A complex electrical chain is called a chain with several closed circuits, with any placement of power supply and consumers in it, which cannot be reduced to a combination of consecutive and parallel connections.
The basic laws for calculating the chains, along with the Ohm law, are two Kirchhoff law, using which, you can find the distribution of currents and stresses on all sections of any complex chain.
In § 2-15, we got acquainted with one method of calculating complex chains, overlay method.
The essence of this method lies in the fact that the current in any branch is an algebraic amount of currents created in it by all alternately acting. d. s. chains.
Consider the calculation of the complex chain by the method of nodal and contour equations or equations according to the laws of Kirchoff.
To find the currents in all branches of chains, it is necessary to know the resistance of the branches, as well as the values \u200b\u200band directions of all e. d. s.
Before drawing up the equations according to the laws of Kirchhoff, it is necessary to arbitrarily set the directions of currents in the branches, showing them on the arrow scheme. If the selected current direction in any branch is opposite to valid, then after solving the equations, this current is obtained with a minus sign.
The number of necessary equations is equal to the number of unknown currents; The number of equations compiled according to the first law of Kirchhoff should be per unit less than the number of chains nodes, the remaining equations are compiled according to the second Circhoff law. In the preparation of equations on the second law of Kirchhoff, the simplest contours should be chosen, and each of them must contain at least one branch that did not enter into previously composed equations.
Calculation of a complex chain using two Kirchhoff Equations Consider on the example.
Example 2-12. Calculate currents in all branches of chain rice. 2-11, if e. d. s. sources, and resistance branches.
Internal resistances of sources neglected.
Fig. 2-11. Complex electrical circuit with two power sources.
The selected arbitrarily directions of currents in the branches are shown in Fig. 2-11.
Since the number of unknown currents is three, then three equations must be made.
At two nodes of the chain, the same nodal equation is necessary. Welcome it for a point in:
4 The second equation will write, bypassing in the direction of the clockwise direction of the contour of the flax,
The third equation will write, bypassing in the direction of the movement of the clockwise contour of the Agvzhza,
Replacing in equations (2-49) and (2-50) alphabetic designations with numerical values, we obtain:
Replacing in the last equation for its expression of equation (2-48), we obtain;
Multiplying equation (2-52a) by 0.3 and resulted with equation (2-51), we obtain.
This article for those who are just starting to study the theory electrical chains. As always, we will not climb the formulas to the Debres, but we will try to explain the basic concepts and the essence of things important for understanding. So, welcome to the world of electrical chains!
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Electrical chains
- This is a combination of devices for which electric current flows.
Consider the simplest electrical chain. What is it from? It has a generator - current source, receiver (for example, light bulb or electric motor), as well as a transmission system (wire). So that the chain becomes exactly a chain, and not a set of wires and batteries, its elements must be interconnected by conductors. Current can flow only on a closed chain. Let's give another definition:
- This is the current source, transmission line and receiver.
Of course, the source, receiver and wires are the easiest option for the elementary electrical circuit. In reality, different chains include many more elements and accessories: resistors, capacitors, switches, ammeters, voltmeters, switches, contact connections, transformers, and so on.
Classification of electrical chains
By destination, electrical chains are:
- Power electric circuits;
- Electrical control chains;
- Electrical measurement chains;
Power chains Designed for transmission and distribution electrical Energy. It is the power chains that lead to the consumer.
Also, the chains are separated by the strength of the current in them. For example, if the current in the circuit exceeds 5 amp, then the power chain. When you click the kettle, included in the socket, you closure the power circuit.
Electrical control chains It is not powerful and are intended to act or change the parameters of the electrical devices and equipment. An example of a control circuit - control equipment, control and alarm.
Electrical circuit measurement Designed to fix changes in the parameters of the electrical equipment.
Calculation of electrical chains
Calculate the chain means finding all currents in it. There are different methods for calculating electrical circuits: the laws of Kirchhof, the contour current method, the method of nodal potentials and others. Consider the use of the contour current method on the example of a particular chain.
First, lay out the contours and denote the current in them. Current direction can be chosen arbitrarily. In our case - clockwise. Then, for each contour, the equation 2 of the 2 Circhoff law. The equations are compiled as: the circuit current is multiplied by the contour resistance, the current of the current of other contours and the overall resisters of these contours are added to the resulting expression. For our scheme:
The resulting system is solved with the standing of the source data of the task. Currents in the branches of the source chain found as an algebraic amount of contour currents
Whatever the chain you need to calculate, our specialists will always help to cope with the tasks. We will find all currents according to the Rules of Kirchhoff and solve any example of transient processes in electrical circuits. Have fun from studying with us!
IN chains direct current Permanent voltages act, constant currents occur and there are only resistive elements (resistance).
