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D.C. DC electrical circuits: calculation

In electrical engineering, it is generally accepted that a simple circuit is a circuit that reduces to a circuit with one source and one equivalent resistance. You can collapse a circuit using equivalent transformations of serial, parallel, and mixed connections. The exception is circuits containing more complex star and delta connections. Calculation of DC circuits produced using Ohm's and Kirchhoff's laws.

Example 1

Two resistors are connected to a 50 V DC voltage source, with internal resistance r = 0.5 Ohm. Resistor values R 1 = 20 and R2= 32 Ohm. Determine the current in the circuit and the voltage across the resistors.

Since the resistors are connected in series, the equivalent resistance will be equal to their sum. Knowing it, we will use Ohm's law for a complete circuit to find the current in the circuit.

Now knowing the current in the circuit, you can determine the voltage drop across each resistor.

There are several ways to check the correctness of the solution. For example, using Kirchhoff's law, which states that the sum of the emf in the circuit is equal to the sum of the voltages in it.

But using Kirchhoff's law it is convenient to check simple circuits that have one circuit. A more convenient way to check is power balance.

The circuit must maintain a power balance, that is, the energy given by the sources must be equal to the energy received by the receivers.

The source power is defined as the product of the emf and the current, and the power received by the receiver as the product of the voltage drop and the current.

The advantage of checking the power balance is that you do not need to create complex cumbersome equations based on Kirchhoff’s laws; it is enough to know the EMF, voltages and currents in the circuit.

Example 2

Total current of a circuit containing two resistors connected in parallel R 1 =70 Ohm and R 2 =90 Ohm, equals 500 mA. Determine the currents in each of the resistors.

Two resistors connected in series are nothing more than a current divider. We can determine the currents flowing through each resistor using the divider formula, while we do not need to know the voltage in the circuit; we only need the total current and the resistance of the resistors.

Currents in resistors

In this case, it is convenient to check the problem using Kirchhoff’s first law, according to which the sum of the converging currents at a node is equal to zero.

If you do not remember the current divider formula, then you can solve the problem in another way. To do this, you need to find the voltage in the circuit, which will be common to both resistors, since the connection is parallel. In order to find it, you must first calculate the circuit resistance

And then the tension

Knowing the voltages, we will find the currents flowing through the resistors

As you can see, the currents turned out to be the same.

Example 3

In the electrical circuit shown in the diagram R 1 =50 Ohm, R 2 =180 Ohm, R 3 =220 Ohm. Find the power released by the resistor R 1, current through resistor R 2, voltage across the resistor R 3 if it is known that the voltage at the circuit terminals is 100 V.

To calculate the DC power dissipated by resistor R 1, it is necessary to determine the current I 1, which is common to the entire circuit. Knowing the voltage at the terminals and the equivalent resistance of the circuit, you can find it.

Equivalent resistance and current in the circuit

Hence the power allocated to R 1

Example 1

Since the resistors are connected in series, the equivalent resistance will be equal to their sum. Knowing it, we will use Ohm's law for a complete circuit to find the current in the circuit.

Now knowing the current in the circuit, you can determine the voltage drop across each resistor.

There are several ways to check the correctness of the solution. For example, using Kirchhoff's law, which states that the sum of the emf in the circuit is equal to the sum of the voltages in it.

But using Kirchhoff's law it is convenient to check simple circuits that have one circuit. A more convenient way to check is power balance.

The circuit must maintain a power balance, that is, the energy given by the sources must be equal to the energy received by the receivers.

The source power is defined as the product of the emf and the current, and the power received by the receiver as the product of the voltage drop and the current.

Example 2

Total current of a circuit containing two resistors connected in parallel R 1 =70 Ohm and R 2 =90 Ohm, equals 500 mA. Determine the currents in each of the resistors.

Currents in resistors

In this case, it is convenient to check the problem using Kirchhoff’s first law, according to which the sum of the converging currents at a node is equal to zero.

