Question

Consider the two-loop circuit shown below:

Ignore the red and pencil markings, just worry about the printed questions

Answer 1

(I₁, I₂) = (1, 1)

Explanation:

You want the matrix version of the given circuit equations, and the solution by matrix methods and by Cramer's rule.

• 15I₁ +5I₂ = 20
• 25I₁ +5I₂ = 30

(a) Matrix equation

The coefficients of the variables fill matrix A; the constants fill matrix (column vector) B:

AI = B

`\left[\begin{array}{cc}15&5\\25&5\end{array}\right]\left[\begin{array}{c}I_1\\I_2\end{array}\right]=\left[\begin{array}{c}20\\30\end{array}\right]`

(b) Matrix algebra

The solution to this matrix equation can be found by left-multiplying both sides by the inverse of matrix A. The inverse of a 2×2 matrix is the transpose of the cofactor matrix, divided by its determinant. It can be written down, as the form is simple: diagonal elements are swapped; off-diagonal elements are negated.

`A^(-1)=(1)/(15(5)-25(5))\left[\begin{array}{cc}5&-5\\-25&15\end{array}\right]=\left[\begin{array}{cc}-0.1&0.1\\0.5&-0.3\end{array}\right]\\\\\textsf{Multiplying by $A^(-1)$, we have ...}\\\\\left[\begin{array}{cc}-0.1&0.1\\0.5&-0.3\end{array}\right]\left[\begin{array}{cc}15&5\\25&5\end{array}\right]\left[\begin{array}{c}I_1\\I_2\end{array}\right]=\left[\begin{array}{cc}-0.1&0.1\\0.5&-0.3\end{array}\right]\left[\begin{array}{c}20\\30\end{array}\right]`

`\left[\begin{array}{cc}1&0\\0&1\end{array}\right]\left[\begin{array}{c}I_1\\I_2\end{array}\right]=\left[\begin{array}{c}(-0.1)(20+(0.1)(30)\\(0.5)(20)+(-0.3)(30)\end{array}\right]\\\\\\\left[\begin{array}{c}I_1\\I_2\end{array}\right]=\left[\begin{array}{c}1\\1\end{array}\right]`

(c) Cramer's rule

Cramer's rule requires we find three determinants. We already found the determinant of the coefficient matrix, above. It is D = -50. The other two are ...

`D_1=\left|\begin{array}{cc}20&5\\30&5\end{array}\right|=(20)(5)-(30)(5)=-50\\\\\\D_2=\left|\begin{array}{cc}15&20\\25&30\end{array}\right|=(15)(30)-(25)(20)=-50\\\\\\I_1=(D_1)/(D)=(-50)/(-50)=1\qquad I_2=(D_2)/(D)=(-50)/(-50)=1\\\\\\\left[\begin{array}{c}I_1\\I_2\end{array}\right]=\left[\begin{array}{c}1\\1\end{array}\right]`