4 Engineering Example 3

4.1 Currents in three loops

In the circuit shown find the currents $\left(i_1,i_2,i_3\right)$ in the loops.

Figure 2

Solution

Loop 1 gives

$2\left(i_1\right)+3\left(i_1-i_2\right)=5\to 5i_1-3i_2=5$

Loop 2 gives

$6\left(i_2-i_3\right)+3\left(i_2-i_1\right)=4\to -3i_1+9i_2-6i_3=4$

Loop 3 gives

$6\left(i_3-i_2\right)+4\left(i_3\right)=6-5\to -6i_2+10i_3=1$

Note that in loop 3, the current generated by the 6 $v$ cell is positive and for the 5 $v$ cell negative in the direction of the arrow.

In matrix form

$\left[\begin{array}{ccc}\hfill 5\hfill & \hfill -3\hfill & \hfill 0\hfill \\ \hfill -3\hfill & \hfill 9\hfill & \hfill -6\hfill \\ \hfill 0\hfill & \hfill -6\hfill & \hfill 10\hfill \end{array}\right]\left[\begin{array}{c}\hfill i_1\hfill \\ \hfill i_2\hfill \\ \hfill i_3\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 5\hfill \\ \hfill 4\hfill \\ \hfill 1\hfill \end{array}\right]$

Solving gives

$i_1=\frac{34}{15},i_2=\frac{19}{9},i_3=\frac{41}{30}$