- The algebraic sum of voltage drops in individual sections of a closed circuit, arbitrarily selected in a complex branched circuit, is equal to the algebraic sum of the EMF in this circuit.
- The algebraic sum of voltage drops in a closed loop is equal to the sum of the effective EMF in this loop. If there are no sources of electromotive force in the circuit, then the total voltage drop is zero.
- The algebraic sum of voltage drops along any closed circuit of an electric circuit is equal to zero.
- The algebraic sum of the voltage drops on passive elements is equal to the algebraic sum of the EMF and the voltages of the current sources acting in this circuit.
Those. the voltage drop across R1 with its own sign plus the voltage drop across R2 with its own sign is equal to the voltage of the emf source 1 with its own sign plus the voltage at the source of the electromotive force 2 with its own sign. The algorithm for placing signs in equations according to Kirchhoff's law is described on a separate page.
Equation for Kirchhoff's second law
Equations can be made according to the second Kirchhoff's law in different ways. The first formula is considered the most convenient.
You can also write equations in this form.
The physical meaning of Kirchhoff's second law
The second law establishes a connection between the voltage drop in a closed section of the electrical circuit and the action of EMF sources in the same closed section. It is related to the concept of electric charge transfer work. If the movement of the charge is carried out in a closed loop, returning to the same point, then the perfect work is zero. Otherwise, the law of conservation of energy would not be fulfilled. This important property of a potential electric field is described by Kirchhoff's 2 law for an electrical circuit.
A current flows through each conductor that makes up an electrical circuit. At the point where the conductors converge, called a node, the rule is true: the total current flowing to it is equal to the amount flowing out.
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In other words, how many charges flow to this point per unit of time, the same amount flows out. If we assume that the incoming will be “+”, and the outgoing one - “-”, then its total value will be zero.
This is the First Kirchhoff Law for an electrical circuit. Its meaning is that the charge does not accumulate.
The Second Law is applicable to an electric branched circuit.
These universal Kirchhoff laws are used very widely, since they allow you to solve many problems. Their advantage is considered to be a simple and understandable formulation, simple calculations.
History
Kirchhoff joined the ranks of German scientists in the nineteenth century, when the country, which was on the verge of an industrial revolution, required the latest technology. Scientists were looking for solutions that could accelerate the development of industry.
They were actively engaged in research in the field of electricity, because they understood that in the future it would be widely used. The problem at that time was not how to compose electrical circuits from possible elements, but in carrying out mathematical calculations. This is where the laws formulated by the physicist appeared. They helped a lot.
The algebraic sum of currents coming to the nodes and outgoing from it is equal to zero. This simultaneously follows from another law - the constancy of energy.
2 wires fit to the node, and one leaves. The value of the current flowing from the node is the same as the sum of it flowing through the other two conductors, i.e. going to him. Kirchhoff's rule explains that, otherwise, the charge would accumulate, but this does not happen. Everyone knows that any complex chain can be easily divided into separate sections.
But, at the same time, it is not easy to determine the path along which it passes. Moreover, in different areas the resistances are not the same, therefore, the distribution of energy will not be uniform.
In accordance with the Second rule of Kirchhoff, the energy of electrons in each of the closed sections of the electrical circuit is equal to zero - the total value of the voltages is always equal to zero in such a circuit. If this rule were violated, the energy of electrons passing through certain sections would decrease or increase. But, this is not observed.
Application
Thus, thanks to these two statements put forward by Kirchhoff, the dependence of currents on voltages in branched sections has been established.
The formula for the First Law is:
For the diagram below, it is true:
I1 - I2 + I3 - I4 + I5 = 0
Positive currents are currents going to the point, and those that leave it are "-".
It is written like this:
- k is the number of EMF sources;
- m - branches of a closed loop;
- Ii, Ri - their i-th resistance and current.
In this diagram: E1 - E2 + E3 = I1R1 - I2R2 + I3R3 - I4R4.
- EMF is received "+" when its direction coincides with the selected bypass direction.
- If the direction of the current and the bypass on the resistor coincide, the voltage will also be positive.
Chain calculation
The method consists in the ability to compose systems of equations, as well as to solve them, to find the currents in each branch (b), and already, knowing them, the ability to find the magnitude of the voltages.