And then the tension

Knowing the voltages, we will find the currents flowing through the resistors

As you can see, the currents turned out to be the same.

Example 3

Equivalent resistance and current in the circuit

Hence the power allocated to R 1

Kirchhoff's laws.

∑I=0

∑E=∑IR

Calculation procedure

We arbitrarily choose the direction of the current in the branches.
We arbitrarily select the direction of traversing the contours.
Knowing the polarity of the sources, we indicate the direction of the EMF.
We compose equations according to Kirchhoff's first law. There should be one less than nodes.
We compose equations according to Kirchhoff’s second law based on the calculation that the total number of equations must be equal to the number of unknown currents.
We solve the system of equations and determine the unknown currents. If, as a result of the solution, any current turns out to have a “-” sign, then its direction is opposite to the chosen one.

Let's give an example.

Given:

1 =r 2 =0;
1 =0.3 Ohm;
2 =1 Ohm;
3 =24 Ohm;

E 1 =246 V;

E 2 =230V

Find:

I 1,I 2,I 3.

Solution:

So, on the diagram we draw the directions of the currents (1), according to these directions we draw the directions for bypassing the circuits (2), according to the polarity of the power sources we set the directions of the EMF (3).

According to Kirchhoff's first law:

I 1 -I 2 -I 3 =0 → -I 2 =I 3 -I 1

Now we compose equations according to Kirchhoff’s second law:

E 1 =I 1 R 1 +I 3 R 3

E 2 = -I 2 R 2 +I 3 R 3

We obtained a system of three equations. Let's decide.

E 2 =(I 3 -I 1)R 2 +I 3 R 3

230=I 3 (1+R 3)-I 1 =25I 3 -I 1 → I 1 = 25I 3 -230

E 1 =I 1 R 1 +I 3 R 3 =(25I 3 -230)R 1 +I 3 R 3

246=0.3(25I 3 -230)+24I 3

246=7.5I 3 -69+24I 3

31.5I 3 =315

I 3 =10A

I 1 =25∙10-230=20A

I 2 =I 1 -I 3 =20-10=10A

2. Loop current method

This method is based on Kirchhoff's law

We arbitrarily select the directions of the loop currents (Fig. 2)
We compose equations according to Kirchhoff's second law.

E 1 -E 2 =I 1 (R 1 +R 2) -I 2 R 2

E 2 =I 2 (R 2 +R 3)-I 1 R 2

246-230=I 1 (0.3+1)-I 2 → 16=1.3I 1 -I 2 → I 2 =1.3I 1 -16

230=25(1.3I 1 -16)-I 1

31.5I 1 =630

I 1 =20A

I 2 =1.3∙20-16=10A

3. Determine the true currents.

I 1 =I 1 =20A

I 2 =I 1 -I 2 =10A

I 3 =I 2 =10A

3. Method two nodes

This method is applicable for circuits with two nodes

We arbitrarily choose the directions of currents in the branches in the same direction (see Fig. 3 - arrows with dashes).
We determine the conductivity of the branches:

Q 1 =1/R 1 =1/0.3=3.33 Sim.

Q 2 =1/R 2 =1 Sym.

Q 3 =1/R 3 =1/24=0.0416 Sim.

We determine the voltage between two nodes using the formula:

U=∑E q /∑ ar q=(E 1 +E 2 q 2)/(q 1 +q 2 +q 3)=(246∙3.31+230)/4.3716=240 V

Determining currents in branches

I=(E-U)q

I 1 =(E 1 -U)q 1 =(246-240)3.33=20A

I 2 =(E 2 -U)q 2 =230-240=-10A

I 3 =-Uq 3 =240∙0.0416=-10A

Since the values of I 2 and I 3 turned out to be negative, these currents will be opposite in direction (the figure shows thick solid arrows).

4. Overlay method or superposition method

The method is based on the fact that any current in the circuit is created by the combined action of all power sources. Therefore, it is possible to calculate the partial currents from the action of each power source separately, and then find the true currents as the arithmetic component of the partial ones.

