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

______________

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.

Circuit with series resistors

In this entry I'm gonna show how to analyze a circuit, just like the previous one but a little more complicated.

Using the previous circuit, we are going to add an extra resistance behind the original, so we can see what happens with the current and with the voltage.

Current will flow in the same sense as in the first circuit, and the polarities of the resistors will be the same. Remember that the current in a resistor enters the positive and out by the negative but remained always positive, in this case it does not chage the value nor the sense when passing through R1 so that the circuit with their respective polarities would be as shown:

One can see that the current flowing through R2 is equal to the current flowing through R1, so in an electrical circuit the current is the same in two or more series resistors(the resistors in series are those that are connected one after the other).

To solve this circuit, we obtain an equivalent resistance of two resistors. For the analysis of electrical circuits when there are two or more resistors in series we must add all the resistances to obtain the total circuit resistance, as Rt--->R=R1+R2+...+Rn

Once we add R1 to R2 we obtain the value of Rt=24Ohm, and we can replace it in the main equation: I=12V/24Ohm=0,5A

One can see that the current value decreased from the original. The more resistance you have, the current value is going to decrease, as there will be more opposition to the flow of it. To obtain the voltage of each element, we must multiply the value of the current through the resistor.

V=I*R=(0,5A)(12Ohm)=6V

V=I*R=(0,5A)(12Ohm)=6V

In the voltage's case, one can see that the voltage is the same in the two resistances, but the voltage was divided from the original, therefore holds that in a series circuit the voltage will be different in each of the elements.

Ohm's law

In this entry I'm going to explain the Ohm's law for the analysis of an electronic circuit.

Ohm's law tells us that the voltage V across a resistor is proportional to the current I flowing through the resistor (R). The way it is the equation with Ohm's law is: V=I*R

Where:

-V is measured in volts and is represented by V, for example, 5volts=5V

-I is the current of the conductor, and is measured in Amperes, represented by A, for example, 10amperes=10A

-R is the resistance of the conductor, measured in Ohm's and denoted by the greek alphabet Omega, for example, 10Ohm's=10Ω

Let's see Ohm's law applied to a simple electrical circuit:

When analyzing an electrical circuit, it must be consider several factors. The polarity of the source, the polarity of the resistor and the current direction. Usually, when an analysis of a circuit is done, it is considered that the current is in clockwise direction, that is:

There are several types of sources for this circuit we will choose the second example of the entry number two. Where the positive power is the largest horizontal bar while the negative will be the smaller horizontal bar. The next thing to do is assign a polarity to the resistance. When a current I passes through a resistor R, it is said that the electric current enters the positive(+) of the resistance and exits through the negative(-). We said that the analysis of a circuit is in the sense of clockwise, so in the circuit, the polarities will be represented as:

Applying the Ohm's law, V=IR, we clear the value of I and we have: I=V/R

We have 12V and 12Ohm, and substituting in the equation: I=12V/12Ohm=1A

If we want to know the voltage of the resistance, since we have the total current value of the circuit, is one ampere(1A), we simply multiply the value of the current flowing through the resistor with the value of it: V=I*R=(1A)(12Ohm)=12V

Conclusion: The voltage provided by a voltage source will be divided equally between the circuit elements.

Units

In the previous entry I forgot to mencione the units, so I put here this table so you can see them and have an idea, because in electrical circuits and in resistance the units can be from (1x10^18) to (1x10^-18):

Electric circuits

As the blog will try to explain how to analyze an electrical circuit, I think we must start by explaining what an electrical circuit is and each one of the part that forms of it.

First an image of an electric circuit and the parts that are on it:

Now I will explain each part of a circuit:

-Voltage(V): When a load is moved through the circuit elements, energy can be transfered. The voltage associated with a circuit element is the energy transferred per unit of charge flowing through an element. Voltage units are Volts(V) which is equivalent to Joules per Coulomb(J/C). If we analyze the electrical circuit from above, have a voltage source, which is providing the energy required for each of the elements of the circuit. For the analysis of electrical circuits on DC, a DC voltage source of independent type can be represented in three ways:

-Intensity or current(I): The electric current is the time it takes an electrical charge to flow through a conductor or a circuit element. The units are Amperes(A) which is equivalent to Coulomb per second(C/s). The current in an electric circuit is represented by a I1 or i1 and its direction will depend on the circuit elements.

-Neutro(0): In order to analyze an electrical circuit we must have one end of the circuit in the physical earth, where the voltage is zero(no value).

-Resistance(R): An electrical resistance is the opposition of a body to the flow of the electrical current. The resistance is measured in Ohms and in an electrical circuit is represented as:

These are the most important parts of a circuit.

Introduction

This blog is being created by Alba García, from the european university of Madrid.

I'm going to use this blog as a tutorial of how to analize electronic circuits.

This is my first entry at the blog so I considered as a important thing, for getting started with circuits, to identify the components that an electronic circuit can have.

For make it easier to understand the elements that a circuit can have I made this table:

(You can click on the table to make it bigger)

This is only a short summary, I will explain each item when I use them.