} = \left[ {\begin{array}{*{20}{c}}{12}\\{ - 7}\\0\\{ - 4}\end{array}} \right]\).

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

So, the resistance vector for loop 2 is

\({r_2} = \left[ {\begin{array}{*{20}{c}}{ - 7}\\{15}\\{ - 6}\\0\end{array}} \right]\).

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

So, the resistance vector for loop 3 is

\({r_3} = \left[ {\begin{array}{*{20}{c}}0\\{ - 6}\\{14}\\{ - 5}\end{array}} \right]\).

In loop 4, current \({I_2}\) is not flowing. Currents \({I_1}\) and \({I_3}\) have negative voltage drops. Current \({I_4}\) has three drops.

So, the resistance vector for loop 4 is

\({r_4} = \left[ {\begin{array}{*{20}{c}}{ - 4}\\0\\{ - 5}\\{13}\end{array}} \right]\).

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{{r_1}}&{{r_2}}&{{r_3}}&{{r_4}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{I_1}}\\{{I_2}}\\{{I_3}}\\{{I_4}}\end{array}} \right] = \left[ v \right]\\\left[ {\begin{array}{*{20}{c}}{12}&{ - 7}&0&{ - 4}\\{ - 7}&{15}&{ - 6}&0\\0&{ - 6}&{14}&{ - 5}\\{ - 4}&0&{ - 5}&{13}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{I_1}}\\{{I_2}}\\{{I_3}}\\{{I_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{40}\\{30}\\{20}\\{ - 10}\end{array}} \right]\end{array}\)

Consider the matrix \(A = \left[ {\begin{array}{*{20}{c}}{12}&{ - 7}&0&{ - 4}&{ - 40}\\{ - 7}&{15}&{ - 6}&0&{ - 30}\\0&{ - 6}&{14}&{ - 5}&{ - 20}\\{ - 4}&0&{ - 5}&{13}&{10}\end{array}} \right]\).

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

\[\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ {\begin{array}{*{20}{c}}{12}&{ - 7}&0&{ - 4}&{ - 40}\end{array};{\rm{ }}\begin{array}{*{20}{c}}{ - 7}&{15}&{ - 6}&0&{ - 30}\end{array};{\rm{ }}\begin{array}{*{20}{c}}0&{ - 6}&{14}&{ - 5}&{ - 20}\end{array};{\rm{ }}\begin{array}{*{20}{c}}{ - 4}&0&{ - 5}&{13}&{10}\end{array}{\rm{ }}} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\]

\(\left[ {\begin{array}{*{20}{c}}{12}&{ - 7}&0&{ - 4}&{ - 40}\\{ - 7}&{15}&{ - 6}&0&{ - 30}\\0&{ - 6}&{14}&{ - 5}&{ - 20}\\{ - 4}&0&{ - 5}&{13}&{10}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0&{ - 11.43}\\0&1&0&0&{ - 10.55}\\0&0&1&0&{ - 8.04}\\0&0&0&1&{ - 5.84}\end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}}{{I_1}}\\{{I_2}}\\{{I_3}}\\{{I_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{11.43}\\{10.55}\\{8.04}\\{5.94}\end{array}} \right]\)

So, the loop currents in the given circuit are \(\left[ {\begin{array}{*{20}{c}}{11.43}\\{10.55}\\{8.04}\\{5.84}\end{array}} \right]\).