Q. 73 - Precalculus Enhanced with Graphing Utilities

Question

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

$$\begin{gathered} I_2= I_1+I_3 \\ 5-3I_1-5I_2=0 \\ 10-5I_2-7I_3=0 \end{gathered}$$

Find the values of currents

Step-by-Step Solution

Answer

The values of current are $\frac{10}{71},\frac{65}{71},\frac{55}{71}$.

Step 1: Given information

We are given a system of equation

$$\begin{gathered} I_2= I_1+I_3 \\ 5-3I_1-5I_2=0 \\ 10-5I_2-7I_3=0 \end{gathered}$$

Step 2: Simplify the equation

Substitute the equation 1 in equation 2 and 3 we get

equation 2 becomes

$$\begin{gathered} 5-3I_1-5(I_1+I_3)=0 \\ 5-8I_1-5I_3=0 \\ 8I_1+5I_3=5 \left(4\right) \end{gathered}$$

Equation 3 becomes

$$\begin{gathered} 10-5(I_1+I_3)=7I_3 \\ 10-5I_1-5I_3=7I_3 \\ 5I_1+12I_3=10 \left(5\right) \end{gathered}$$

Multiply equation 4 by -5 and equation 5 by 8

And then add them

We get,

$-40I_1-25I_3=-25+40I_1+96I_3=8071I_3=55$

Therefore $I_3=\frac{55}{71}$

Step 3: Find the remaining values

We get,

$$\begin{gathered} 8I_1+5I_3=5 \\ 8I_1+5\left(\frac{55}{71}\right)=5 \\ {I}_1=\frac{10}{71} \end{gathered}$$

We also have

$$\begin{gathered} I_2=I_1+I_3 \\ {I}_2=\frac{10}{71}+\frac{55}{71} \\ {I}_2=\frac{65}{71} \end{gathered}$$

Step 4: Conclusion

The values of current are $\frac{10}{71},\frac{65}{71},\frac{55}{71}$.