Simply put, the number of branches must match the unknowns in the system. First, they are written down based on the first rule: their number is identical with the number of nodes.
But, (y - 1) expressions will be independent. This is ensured by a choice, but it happens so that they (the next one with adjacent ones) differ by at least one branch.
An independent contour is considered to contain one (or more) branches, which are not included in the others.
As an example, consider the following scheme:
She will restrain:
knots – 4;
branches –6.
According to the First Law, three expressions are written, i.e. y - 1 = 4 - 1 = 3.
And the same amount on the basis of the Second, since b - y + 1 = 6 - 4 + 1 = 3.
In the branches, a plus direction and a bypass path are chosen (in our case, clockwise).
It turns out:
It remains to solve the resulting system with respect to currents, realizing that when in the process of solving it turns out to be negative, this indicates that it will be directed in the opposite direction.
Kirchhoff's rule for sinusoidal currents
The rules for sinusoidal are the same as for DC current. True, the values of voltages with complex currents are taken into account.
The first one sounds:"In the electrical circuit, the sum of the algebraic complex currents in the node is equal to zero."
The second rule looks like this:“The algebraic sum of the EMF complex in a closed circuit is equal to the sum of the algebraic values of the complex voltages available on the passive components of this circuit.
Video: Kirchhoff's Laws
In complex electrical circuits, that is, where there are several different branches and several sources of EMF, there is also a complex distribution of currents. However, with the known values of all EMF and resistances of the resistive elements in the circuit, we can clean out the values of these currents and their direction in any circuit of the circuit using the first and second Kirchhoff's laws... I summarized the essence of Kirchhoff's laws in my textbook on electronics, on the pages of the site http: //www.site.
You can see an example of a complex electrical circuit in Figure 1.
Figure 1. Complex electrical circuit.
Kirchhoff's laws are sometimes called Kirchhoff rules especially in older literature.
So, to begin with, let me remind you the essence of the first and second Kirchhoff's laws, and then we will consider examples of calculating currents, voltages in electrical circuits, with practical examples and answers to questions that were asked to me in the comments on the site.
Formulation # 1: The sum of all currents flowing into a node is equal to the sum of all currents flowing out of a node.
Formulation No. 2: The algebraic sum of all currents in a node is zero.
Let me explain the first Kirchhoff's law using the example of Figure 2.
Figure 2. Electrical circuit assembly.
Here the current I 1 is the current flowing into the node, and the currents I 2 and I 3- currents flowing from the node. Then, applying formulation number 1, you can write:
I 1 = I 2 + I 3 (1)
To confirm the validity of the wording No. 2, we will transfer the currents I 2 and I 3 to the left of the expression (1) , thus we get:
I 1 - I 2 - I 3 = 0 (2)
Minus signs in an expression (2) and mean that currents flow out of the node.
The signs for the inflowing and outgoing currents can be taken arbitrarily, however, in general, the inflowing currents are always taken with the "+" sign, and the outgoing currents with the "-" sign (for example, how it happened in the expression (2) ).
You can watch a separate video tutorial on the first Kirchoff's law in the VIDEO LESSONS section.
Formulation: The algebraic sum of the EMF acting in a closed loop is equal to the algebraic sum of the voltage drops across all resistive elements in this loop.
Here, the term "algebraic sum" means that both the value of the EMF and the value of the voltage drop across the elements can be both with the sign "+" and with the sign "-". In this case, the sign can be determined by the following algorithm:
1. Select the direction of the contour bypass (two options, either clockwise or counterclockwise).
2. Arbitrarily choose the direction of currents through the circuit elements.
3. We place signs for EMF and voltages falling on the elements according to the rules:
EMFs creating a current in the circuit, the direction of which coincides with the direction of bypassing the circuit, are recorded with a "+" sign, otherwise the EMFs are recorded with a "-" sign.
The voltages falling on the circuit elements are recorded with a "+" sign if the current flowing through these elements coincides in direction with the bypass of the circuit, otherwise the voltages are recorded with a "-" sign.
