2 Engineering Example 2

2.1 Currents in two loops

In the circuit shown find the currents $\left(i_1,i_2\right)$ in the loops.

Figure 1

Solution

We note that the current across the $3\Omega$ resistor (travelling top to bottom in the diagram) is given by $\left(i_1-i_2\right)$ . With this proviso we can apply Kirchhoff’s law:

In the left-hand loop $3\left(i_1-i_2\right)+5i_1=3\to 8i_1-3i_2=3$

In the right-hand loop $3\left(i_2-i_1\right)+6i_2=4\to -3i_1+9i_2=4$

In matrix form:

$\left[\begin{array}{cc}\hfill 8\hfill & \hfill -3\hfill \\ \hfill -3\hfill & \hfill 9\hfill \end{array}\right]\left[\begin{array}{c}\hfill i_1\hfill \\ \hfill i_2\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 4\hfill \end{array}\right]$

The inverse of $\left[\begin{array}{cc}\hfill 8\hfill & \hfill -3\hfill \\ \hfill -3\hfill & \hfill 9\hfill \end{array}\right]$ is $\frac{1}{63}\left[\begin{array}{cc}\hfill 9\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 8\hfill \end{array}\right]$ so solving gives

$\left[\begin{array}{c}\hfill i_1\hfill \\ \hfill i_2\hfill \end{array}\right]=\frac{1}{63}\left[\begin{array}{cc}\hfill 9\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 8\hfill \end{array}\right]\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 4\hfill \end{array}\right]=\frac{1}{63}\left[\begin{array}{c}\hfill 39\hfill \\ \hfill 41\hfill \end{array}\right]$

so $i_1=\frac{39}{63}{i}_2=\frac{41}{63}$