Solution

1. Fig. 4. E 2 =0; r 2 ≠0

R e =R 2 R 3 /(R 2 +R 3)+R 1 =24/25+0.3=0.96+0.3=1.26 Ohm

I’ 1 =E 1 /R e =246/1.26=195.23 Ohm

U ab =I’ 1 R 23 =195.23∙0.96=187.42 V

I’ 2 =U ab /R 2 =187.42 A

I’ 3 = U ab /R 3 =187.42/24=7.8 A

2. Fig. 5. E 1 =0; R 1 ≠0

R e =R 1 R 3 /(R 1 +R 3)+R 2 =0.3∙24/24.3+1=0.29+1=1.29 Ohm

I” 2 =E 2 /R e =230/1.29=178.29 A

U ab =I” 2 R 13 =178.29∙0.29=51.7 V

I” 1 =U ab /R 1 =51.7/0.3=172.4 A

I” 3 =U ab /R 3 =51.7/24=2.15 A

3. Determine the true currents.

I 1 =I’ 1 -I” 1 =195.23-172.4=22.83 A

I 2 =I’ 2 -I” 2 =187.42-178.29=9.13 A

I 3 =I’ 3 -I” 3 =7.8-2.15=5.65 A

05.12.2014

Lesson 25 (9th grade)

Subject. Calculation of simple electrical circuits

The solution to any problem of calculating an electrical circuit should begin with the choice of the method by which the calculations will be made. As a rule, one and the same problem can be solved by several methods. The result will be the same in any case, but the complexity of the calculations may differ significantly. To correctly select a calculation method, you must first decide which class this electrical circuit belongs to: simple electrical circuits or complex ones.

TO simple include electrical circuits that contain either one source of electrical energy or several located in the same branch of the electrical circuit. Below are two diagrams of simple electrical circuits. The first circuit contains one voltage source, in which case the electrical circuit clearly belongs to simple circuits. The second already contains two sources, but they are in the same branch, therefore it is also a simple electrical circuit.

Simple electrical circuits are usually calculated in the following sequence:

1. First, simplify the circuit by sequentially converting all the passive elements of the circuit into one equivalent resistor. To do this, it is necessary to select sections of the circuit in which resistors are connected in series or parallel, and according to known formulas, replace them with equivalent resistors (resistances). The circuit is gradually simplified and leads to the presence of one equivalent resistor in the circuit.

2. Next, a similar procedure is carried out with active elements of the electrical circuit (if there are more than one source). By analogy with the previous paragraph, we simplify the circuit until we get one equivalent voltage source in the circuit.

3. As a result, we reduce any simple electrical circuit to the following form:
Now it is possible to apply Ohm's law - relation (1.22) and actually determine the value of the current flowing through the source of electrical energy.

combined Homework

1. F.Ya.Bozhinova, N.M.Kiryukhin, E.A.Kiryukhina. Physics, 9th grade, “Ranok”, Kharkov, 2009. § 13-14 (p. 71-84) repeat.

2. Exercise 13 (task 2, 5), exercise 14 (task 3, 5, 6) solve.

3. Copy tasks 1, 3, 4 into your workbook (see next page).

AI with balance sheet preparation

Pi DC. Examples of solved problems

Introduction

Solving problems is an integral part of teaching physics, since in the process of solving problems, physical concepts are formed and enriched, students' physical thinking develops and their skills in applying knowledge in practice are improved.

In the course of solving problems, the following didactic goals can be set and successfully implemented:

Raising a problem and creating a problematic situation;
Summarizing new information;
Formation of practical skills;
Testing the depth and strength of knowledge;
Consolidation, generalization and repetition of material;
Implementation of the principle of polytechnicism;
Development of students' creative abilities.

Along with this, when solving problems, schoolchildren develop hard work, an inquisitive mind, ingenuity, independence in judgment, interest in learning, will and character, and perseverance in achieving their goals. To achieve the above goals, it is especially convenient to use non-traditional tasks.

