Chapter 11.2, Problem 83AYU

To determine

The currents $I_1,I_2,I_3,I_4$ by using the system of equations

$\{\begin{array}{l}{I}_1+I_2+I_4=I_3\\ -200I_1-40+80I_2=0\\ 360-20I_3-80I_2+40=0\\ -80+70I_4-200I_1=0\end{array}$. The circuit is,

Answer to Problem 83AYU

Solution:

The currents are $I_1=1,I_2=3,I_3=8,I_4=4$.

Explanation of Solution

Given information:

The system of equation is

$\{\begin{array}{l}{I}_1+I_2+I_4=I_3\\ -200I_1-40+80I_2=0\\ 360-20I_3-80I_2+40=0\\ -80+70I_4-200I_1=0\end{array}$

Explanation:

The given system of equations can be written as,

$\{\begin{array}{l}{I}_1+I_2-I_3+I_4=0\\ -200I_1+80I_2=40\\ -80I_2-20I_3=-400\\ -200I_1+70I_4=80\end{array}$

The system has four equation and four variables.

Therefore, the augmented matrix of the system of the equation is $4\times 5$ matrix as below,

$\left[\begin{array}{rrrrr}\hfill 1& \hfill 1& \hfill -1& \hfill 1& \hfill 0\\ \hfill -200& \hfill 80& \hfill 0& \hfill 0& \hfill 40\\ \hfill 0& \hfill -80& \hfill -20& \hfill 0& \hfill -400\\ \hfill -200& \hfill 0& \hfill 0& \hfill 70& \hfill 80\end{array}\right]$

By performing the row operations $\frac{1}{40}{r}_2,\frac{1}{20}{r}_3,\frac{1}{10}{r}_4$,

$\Rightarrow \left[\begin{array}{rrrrr}\hfill 1& \hfill 1& \hfill -1& \hfill 1& \hfill 0\\ \hfill -5& \hfill 2& \hfill 0& \hfill 0& \hfill 1\\ \hfill 0& \hfill -4& \hfill -1& \hfill 0& \hfill -20\\ \hfill -20& \hfill 0& \hfill 0& \hfill 7& \hfill 8\end{array}\right]$

Now, perform $R_2=r_2+5r_1,r_4=r_4+20r_1$,

$\Rightarrow \left[\begin{array}{rrrrr}\hfill 1& \hfill 1& \hfill -1& \hfill 1& \hfill 0\\ \hfill 0& \hfill 7& \hfill -5& \hfill 5& \hfill 1\\ \hfill 0& \hfill -4& \hfill -1& \hfill 0& \hfill -20\\ \hfill 0& \hfill 20& \hfill -20& \hfill 27& \hfill 8\end{array}\right]$

Now perform $R_2=\frac{1}{7}\cdot r_2$,

$\Rightarrow \left[\begin{array}{rrrrr}\hfill 1& \hfill 1& \hfill -1& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill \frac{-5}{7}& \hfill \frac{5}{7}& \hfill \frac{1}{7}\\ \hfill 0& \hfill -4& \hfill -1& \hfill 0& \hfill -20\\ \hfill 0& \hfill 20& \hfill -20& \hfill 27& \hfill 8\end{array}\right]$

Now, $R_3=r_3+4r_2,R_4=r_4-20r_2$,

$\Rightarrow \left[\begin{array}{rrrrr}\hfill 1& \hfill 1& \hfill -1& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill \frac{-5}{7}& \hfill \frac{5}{7}& \hfill \frac{1}{7}\\ \hfill 0& \hfill 0& \hfill \frac{-27}{7}& \hfill \frac{20}{7}& \hfill \frac{-136}{7}\\ \hfill 0& \hfill 0& \hfill \frac{-40}{7}& \hfill \frac{89}{7}& \hfill \frac{36}{7}\end{array}\right]$

Now perform $R_3=-\frac{7}{27}\cdot r_3$,

$\Rightarrow \left[\begin{array}{rrrrr}\hfill 1& \hfill 1& \hfill -1& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill \frac{-5}{7}& \hfill \frac{5}{7}& \hfill \frac{1}{7}\\ \hfill 0& \hfill 0& \hfill 1& \hfill -\frac{20}{27}& \hfill \frac{136}{27}\\ \hfill 0& \hfill 0& \hfill \frac{-40}{7}& \hfill \frac{89}{7}& \hfill \frac{36}{7}\end{array}\right]$

Now, $R_4=r_4+\frac{40}{7}{r}_3$,

$\Rightarrow \left[\begin{array}{rrrrr}\hfill 1& \hfill 1& \hfill -1& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill \frac{-5}{7}& \hfill \frac{5}{7}& \hfill \frac{1}{7}\\ \hfill 0& \hfill 0& \hfill 1& \hfill -\frac{20}{27}& \hfill \frac{136}{27}\\ \hfill 0& \hfill 0& \hfill 0& \hfill \frac{1603}{189}& \hfill \frac{6412}{189}\end{array}\right]$

Now perform $R_4=\frac{189}{1603}{R}_4$,

$\Rightarrow \left[\begin{array}{rrrrr}\hfill 1& \hfill 1& \hfill -1& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill \frac{-5}{7}& \hfill \frac{5}{7}& \hfill \frac{1}{7}\\ \hfill 0& \hfill 0& \hfill 1& \hfill -\frac{20}{27}& \hfill \frac{136}{27}\\ \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 4\end{array}\right]$

Now $R_1=R_1-R_4,R_2=R_2-\frac{5}{7}{R}_4,R_3=R_3+\frac{20}{27}{R}_4$,

$\Rightarrow \left[\begin{array}{rrrrr}\hfill 1& \hfill 1& \hfill -1& \hfill 0& \hfill -4\\ \hfill 0& \hfill 1& \hfill \frac{-5}{7}& \hfill 0& \hfill \frac{-19}{7}\\ \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 8\\ \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 4\end{array}\right]$

Now $R_1=R_1+R_3,R_2=R_2+\frac{5}{7}{R}_3$,

$\Rightarrow \left[\begin{array}{rrrrr}\hfill 1& \hfill 1& \hfill 0& \hfill 0& \hfill 4\\ \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 3\\ \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 8\\ \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 4\end{array}\right]$

Now $R_1=R_1-R_2$,

$\Rightarrow \left[\begin{array}{rrrrr}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 1\\ \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill 3\\ \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 8\\ \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 4\end{array}\right]$

$\Rightarrow I_1=1,I_2=3,I_3=8,I_4=4$

Therefore, the currents are $I_1=1,I_2=3,I_3=8,I_4=4$.