An ideal source of voltage The source is called, the voltage on the clips of which created by the internal electromotive force (EMF), depends on the current generated by it in the current (Fig. 6.1a). At the same time there is equality. The voltamper characteristic of the ideal voltage source is shown in Fig. 6.1b.
An ideal source of current They call the source that gives up a current that does not depend on the voltage on the source clips, fig. 6.2a. Its voltamper characteristic is shown in Fig. 6.2b.
IN resistance The connection between voltage and current is determined by the law of OMA as
An example of an electrical circuit is shown in Fig. 6.3. It is allocated in it branchesconsisting of a sequential connection of several elements (source E and resistance) or one element (s) and knots - Points of connecting three and more branches marked with bold points. The considered example has branches and nodes.
In addition, the chains are allocated independent closed contoursnot containing ideal current sources. Their number is equal. In the example in fig. 6.3 their number, for example, contours with branches E and shown in Fig. 6.3 ovals with arrows indicating positive direction Contour bypass.
The connection of currents and stresses in the chain is determined by the laws of Kirchhoff.
First Kirchhoff's law: the algebraic amount of currents convergent in the electrical circuit unit is zero,
The currents flowing into the node have a plus sign, and flowing minus.
The second law of Kirchhoff: The algebraic amount of stresses on the elements of a closed independent circuit is equal to the algebraic amount of EMF of ideal voltage sources included in this circuit,
Voltages and EMFs are taken with a plus sign if their positive directions coincide with the direction of circuit bypass, otherwise a minus sign is used.
For shown in Fig. 6.3 Example According to Ohm's law, we obtain the subsystem of component equations
According to the laws of Kirchhof, the subsystem of topological equations of the chain has the form
Calculation based on the Law Ohm
This method is convenient for calculating relatively simple chains with one source of the signal. It implies the calculation of the resistance of the chain areas for which the knowledge is known
current (or voltage), followed by the definition of an unknown voltage (or current). Consider an example of calculating the chain, the diagram of which is shown in Fig. 6.4, with a current of the ideal source A and resistances Ohm, Ohm. It is necessary to determine the currents of the branches and, as well as voltages on the resistances, and.
Known source current, then you can calculate the resistance of the chain relative to the clips of the current source (parallel resistance connection and sequentially connected
Fig. 6.4 resistance and),
Voltage at the current source (on the resistance) is equal
Then you can find the currents of the branches.
The results obtained can be checked using the first Kirchhoff law in the form. Substituting the calculated values, we get a, which coincides with the current source current.
Knowing currents of branches, it is not difficult to find voltages on the resistances (the value has already been found)
According to the second law of Kirchhoff. Folding the results obtained, we are convinced of it.
Calculation of the chain according to the Kirchhoff equations
We will calculate currents and stresses in the chain shown in Fig. 6.3 at and. The chain is described by the system of equations (6.4) and (6.5), from which the branches of the branches are obtained
From the first equation express, and from the third
Then from the second equation we get
and, therefore
From the equations of the law of the Ohm
For example, for the chain in Fig. 6.3 In general, we get
Substituting B. left part equality (6.11) previously obtained expressions for currents, we get
what corresponds to the right part of the expression (6.11).
Similar calculations can be done for the chain in Fig. 6.4.
The condition balance sheet allows you to additionally monitor the correctness of the calculations.
The presentation of methods for calculating and analyzing electrical circuits is usually reduced to finding currents of branches with known values \u200b\u200bof EDC and resistance.
The methods of calculating and analyzing the electrical circuits of DC are suitable for both alternating current circuits.
2.1 Equivalent Resistance Method
(Method of coagulation and deployment of the chain).
This method is used only for electrical circuits containing one power source. To calculate, individual sections of the scheme containing consecutive or parallel branches are simplified by replacing them with equivalent resistances. Thus, the chain is carved to one equivalent resistance of the circuit connected to the power source.
Then the current of the branch containing the EMF is determined, and the diagram unfolds in the reverse order. At the same time, the incidence of stresses and currents of the branches are calculated. So, for example, in Scheme 2.1 BUT Resistance R.3 and R.4 included sequentially. These two resistances can be replaced with one equivalent.
R.3,4 = R.3 + R.4
After such a replacement, a simpler scheme is obtained (Fig.2.1 B. ).
Here you should pay attention to possible mistakes In determining the method of resistance connections. For example resistance R.1 and R.3 Cannot be considered connected consistently, as well as resistance R.2 and R.4 Cannot be considered connected in parallel, because it does not correspond to the main features of the serial and parallel connection.
Figure 2.1 to the calculation of the electrical circuit method
Equivalent resistances.
