Question

By using Kirchhoff's Laws, it can be shown that the currents I1, I2, and I3 that pass through the three branches of the circuit in the figure satisfy the given linear system. Solve the system to find I1, I2, and I3. (Round your answers to three decimal places.)

I(sub)1 + I(sub)2 − I3 = 0
8I(sub)1 − 4I(sub)2 = 2
4I(sub)2 + 2I(sub)3 = 8
R = 8, R2 = 4, R3 = 2
V = 2, V(sub)2 = 8

Answer 1

Solve the second equation for I_1 and the third equation for I_3. Then plug those into the first equation. From here you can solve for I_2. Then you can plug that in to equation 2 to solve for I_1. Then plug in to equation 3 to solve for I_3.

Answer 2

not sure how to go about doing that. do you know what Isub1 Isub2 and Isub3 egual?

Answer 3

So, starting with equation two you have:
\[8I_1-4I_2=2 \rightarrow 8I_1=2+4I_2 \rightarrow I_1=\frac{1}{4}+\frac{1}{2}I_2\]
Then the third equation:
\[4I_2+2I_3=8 \rightarrow 2I_3=8-4I_2 \rightarrow I_3=4-2I_2\]
Now plugging that into the first you have:
\[\frac{1}{4}+\frac{1}{2}I_2+I_2-4+2I_2=0\]
Now solve for I_2. Once you get that, plug it into the above equations to solve for I_1 and I_3. Does that make more sense?

Answer 4

somewhat. do you mind displaying the other two because i am crunched for time and i have quite a few others to do and the previous model you showed helps alot

Answer 5

Okay. Going off the last equation and solving for I_2. You get I_2=15/14 (i'll leave the algebra to you)
From there you can simply plug in the first equation.
\[I_3=4-2 \frac{15}{14} \rightarrow \frac{28}{7}-\frac{15}{7}=\frac{13}{7}\]
Then:
\[I_1=\frac{1}{4}+\frac{1}{2}\frac{15}{14} \rightarrow \frac{7}{28}+\frac{15}{28}=\frac{22}{28}=\frac{11}{14}\]
So you have:
\[I_2=\frac{15}{14};I_1= \frac{11}{14};I_3=\frac{13}{7}\]

Answer 6

as long as my algebra is right.