Chapter 11.1, Problem 74AYU

To determine

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

  $\begin{array}{c}{I}_2=I_1+I_2\\ 4I_3+6I_2=8\\ 4+6I_2=8I_1\end{array}$

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

Answer to Problem 74AYU

Solution: $l_1=\frac{11}{13},I_2=\frac{6}{13}{J}_3=\frac{17}{13}$

Explanation of Solution

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

  $\begin{array}{l}4I_3+6I_2=8-----2\hfill \\ 4+6I_2=8I_1-----3\hfill \end{array}$

Calculation: Substituting equation 1 in equation $2,$ we have

  $4\left(l_1+I_2\right)+6I_2=8-----2$

  $4l_1+10I_2=8-----4$

  $8I_1-6I_2=4-----3$

Multiplying equation 4 by 2, we have

  $8l_1+20I_2=16-----4$

  $8l_1-6l_2=4-----3$

Adding both the equation above, we have

  $\begin{array}{l}26I_2=12\hfill \\ I_2=\frac{6}{13}\hfill \end{array}$

Using equation 3 we have

  $8l_1-6I_2=4$

  $\begin{array}{c}8I_1-6\left(\frac{6}{13}\right)=4\\ 8I_1=\frac{88}{13}\\ I_1=\frac{11}{13}\end{array}$

Using equation 1 we have

  $I_3=I_1+I_2$

  $I_3=I_1+I_2$

  $I_3=\frac{11}{13}+\frac{6}{13}$

  $I_3=\frac{17}{13}$