Between resistances R.1 and R.2 , at point IN, there is a branch with a current I.2 . Therefore current I.1 Will not be equal to the current I.3 Thus resistance R.1 and R.3 Cannot be considered enabled sequentially. Resistance R.2 and R.4 On the one hand are attached to the common point. D.and on the other hand - to different points IN and FROM. Consequently, the voltage attached to the resistance R.2 and R.4 It is impossible to be considered in parallel.
After replacing the resistance R.3 and R.4 Equivalent resistance R.3,4 and simplifying the scheme (Fig. 2.1 B.), it is more clearly seen that resistance R.2 and R.3,4 It is connected in parallel and they can be replaced by one equivalent, based on the fact that, with a parallel combination of branches, the total conductivity is equal to the amount of the velocities of the branches:
GBD.= G.2 + G.3,4 , Or = + From
RBD.=
And get an even simpler diagram (Fig. 2.1, IN). In it resistance R.1 , RBD., R.5 Connected sequentially. Replacing these resistance to one equivalent resistance between points A. and F., get the simplest scheme (Figure 2.1, G.):
Raf.= R.1 + RBD.+ R.5 .
In the resulting scheme, you can determine the current in the chain:
I.1
=
.
Currents in other branches are not difficult to determine moving from the scheme to the diagram in the reverse order. From the scheme in Figure 2.1 INYou can determine the voltage drop on the plot B., D. Chains:
UBD.= I.1 · RBD.
Knowing the voltage drop on the plot between the dots B. and D. You can calculate Toki. I.2 and I.3 :
I.2 = , I.3 =
Example 1. Let (Fig 2.1 BUT) R.0 \u003d 1 ohms; R.1 \u003d 5 ohms; R.2 \u003d 2 ohms; R.3 \u003d 2 ohms; R.4 \u003d 3 ohms; R.5 \u003d 4 ohms; E.\u003d 20 V. Find currents of the branches, make up the balance of capacity.
Equivalent resistance R.3,4 Equal to the amount of resistance R.3 and R.4 :
R.3,4 = R.3 + R.4 \u003d 2 + 3 \u003d 5 ohms
After replacement (Figure 2.1 B.) Calculate the equivalent resistance of two parallel branches R.2 and R.3,4 :
RBD.= \u003d\u003d 1,875 ohms,
And the scheme is still simplified (Fig. 2.1 IN).
Calculate the equivalent resistance of the entire chain:
R.Ek.= R.0 + R.1 + RBD.+ R.5 \u003d 11.875 ohms.
Now you can calculate the total circuit current, i.e., generated by the source of energy:
I.1 \u003d \u003d 1.68 A.
Voltage drop on the plot BD. It will be:
UBD.= I.1 · RBD.\u003d 1.68 · 1,875 \u003d 3.15 V.
I.2 = = \u003d 1.05 A;I.3 \u003d\u003d\u003d 0.63 A.
Make a capacity balance:
E · I1 \u003d i12.· (R0 + R1 + R5) + i22· R2 + i32.· R3.4,
20 · 1.68 \u003d 1.682 · 10 + 1,052 · 3 + 0.632 · 5,
33,6=28,22+3,31+1,98 ,
The minimum discrepancy is due to rounding when calculating currents.
In some schemes, it is impossible to highlight the resistance to the included between themselves sequentially or in parallel. In such cases, it is better to use other universal methodswhich can be applied to calculate the electrical circuits of any complexity and configuration.
2.2 Method of Kirchhoff laws.
The classic method for calculating the complex electrical circuits is the direct application of Kirchhoff's laws. All other methods for calculating electrical chains proceed from these fundamental electrical engineering laws.
Consider the application of the laws of Kirchhoff to determine the currents of the complex chain (Fig. 2.2) if its EMF and resistance are specified.
Fig. 2.2. To the calculation of a complex electrical circuit for
Definitions of currents according to the laws of Kirchhoff.
The number of independent circuit currents is equal to the number of branches (in our case M \u003d 6). Therefore, to solve the problem, it is necessary to compile a system of six independent equations, together according to the first and second laws of Kirchhoff.
The number of independent equations compiled according to the first law of Kirchhoff always per unit less than nodes,T. K. The sign of independence is the presence of at least one new current in each equation.
Since the number of branches M. Always more than knots TO, The missing number of equations is compiled according to the second law of Kirchoff for closed independent contours,That is, in order for each new equation, at least one new branch.
In our example, the number of nodes is four - A., B., C., D.Therefore, we will make only three equations according to the first law of Kirchhoff, for any three nodes:
For node A: I1 + i5 + i6 \u003d 0
For node B: i2 + i4 + i5 \u003d 0
For node C: i4 + i3 + i6 \u003d 0
According to the second law of Kirchhoff, we also need to create three equations:
For contour A., C., B, A:I.5 · R.5 — I.6 · R.6 — I.4 · R.4 =0
For contour D.,A.,IN,D.: I.1 · R.1 — I.5 · R.5 — I.2 · R.2 \u003d E1-E2
For contour D., In, s,D.: I.2 · R.2 + I.4 · R.4 + I.3 · R.3 \u003d E2.