For example, consider the circuit shown in Figure 3, and write the expression according to the second Kirchhoff's law, traversing the circuit clockwise, and choosing the direction of currents through the resistors, as shown in the figure.
Figure 3. Electrical circuit to clarify the second Kirchhoff's law.
E 1 - E 2 = -UR 1 - UR 2 or E 1 = E 2 - UR 1 - UR 2 (3)
Calculations of electrical circuits using Kirchhoff's laws.
Now let's look at a variant of a complex chain, and I will tell you how to apply Kirchhoff's laws in practice.
So, in Figure 4 there is a complex circuit with two sources of EMF of magnitude E 1 = 12 in and E 2 = 5 in, with internal resistance of sources r 1 = r 2 = 0.1 Ohm working for the total load R = 2 ohms... How the currents will be distributed in this circuit, and what values they have, we have to find out.
Figure 4. An example of calculating a complex electrical circuit.
Now, according to the first Kirchhoff's law for node A, we compose the following expression:
I = I 1 + I 2,
because I 1 and I 2 flow into a node A, and the current I flows from it.
Using Kirchhoff's second law, we write down two more expressions for the outer contour and the inner left contour, choosing the clockwise direction of the traversal.
For the outer contour:
E 1 -E 2 = Ur 1 - Ur 2 or E 1 -E 2 = I 1 * r 1 - I 2 * r 2
For the inner left path:
E 1 = Ur 1 + UR or E 1 = I 1 * r 1 + I * R
So, we got a system of three equations with three unknowns:
I = I 1 + I 2;
E 1 -E 2 = I 1 * r 1 - I 2 * r 2;
E 1 = I 1 * r 1 + I * R.
Now let's substitute the values of voltages and resistances known to us into this system:
I = I 1 + I 2;
7 = 0.1I 1 - 0.1I 2;
I 2 = I - I 1;
I 2 = I 1 - 70;
12 = 0.1I 1 + 2I.
The next step is to equate the first and second equations and obtain a system of two equations:
I - I 1 = I 1 - 70;
12 = 0.1I 1 + 2I.
We express from the first equation the value of I
I = 2I 1 - 70;
And we substitute its value into the second equation
12 = 0.1I 1 + 2 (2I 1 - 70).
We solve the resulting equation
12 = 0.1I 1 + 4I 1 - 140.
12 + 140 = 4.1I 1
I 1 = 152 / 4.1
I 1 = 37.073 (A)
Now into expression I = 2I 1 - 70 substitute the value
I 1 = 37.073 (A) and get:
I = 2 * 37.073 - 70 = 4.146 A
Well, according to the first Kirchhoff's law, the current I 2 = I - I 1
I 2 = 4.146 - 37.073 = -32.927
Sign "minus" for current I 2 means that we did not correctly choose the direction of the current, that is, in our case, the current I 2 flows out of the node A .
Now the obtained data can be checked in practice or this circuit can be simulated, for example, in the Multisim program.
You can see a screenshot of the circuit simulation for testing Kirchhoff's laws in Figure 5.
Figure 5. Comparison of the results of calculation and simulation of the circuit.
To consolidate the result, I propose to watch the video I have prepared:
Kirchhoff's laws – rules that show how currents and voltages in electrical circuits relate. These rules were formulated by Gustav Kirchhoff in 1845. In the literature, they are often called Kirchhoff's laws, but this is not true, since they are not laws of nature, but were derived from Maxwell's third equation with a constant magnetic field. But nevertheless, the first name is more familiar to them, therefore we will call them, as is customary in the literature - Kirchhoff's laws.
Kirchhoff's first law - the sum of the currents converging at the node is equal to zero.
Let's figure it out. A node is a point that connects branches. A branch is a section of a chain between nodes. The figure shows that current i enters the node, and currents i leave the node 1 and i 2 ... We compose an expression according to the first Kirchhoff's law, taking into account that the currents entering the node have a plus sign, and the currents outgoing from the node have a minus i-i 1 -i 2 = 0. The current i, as it were, spreads by two smaller currents and is equal to the sum of the currents i 1 and i 2 i = i 1 + i 2 ... But if, for example, the current i 2 entered the node, then the current I would be defined as i = i 1 -i 2 ... It is important to consider the signs when drawing up the equation.
