Q. 84 - Precalculus Enhanced with Graphing Utilities

Question

Electricity: Kirchhoff’s Rules An application of Kirchhoff’s Rules to the circuit shown results in the following system of equations:

$\left\{\begin{array}{c}{I}_1=I_3+I_2\\ 24-6I_1-3I_3=0\\ 12+24-6I_1-6I_2=0\end{array}\right.$

Find the currents $I_1, I_{2,} \mathrm{and} I_3.$

Step-by-Step Solution

Answer

The value of the currents are $I_1 =\frac{7}{2}, I_2=\frac{5}{2}, I_3=1.$

Step 1. Given Information

The given system of the equations are

$\left\{\begin{array}{c}{I}_1=I_3+I_2\\ 24-6I_1-3I_3=0\\ 12+24-6I_1-6I_2=0\end{array}\right.$

We have to find the currents $I_1, I_2, a\mathrm{nd} I_3.$

Step 2. Rewrite the system

By rewriting the system we get,

$\left\{\begin{array}{c}{I}_1-I_3-I_2=0\\ 2I_1+0I_2+I_3=8\\ I_1+I_2+0I_3=6\end{array}\right.$

Step 3. Solve

Let's write the augmented matrix that represents the system  

$$\begin{gathered} \left[\begin{array}{ccccc}1& -1& -1& |& 0\\ 2& 0& 1& |& 8\\ 1& 1& 0& |& 6\end{array}\right] \\ {R}_2=r_2-2r_1 \\ \left[\begin{array}{ccccc}1& -1& -1& |& 0\\ 0& 2& 3& |& 8\\ 1& 1& 0& |& 6\end{array}\right] \\ {R}_3=r_3-r_1 \\ \left[\begin{array}{ccccc}1& -1& -1& |& 0\\ 0& 2& 3& |& 8\\ 0& 2& 1& |& 6\end{array}\right] \\ {R}_2=\frac{1}{2}{r}_2 \\ \left[\begin{array}{ccccc}1& -1& -1& |& 0\\ 0& 1& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& |& 4\\ 0& 2& 1& |& 6\end{array}\right] \\ {R}_3=r_3-2r_2 \\ \left[\begin{array}{ccccc}1& -1& -1& |& 0\\ 0& 1& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& |& 4\\ 0& 0& -2& |& -2\end{array}\right] \end{gathered}$$

Step 4. Solve

By proceeding with the calculation further

$$\begin{gathered} R_3=-\frac{1}{2}{r}_3 \\ \left[\begin{array}{ccccc}1& -1& -1& |& 0\\ 0& 1& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& |& 4\\ 0& 0& 1& |& 1\end{array}\right] \end{gathered}$$

The matrix is now in row echelon form and  by using the row-echelon form we can represent the final matrix system as

$\left\{\begin{array}{c}{I}_1-I_2-I_3=0 \left(1\right)\\ I_2+\frac{3}{2}{I}_3=4 \left(2\right)\\ I_3=1 \left(3\right)\end{array}\right.$

Step 5. Solve for I 2

To find $I_2$ we will substitute $I_3$ in the equation in (2)  

$$\begin{gathered} I_2+\frac{3}{2}\left(1\right)=4 \\ {I}_2=4-\frac{3}{2} \\ {I}_2=\frac{5}{2} \end{gathered}$$

Step 7. Solve for I 1

To find $I_1,$ we will substitute $I_2 \mathrm{and} I_3$ in the equation in (1)  

$$\begin{gathered} I_1-\left(\frac{5}{2}\right)-1=0 \\ {I}_1=\frac{7}{2} \end{gathered}$$

Therefore, the value of currents are $I_1=\frac{7}{2}, I_2=\frac{5}{2}, \mathrm{and} I_3=1.$