Chapter 11.1, Problem 73AYU

To determine

To find: An application of Kirchhoff s Rules to the circuit shown results in the following system of equations:

  $\begin{array}{c}{I}_2=I_1+I_3\\ 5-3I_1-5I_2=0\\ 10-5I_2-7I_3=0\end{array}$

Find the currents $I_1,I_2$ and $I_3$ .

  

Answer to Problem 73AYU

Solution: $I_1=\frac{10}{71},I_2=\frac{65}{71},I_3=\frac{55}{71}$

Explanation of Solution

Given: $I_2=I_1+I_3----1$

  $\begin{array}{c}5-3I_1-5I_2=0----2\\ 10-5I_2-7I_3=0----3\end{array}$

Calculation:

  $I_2=I_1+I_3\to I_1=I_2-I_3$

  $5-3I_1-5I_2=0\to 3\left(I_2-I_3\right)+5I_2=5\to 8I_2-3I_3=5$

  $10-5I_2-7I_3=0\to 5I_2+7I_3=10$

  $\begin{array}{l}8I_2-3I_3=5----4\hfill \\ 5I_2+7I_3=10----5\hfill \end{array}$

Multiplying equation 4 by 7 and equation 5 by $3,$ we have

  $56I_2-21I_3=35$

  $15I_2+21I_3=30$

Adding both the equation above, we have

  $71I_2=65$$I_2=\frac{65}{71}$

Using equation 4 we have

  $\begin{array}{c}8I_2-3I_3=5\\ 8\left(\frac{65}{71}\right)-3I_3=5\\ \frac{520}{71}-3I_3=5\\ 3I_3=\frac{165}{71}\\ I_3=\frac{55}{71}\end{array}$

Using equation 1 we have

  $I_2=I_1+I_2$

  $\begin{array}{r}\hfill \frac{65}{71}=I_1+\frac{55}{71}\\ \hfill I_1=\frac{10}{71}\end{array}$