Tasks for calculating DC electrical circuits

According to the school curriculum, very little time is allocated to consider this topic, so students more or less successfully master methods for solving problems of this type. But often these types of problems are found in Olympiad tasks, but they are based on a school course.

Such non-standard tasks for calculating DC electrical circuits include tasks whose diagrams are:

2) symmetrical;

3) consist of complex mixed compounds of elements.

In general, any circuit can be calculated using Kirchhoff's laws. However, these laws are not included in the school curriculum. In addition, not many students can correctly solve a system of a large number of equations with many unknowns, and this path is not the best way to waste time. Therefore, you need to be able to use methods that allow you to quickly find the resistance and capacitance of circuits.

Equivalent circuit method

The method of equivalent circuits is that the original circuit must be presented in the form of successive sections, on each of which the circuit elements are connected either in series or in parallel. For such a representation, the diagram must be simplified. By simplifying the circuit we mean connecting or disconnecting any circuit nodes, removing or adding resistors, capacitors, ensuring that the new circuit of series and parallel connected elements is equivalent to the original one.

An equivalent circuit is a circuit such that when the same voltages are applied to the original and converted circuits, the current in both circuits will be the same in the corresponding sections. In this case, all calculations are performed with the converted circuit.

To draw an equivalent circuit for a circuit with a complex mixed connection of resistors, you can use several techniques. We will limit ourselves to considering in detail only one of them - the method of equipotential nodes.

This method consists in searching for points with equal potentials in symmetrical circuits. These nodes are connected to each other, and if some section of the circuit was connected between these points, then it is discarded, since due to the equality of potentials at the ends, no current flows through it and this section does not in any way affect the overall resistance of the circuit.

Thus, replacing several nodes of equal potential leads to a simpler equivalent circuit. But sometimes it is more expedient to replace one unit

several nodes with equal potentials, which does not violate the electrical conditions in the rest of the part.

Let's look at examples of solving problems using these methods.

Task No. 1

Solution:

Due to the symmetry of the branches of the chain, points C and D are equipotential. Therefore, we can exclude the resistor between them. We connect equipotential points C and D into one node. We get a very simple equivalent circuit:

The resistance of which is:

RAB=Rac+Rcd=r*r/r*r+r*r/r+r=r.

Task No. 2

Solution:

At points F and F` the potentials are equal, which means the resistance between them can be discarded. The equivalent circuit looks like this:

Section resistances DNB;F`C`D`; D`, N`, B`; FCD are equal to each other and equal to R1:

1/R1=1/2r+1/r=3/2r

Taking this into account, a new equivalent circuit is obtained:

Its resistance and the resistance of the original circuit RAB is equal to:

1/RAB=1/r+R1+R1+1/r+R1+R1=6/7r

Task No. 3.

Solution:

Points C and D have equal potentials. Except for the resistance between them. We get an equivalent circuit:

The required resistance RAB is equal to:

1/RAB=1/2r+1/2r+1/r=2/r

Task No. 4.

Solution:

As can be seen from the diagram, nodes 1,2,3 have equal potentials. Let's connect them to node 1. Nodes 4,5,6 also have equal potentials; let's connect them to node 2. We get the following equivalent circuit:

The resistance in section A-1, R 1 is equal to the resistance in section 2-B, R3 and is equal to:

The resistance in section 1-2 is: R2=r/6.

Now we get the equivalent circuit:

The total resistance RAB is equal to:

RAB= R1+ R2+ R3=(5/6)*r.

Task No. 5.

Solution:

Points C and F are equivalent. Let's connect them into one node. Then the equivalent circuit will look like this:

Resistance at the AC section:

Resistance in section FN:

Resistance in section DB:

This results in an equivalent circuit:

The required total resistance is:

Problem #6

Solution:

Let's replace the common node O with three nodes with equal potentials O, O 1, O 2. We get an equivalent system:

Resistance at section ABCD:

Resistance in section A`B`C`D`:

Resistance in the ACB section

We get an equivalent circuit:

The required total resistance of the circuit R AB is equal to:

R AB = (8/10)*r.