The first Kirchhoff's law is a consequence of the law of conservation of electricity: the charge arriving at a node over a certain period of time is equal to the charge leaving the node during the same time interval, i.e. the electric charge in the node does not accumulate and does not disappear.
Kirchhoff's second law – the algebraic sum of the EMF acting in a closed loop is equal to the algebraic sum of the voltage drops in this loop.
Voltage is expressed as the product of current and resistance (Ohm's Law).
This law also has its own rules for application. To begin with, you need to set the direction of the contour traversal with the arrow. Then sum up the EMF and voltages, respectively, taking with a plus sign if the value coincides with the direction of the bypass and minus if it does not coincide. Let's make an equation according to the second Kirchhoff's law, for our scheme. We look at our arrow, E 2 and E 3 coincide with it in the direction, which means the plus sign, and E 1 is directed in the opposite direction, which means the minus sign. Now we look at the voltages, the current I 1 coincides in the direction of the arrow, and the currents I 2 and I 3 are opposite. Hence:
-E 1 + E 2 + E 3 = I 1 R 1 -I 2 R 2 -I 3 R 3
Based on Kirchhoff's laws, methods for analyzing alternating sinusoidal current circuits were compiled. The loop current method is a method based on the application of the second Kirchhoff's law and the method of nodal potentials based on the application of the first Kirchhoff's law.
Two Kirchhoff's laws, together with Ohm's law, make up three laws with which you can determine the parameters of an electrical circuit of any complexity. We will check Kirchhoff's laws using examples of the simplest electrical circuits, which will not be difficult to assemble. To do this, you will need several, a pair of power sources, which are suitable for galvanic cells (batteries) and a multimeter.
Kirchhoff's first law says that the sum at any node in an electrical circuit is zero. There is another formulation, similar in meaning: the sum of the values of the currents entering the node is equal to the sum of the values of the currents leaving the node.
Let's take a closer look at what has been said. A node is the junction of three or more conductors.
The current that flows into the node is indicated by an arrow directed towards the node, and the current leaving the node is indicated by an arrow directed away from the node.
According to the first Kirchhoff's law
We conditionally assigned the “+” sign to all incoming currents, and “-” to all outgoing ones. Although this is not essential.
1 Kirchhoff's law is consistent with the law of conservation of energy, since electric charges cannot accumulate in the nodes, therefore, the charges coming to the node leave it.
A simple circuit consisting of a power supply with a voltage of 3 V (two series-connected batteries of 1.5 V each), three resistors of different ratings: 1 kOhm, 2 kOhm, 3.2 kOhm (you can use resistors of any other ratings). We will measure the currents with a multimeter in the places indicated by the ammeter.
If we add up the readings of three ammeters taking into account the signs, then, according to the first Kirchhoff's law, we should get zero:
I 1 - I 2 - I 3 = 0.
Or the readings of the first ammeter A1 will be equal to the sum of the readings of the second A2 and third A3 ammeters.
The second Kirchhoff's law is perceived by novice radio amateurs much more difficult than the first. However, now you will see that it is quite simple and understandable if you explain it in normal words, and not in abstruse terms.
Simplified 2 Kirchhoff's law says: the sum of the EMF in a closed loop is equal to the sum of the voltage drops
ΣE = ΣIR
Let us analyze the simplest case of this law using the example of a 1.5 V battery and one resistor.
Since there is only one resistor and one battery, the EMF of a 1.5 V battery will be equal to the voltage drop across the resistor.
If we take two resistors of the same value and connect them to a battery, then 1.5 V will be distributed equally across the resistors, that is, 0.75 V.
If we take three resistors of the same value again, for example, 1 kΩ each, then the voltage drop across them will be 0.5 V.
The formula will write it as follows:
Let's consider a conditionally more complex example. Let's add another power supply E2 to the last circuit, with a voltage of 4.5 V.
Please note that both sources are connected in series and in accordance, that is, the plus of one battery is connected to the minus of the other battery, or vice versa. With this method of connecting galvanic cells, their electromotive forces add up: E1 + E2 = 1.5 + 4.5 = 6 V, and the voltage drop across each resistance is 2 V. The formula describes this as follows:
