Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 11/Solution 1

By Kirchoff's voltage law, the sum of the voltage drops across the left and right loop must be zero. This leads to the equations

$\begin{array}{rl}2i_1+1(i_1-i_2)& =9\\ 1(i_1-i_2)-3i_2& =14\end{array}$

We solve this system using row reduction. This system is represented by the augmented matrix

$\left[\begin{array}{ccc}3& -1& 9\\ -1& 4& -14\end{array}\right]$.

We add $1/3$ of the first row to the second row:

$\left[\begin{array}{ccc}3& -1& 9\\ 0& 11/3& -11\end{array}\right]$.

Now, we multiply the second row by $3/11$:

$\left[\begin{array}{ccc}3& -1& 9\\ 0& 1& -3\end{array}\right]$.

We add the second row to the first row:

$\left[\begin{array}{ccc}3& 0& 6\\ 0& 1& -3\end{array}\right]$.

Finally, we divide the first row by 3:

$\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& -3\end{array}\right]$.

Therrefore, we get that $i_1=2\text{ Amperes}$ and that $i_2=-3\text{ Amperes}$.

We now verify that this answer makes sense. The direction of $i_2$ is opposite the direction of the current, because the current is running from high voltage to low voltage. Therefore, we expect that $i_2$ should be negative and $i_1$ should be positive. This is confirmed by our calculations.

Answer: $i_1=2\text{A},i_2=-3\text{A}$