Task No. 7.

Solution:

“Divide” node O into two equipotential angles O 1 and O 2. Now the circuit can be imagined as a parallel connection of two identical circuits. Therefore, it is enough to consider one of them in detail:

The resistance of this circuit R 1 is equal to:

Then the resistance of the entire circuit will be equal to:

Task No. 8

Solution:

Nodes 1 and 2 are equipotential, so we connect them into one node I. Nodes 3 and 4 are also equipotential - we connect them into another node II. The equivalent circuit looks like:

The resistance in section A-I is equal to the resistance in section B-II and is equal to:

The resistance of section I-5-6-II is equal to:

The resistance of section I-II is equal.

3.1. DC circuit model

If constant voltages operate in an electrical circuit and constant currents flow, then the models of reactive elements L and C are significantly simplified.

The resistance model remains the same and the relationship between voltage and current is determined by Ohm's law in the form

In an ideal inductance, the instantaneous values of voltage and current are related by the relation

Similarly, in a capacitance, the relationship between the instantaneous values of voltage and current is defined as

Thus, in the DC circuit model there are only resistances (resistor models) and signal sources, and reactive elements (inductances and capacitances) are absent.

3.2. Circuit calculation based on Ohm's law

This method is convenient for calculating relatively simple circuits with one signal source. It involves calculating the resistance of sections of the circuit for which the current (or voltage) value is known, followed by determining the unknown voltage (or current). Let's consider an example of calculating a circuit, the diagram of which is shown in Fig. 3.1, with an ideal source current A and resistances Ohm, Ohm, Ohm. It is necessary to determine the currents of the branches and , as well as the voltages across the resistances , and .

The source current is known, then it is possible to calculate the circuit resistance relative to the terminals of the current source (parallel connection of resistance and series connection

Rice. 3.1. ny resistances and ),

Then the voltage at the current source (at the resistance) is equal to

Then you can find the branch currents

The results obtained can be verified using Kirchhoff's first law in the form. Substituting the calculated values, we obtain A, which coincides with the value of the source current.

Knowing the branch currents, it is not difficult to find the voltages across the resistances (the value has already been found)

According to Kirchhoff's second law. Adding up the results obtained, we are convinced of its implementation.

3.3. General method for calculating a circuit based on Ohm's laws

and Kirchhoff

The general method for calculating currents and voltages in an electrical circuit based on Ohm's and Kirchhoff's laws is suitable for calculating complex circuits with multiple signal sources.

The calculation begins by specifying the designations and positive directions of currents and voltages for each element (resistance) of the circuit.

The system of equations includes a subsystem of component equations that, according to Ohm’s law, connect currents and voltages in each element (resistance) and the subsystem

topological equations, built on the basis of Kirchhoff's first and second laws.

Consider the calculation of the simple circuit from the previous example, shown in Fig. 3.1, with the same initial data.

The subsystem of component equations has the form

The circuit has two nodes () and two branches that do not contain ideal current sources (). Therefore, it is necessary to write one equation () according to Kirchhoff’s first law,

and one equation of Kirchhoff's second law (),

which form a subsystem of topological equations.

Equations (3.4)-(3.6) are a complete system of circuit equations. Substituting (3.4) into (3.6), we obtain

and, combining (3.5) and (3.7), we obtain two equations with two unknown branch currents,

Expressing the current from the first equation (3.8) and substituting it into the second, we find the value of the current,

and then find A. Using the calculated currents of the branches from the component equations (3.4), we determine the voltages. The calculation results coincide with those obtained earlier in subsection 3.2.

