Q. 74 - Precalculus Enhanced with Graphing Utilities

Question

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

$$\begin{gathered} I_3=I_1+I_2 \\ 8 = 4I_3+6I_2 \\ 8I_1=4+6I_2 \end{gathered}$$

Find the currents 

Step-by-Step Solution

Answer

The values of current are $\frac{11}{13},\frac{6}{13},\frac{17}{13}$ respectively.

Step 1: Given information

We are given a system of equation

$$\begin{gathered} I_3=I_1+I_2 \\ 8 = 4I_3+6I_2 \\ 8I_1=4+6I_2 \end{gathered}$$

Step 2: Simplify the equations

Substitute equation 1 in equation 2 we get

Equation 2 becomes

$$\begin{gathered} 4(I_1+I_2)+6I_2=8 \\ 4I_1+10I_2=8 \left(4\right) \end{gathered}$$

Now we multiply equation 4 by -2 and add it to equation 3

We get,

$-8I_1-20I_2=-16+8I_1-6I_2=4-26I_2=-12$

hence $I_2=\frac{12}{26}$

Now we find the remaining values

$$\begin{gathered} 4I_1+10I_2=8 \\ 4I_1+10·\frac{6}{13}=8 \\ {I}_1=\frac{11}{13} \end{gathered}$$

Now also,

$$\begin{gathered} I_3=I_1+I_2 \\ {I}_3=\frac{11}{13}+\frac{6}{13} \\ {I}_3=\frac{17}{13} \end{gathered}$$

Step 3: Conclusion

The values of current are $\frac{11}{13},\frac{6}{13},\frac{17}{13}$respectively