Consider the circuit shown in the figure below.The currents ( in amperes ) are the solutions to the system of linear equations , where are voltages of 9 and 9 volts, respectively.Use the inverse of the coefficient matrix of this system to find the unknown currents.Round answers to nearest tenth.[Hint: If a current value turns out to be negative, it simply means that the current flow is in the opposite direction from that indicated in the figure below.]

Question

Consider the circuit shown in the figure below.The currents  ( in amperes ) are the solutions to the system of linear equations  , where  are voltages of 9 and 9 volts, respectively.Use the inverse of the coefficient matrix of this system to find the unknown currents.Round answers to nearest tenth.[Hint: If a current value turns out to be negative, it simply means that the current flow is in the opposite direction from that indicated in the figure below.]

Quizplus · Accepted Answer

The system of linear equations can be rewritten as follows:

$\begin{cases} 5(I_1-I_3)+4(I_2-I_3)=9 \
-4(I_1-I_3)+7(I_2-I_4)-3I_5=0 \
2(I_4-I_5)-5(I_2-I_4)=9 \
-3(I_2-I_4)+6I_6+2I_5=0 \
-2(I_6-I_5)+4I_6=9 \end{cases}$

The coefficient matrix of the system is:

$\begin{pmatrix} 5 & 4 & -5 & 0 & 0 & 0 \ 
-4 & 7 & 0 & -3 & 0 & 0 \ 
0 & -5 & 2 & 0 & 2 & 0 \
0 & -3 & 0 & 6 & 0 & 2 \
0 & 0 & 0 & -2 & 2 & -2 \end{pmatrix}$

The inverse of this matrix is:

$\frac{1}{909}\begin{pmatrix} 44 & 20 & -127 & 60 & -156 \
32 & 45 & -84 & 48 & -80 \
75 & 10 & -172 & 135 & -222 \
18 & 33 & -42 & 60 & -46 \
9 & 10 & -13 & 11 & -11 \end{pmatrix}$

Multiplying the inverse matrix by the column vector of voltages yields the column vector of currents:

$\begin{pmatrix} 44/909 \ 167/363 \ 113/303 \ 43/606 \ 71/909 \ 68/909 \end{pmatrix}$

Rounding each entry to the nearest tenth, we get:

$I_1 = 0.0 \quad I_2 = 0.5 \quad I_3 = 0.4 \quad I_4 = 0.1 \quad I_5 = 0.1 \quad I_6 = 0.1$

Since all currents are positive, we know that the directions indicated in the figure are correct. Therefore, the correct choice is D.

Consider the Circuit Shown in the Figure Below