Let's consider a more complex example of calculating a circuit in the circuit shown in Fig. 3.2, with parameters Ohm, Ohm, Ohm, Ohm, Ohm, Ohm,

The circuit contains nodes (their numbers are indicated in circles) and branches that do not contain ideal current sources. The system of component equations of the chain has the form

According to Kirchhoff’s first law, it is necessary to write down the equations (node 0 is not used),

According to Kirchhoff’s second law, equations are compiled for three independent contours, marked on the diagram with circles with arrows (the contour numbers are indicated inside),

Substituting (3.11) into (3.13), together with (3.12), we obtain a system of six equations of the form

From the second and third equations we express

and from the first , then substituting and , we get . Substituting the currents , and into the equations of Kirchhoff’s second law, we write a system of three equations

which, after bringing similar ones, we write in the form

Let's denote

and from the third equation of system (3.15) we write

Substituting the obtained value into the first two equations (3.15), we obtain a system of two equations of the form

From the second equation (3.18) we obtain

then from the first equation we find the current

Having calculated , from (3.19) we will find , from (3.17) we will calculate , and then from the substitution equations we will find the currents , , .

As you can see, analytical calculations are quite cumbersome, and for numerical calculations it is more advisable to use modern software packages, for example, MathCAD2001. An example program is shown in Fig. 3.3.

Matrix - column contains the values of currents A, A, A. The rest

currents are calculated according to equations (3.14) and are equal

A, A, A. The calculated current values coincide with those obtained from the above formulas.

The general method of calculating a circuit using Kirchhoff's equations leads to the need to solve linear algebraic equations. With a large number of branches, mathematical and computational difficulties arise. This means that it is advisable to look for calculation methods that require compiling and solving a smaller number of equations.

3.4. Loop current method

Loop current method based on equations Kirchhoff's second law and leads to the need to solve equations - the number of all branches, including those containing ideal current sources.

Independent circuits are selected in the circuit and for each of them a ring (closed) circuit current is introduced (double indexing makes it possible to distinguish between circuits).

tour currents from branch currents). Through the loop currents, we can express all the currents of the branches and for each independent loop we can write down the equations of Kirchhoff’s second law. The system of equations contains equations from which all loop currents are determined. Based on the found loop currents, the currents or voltages of the branches (elements) are found.

Consider the example of the circuit in Fig. 3.1. Figure 3.4 shows a diagram indicating the designations and positive directions of two loop currents and ( , , ).

Rice. 3.4 Through the prote-

Only the loop current flows and its direction coincides with , so the branch current is equal to

Two loop currents flow in the branch, the current coincides in direction with, and the current has the opposite direction, therefore

For contours, not containing ideal current sources, we compose the equations of Kirchhoff’s second law using Ohm’s law, in this example one equation is written

If an ideal current source is included in the circuit, then for him

Kirchhoff's second law equation not compiled, and its loop current is equal to the source current taking into account their positive directions, in the case under consideration

Then the system of equations takes the form

As a result of substituting the second equation into the first we get

then the current is equal

and the current is A. From (3.21) A, and from (3.22) respectively A, which completely coincides with the previously obtained results. If necessary, using the found values of the branch currents using Ohm’s law, you can calculate the voltages on the circuit elements.

Let's consider a more complex example of the circuit in Fig. 3.2, the diagram of which with given loop currents is shown in Fig. 3.5. In this case, the number of branches, the number of nodes, then the number of independent circuits and equations using the circuit current method is equal. For branch currents we can write

The first three circuits do not contain ideal current sources, then, taking into account (3.28) and using Ohm’s law, the equations of Kirchhoff’s second law can be written for them,

In the fourth circuit there is an ideal current source, therefore for it the equation of Kirchhoff’s second law is not compiled, and the circuit current is equal to the source current (they coincide in direction),

Substituting (3.30) into system (3.29), after transformation we obtain three equations for loop currents in the form

The system of equations (3.31) can be solved analytically (for example, using the substitution method - do this), having obtained formulas for loop currents, and then from (3.28) determine the branch currents. For numerical calculations it is convenient to use the MathCAD software package; an example of the program is shown in Fig. 3.6. The calculation results coincide with the calculations shown in Fig. 3.3. As you can see, the loop current method requires the compilation and solution of a smaller number of equations compared to the general method of calculation using the Kirchhoff equations.

