Kirchhoff's Voltage Law and Kirchoff's Current Law (part two)

In this part two of the Kirchhoff's law we are going to analyze the same circuit but this time by Kirchhoff's voltage law(LVK).

A mesh is a closed circuit, so this circuit is divided into two meshes:

Using the LVK we have that: 12V+(12Ohm)I2=0

And in the mesh number 2: (12Ohm)I2+(12Ohm)I3=0

For understanding how to get to this values we need to look at mesh number 1 first, I begin analyzing the voltage source and the source is taken in this case by the negative, now, if I continue to remember that the R2 current enters the positive, so it is considered positive. I1 is the current flowing through R2. The next step is to consider I2, I2 enters on the positive side of R1 SO that is why it is: + (12Ohm) I2.

Now we are going to the mesh number two. It is (-12Ohm)I2 because we are saying that the low current I2 goes from node A to R1 and therefore polarized as:

If we simplify: I1=I2+I3

-12+12*I1+12*I2=0

-12*I2+12*I3=0

Now we have three equations with three unknowns, so we proceed to solve the system.

I1=I2+I3 (1)

12+12*I1+12*I2=0 (2)

-12*I2+12*I3=0 (3)

Substituting (1) in equation (2):

12*(I2+I3)+12*I2=12

24*I2+12*I3=12 (4)

Now we simplify by reduction the equations number 3 and 4:

+12*I2-12*I3=0

24*I2+12*I3=12

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36*I2=12

Once we have the system reducted, we have and equation with only one unknown, that is I2, we obtain the value of I2: I2=12/36=0,33A

We solve the equation number (3), and now we have I3 also: I3=12I2/12=I2=0,33A

And for ending, we substitute all data in the first equation: I1=I2+I3=0,33+0,33=0,66A