### 2 Engineering Example 2

#### 2.1 Currents in two loops

In the circuit shown find the currents $\left({i}_{1},\phantom{\rule{1em}{0ex}}{i}_{2}\right)$ in the loops.

Figure 1

##### Solution

We note that the current across the $3\phantom{\rule{1em}{0ex}}\Omega$ 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 $3\left({i}_{1}-{i}_{2}\right)+5{i}_{1}=3\phantom{\rule{1em}{0ex}}\to \phantom{\rule{1em}{0ex}}8{i}_{1}-3{i}_{2}=3$

In the right-hand loop $3\left({i}_{2}-{i}_{1}\right)+6{i}_{2}=4\phantom{\rule{1em}{0ex}}\to \phantom{\rule{1em}{0ex}}-3{i}_{1}+9{i}_{2}=4$

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

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{c}\hfill {i}_{1}\hfill \\ \hfill {i}_{2}\hfill \end{array}\right]=\frac{1}{63}\left[\begin{array}{cc}\hfill 9\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 8\hfill \end{array}\right]\left[\begin{array}{c}\hfill 3\hfill \\ \hfill 4\hfill \end{array}\right]=\frac{1}{63}\left[\begin{array}{c}\hfill 39\hfill \\ \hfill 41\hfill \end{array}\right]$

so ${i}_{1}=\frac{39}{63}\phantom{\rule{1em}{0ex}}{i}_{2}=\frac{41}{63}$