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

Kirchhoff's Voltage Law and Kirchoff's Current Law

If we have several sections of an electrical circuit, which would have to go obtaining equivalent resistance of each section, would complicate the analysis of Ohm's law. For such cases we use the electric circuit analysis by the law of Kirchhoff's voltage and current.

-Kirchhoff's law of voltages: the sum of the voltages in a closed mesh equals zero. That is:

Where E is summation, and n is the corresponding value for each element of the mesh. -Kirchhoff's Current Law: The sum of the currents entering a node equals the sum of the currents leaving. That is

Coming back to the circuit:

Let's look first by Kirchhoff's Current Law. The LCK tells us that the currents entering a node are equal to the sum of the currents leaving.In this circuit, specifically in R1, there are two nodes, the node above which we will call A and the node below which we will call B.

To solve the circuit can be considered only one node, in this case we will choose the node A. Node A is affected by three currents are those of R2, R1 and R3. So we can say that:

Then, we say that enters a current I1 to node A and leaving two currents I2 and I3, so I1 = I2 + I3.The real meaning of the currents depend on the value obtained at the end of each. If the currents are negative, then the flow of them, if they come, would actually be leaving, and if they leave, it would be coming, that is the opposite of the original analysis.

Multisim

All I'm explaining in the blog is by hand calculations, but if yoy want to operate with larger circuits or more complex ones you can use a computer program called multisim, which is specially designed for that, I leave here some youtube videos that explain how to use it, and in the right sidebar of the blog is another video in case you want to take a look.

Mixed circuit (circuit with resistors in series and parallel)

A mixed circuit is formed by components (voltage sources and / or current, resistors, inductors or capacitors) connected in series and parallel. Consider a simple example:

In this example we can not add R2 to R1 because R1 is being affected by R3 and R1 is in parallel with R3. We will reduce the circuit, taking an equivalent resistance between R1 and R3

And to get the total resistance(Rt), we add Req to R2: Rt=R2+Req=12Ohm+6Ohm=18Ohm

Using Rt we can obtain the total current: It=V/Rt=12V/18Ohm=0,66A

Now we have obtained the current value for all the circuit. For obtaining the individual value of the currents and voltages of each element, we apply the same method:

V(r2)=(12Ohm)(0,66A)=7,92V

V(req)=(6Ohm)(0,66A)=3,96V

We divide this voltage between resistors R1 and R3: I=3,96V/12Ohm=0,33A ; Since R1 and R3 are equal, the current will be the same in R3.

Conclusions: Of the total current(0,66A) it can be observed that it is divided between R1 and R3(0,33A), and we also can see that the current is distributed between the circuit elements.

Circuit with parallel resistors

So far I've only worked circuits with resistors in series, but there is another way to place the resistance, named resistors in parallel, there you have an example in the following image:

This particular class of circuits are solved in a different way because the resistors are in parallel with each other. We need to obtain the total circuit resistance, in case one or more resistor are parallel we must do the sum of the inverse of the sum of the inverses of each of the resistors:

So the total resistance of the circuit will be:

Applying Ohm's law: It=12V/8Ohm=1,5A

But this current value corresponds to the total value of the circuit, to get the individual current from the two resistances(12Ohm and 24Ohm), we first multiply the total current value by the total resistance: V=I*Rt=(1,5A)(8Ohm)=12V

Now we will divide the 12V between the two resistors individually to obtain the current value:

I=12V/12Ohm=1A

I=12V/12Ohm=0,5A

Conclusions: The voltage will be the same between the elements situated in parallel, and the current will be divided in equal parts between the circuit elements.