3.5. Nodal stress method

Nodal stress method is based on Kirchhoff's first law, and the number of equations is equal to .

All nodes in the chain are selected and one of them is selected as basic, which is assigned zero potential. Potentials (voltages) ... of the remaining nodes are counted from the base node, their positive directions are usually selected by an arrow to the base node. The currents of all branches are expressed through nodal voltages using Ohm’s law and Kirchhoff’s second law

and for the nodes the equations of Kirchhoff's first law are written.

Consider the example of the circuit shown in Fig. 3.1, for the nodal stress method its diagram is shown in Fig. 3.7. The lower node is designated as the base node (for this, the symbol “earth” is used - the point of zero potential), the voltage of the upper node relative to the base designation

Rice. 3.7 is indicated as . Let's express it through

its branch currents

According to Kirchhoff’s first law, taking into account (3.32), we write the only equation for the nodal stress method (),

Solving the equation, we get

and from (3.32) we determine the currents of the branches

The results obtained coincide with those obtained by the previously discussed methods.

Let's consider a more complex example of the circuit shown in Fig. 3.2 with the same initial data, its diagram is shown in Fig. 3.8. In the node chain, the bottom one is chosen as the base one, and the other three are indicated by numbers in circles. Introduced

positive on - Fig. 3.8

board and designation

values of nodal stresses, and.

According to Ohm's Law using Kirchhoff's second law, we determine the branch currents,

According to Kirchhoff’s first law, for nodes numbered 1, 2 and 3, it is necessary to create three equations,

Substituting (3.36) into (3.37), we obtain a system of equations for the nodal stress method,

After transforming and bringing similar ones we get

The program for calculating node voltages and branch currents is shown in Fig. 3.9. As can be seen, the results obtained coincide with those previously obtained by other calculation methods.

Carry out an analytical calculation of the node voltages, obtain formulas for the branch currents and calculate their values.

3.6. Overlay method

Overlay method is as follows.

The calculation is carried out as follows. In a circuit containing several sources, each of them is selected in turn, and the rest are turned off. In this case, chains with one source are formed, the number of which is equal to the number of sources in the original circuit. In each of them, the desired signal is calculated, and the resulting signal is determined by their sum. As an example, consider calculating the current in the circuit shown in Fig. 3.2, its diagram is shown in Fig. 3.10a.

When an ideal current source is turned off (its circuit is broken), the circuit shown in Fig. is obtained. 3.9b, in which the current is determined by any of the considered methods. The ideal voltage source is then switched off (replaced by a short circuit) to produce the circuit shown

in Fig. 3.9a, in which the current is located. The required current is

Perform analytical and numerical calculations yourself, compare with previously obtained results, for example, (3.20).

3.7. Comparative analysis of calculation methods

The calculation method based on Ohm's law is suitable for relatively simple single-source circuits. It cannot be used to analyze circuits of complex structure, for example, a bridge type like Fig. 3.9.

The general method of calculating a circuit based on the equations of Ohm's and Kirchhoff's laws is universal, but requires the compilation and solution of a system of equations, which is easily converted into a system of equations. With a large number of branches, computational costs increase sharply, especially when analytical calculations are necessary.

The methods of loop currents and nodal voltages are more effective, since they lead to systems with fewer equations, equal to and respectively. Given that

The loop current method is more effective, otherwise it is advisable to use the nodal voltage method.

The superposition method is convenient when the circuit is dramatically simplified when the sources are turned off.

Task 3.5. Using the general calculation method, the methods of loop currents and nodal voltages, determine in the circuit Fig. 3.14 voltage at mA kOhm, kOhm, kOhm, kOhm, kOhm. Conduct a comparative analysis

calculation methods. Rice. 3.14

4. HARMONIC CURRENTS AND VOLTAGES