Solving a system of six equations, you can find currents of all sections of the scheme.
If, in solving these equations, currents of individual branches will turn out negative, this will indicate that the actual direction of currents is opposite to an arbitrarily selected direction, but the current value will be correct.
Clarify now the procedure for calculating:
1) to arbitrarily choose and apply the positive directions of currents of the branches on the scheme;
2) compile a system of equations on the first law of Kirchoff - the number of equations per unit less than nodes;
3) arbitrarily choose the direction of accounting for independent contours and make a system of equations on the second Circhoff law;
4) decide general System equations, calculate currents, and, in case of obtaining negative results, change the directions of these currents.
Example 2.. Let in our case (Fig. 2.2.) R.6 = ∞ that is equivalent to the cliff of this section of the chain (Fig. 2.3). We define currents of the branches of the remaining chain. Calculate the balance of power, if E.1 =5 IN, E.2 =15 B, R.1 \u003d 3 ohms, R.2 = 5 ohms R. 3 =4 Ohm R. 4 =2 Ohm R. 5 =3 Ohm.
Fig. 2.3 Scheme to solve the problem.
Decision. 1. We choose an arbitrarily direction of the branches of the branches, we have three: I.1 , I.2 , I.3 .
2. We will make only one independent equation according to the first law of Kirchoff, because in the scheme only two nodes IN and D..
For node IN: I.1 + I.2 — I.3 \u003d O.
3. Select independent contours and the direction of their bypass. Let the outlines of the Davd and CVTF be clocked clockwise:
E1-E2 \u003d i1 (R1 + R5) - I2 · R2,
E2 \u003d i2.· R2 + i3.· (R3 + R4).
We substitute the resistance and emf values.
I.1 + I.2 — I.3 =0
I.1 +(3+3)- I.2 · 5=5-15
I.2 · 5+ I.3 (4+2)=15
Deciding the system of equations, calculate the currents of the branches.
I.1 =- 0.365A. ; I.2 = I.22 — I.11 = 1,536A. ; I.3 \u003d 1,198a.
How to verify the correctness of the decision to make the balance of capacity.
Σ EIII \u003d.Σ IY2 · RY
E1 · I1 + E2 · I2 \u003d I12 · (R1 + R5) + I22 · R2 + i32 · (R3 + R4);
5 (-0.365) + 15 · 1,536 \u003d (-0.365) 2 · 6 + 1,5632 · 5 + 1,1982 · 6
1,82 + 23,44 = 0,96 + 12,20 + 8,60
21,62 ≈ 21,78.
Difference is insignificant, therefore the solution is true.
One of the main disadvantages of this method is a large number of equations in the system. More economical when computing is The method of contour currents.
2.3 The method of contour currents.
When calculating Using contour currents It is believed that in each independent circuit flows its (conditional) CONTURRY TOK.. Equations are relative to contour currents according to the second law of Kirchhoff. Thus, the number of equations is equal to the number of independent contours.
Real currents of branches are defined as an algebraic amount of contour currents of each branch.
Consider, for example, a scheme of Fig. 2.2. Throw it into three independent contours: FROM YOU; AUD.BUT; Sun.D.IN And we agree that for each of them passes its contour, respectively I.11 , I.22 , I.33 . The direction of these currents will choose in all circuits the same clockwise as shown in the figure.
By comparing the contour currents of the branches, it can be established that according to external branches, real currents are equal to contour, and in the inner branches they are equal to the sum or difference in contour currents:
I1 \u003d i22, i2 \u003d i33 - i22, i3 \u003d i33,
I4 \u003d i33 - i11, i5 \u003d i11 - i22, i6 \u003d - i11.
Consequently, according to the known contour currents, the scheme can easily determine the valid currents of its branches.
To determine the contour currents of this scheme, it suffices to be only three equations for each independent circuit.
By making the equation for each contour, it is necessary to take into account the effect of adjacent circuits of currents on adjacent branches:
I11 (R5 + R6 + R4) - I22 · R5 - i33 · R4 \u003d O,
I22 (R1 + R2 + R5) - I11 · R5 - I33 · R2 \u003d E1 - E2,
I.33 (R.2 + R.3 + R.4 ) — I.11 · R.4 — I.22 · R.2 = E.2 .
So, the procedure for calculating the contour currents is performed in the following sequence:
1. Install independent contours and select contour currents in them;
2. Denote the currents of the branches and arbitrarily give them directions;
3. establish the relationship of the valid currents of branches and contour currents;
4. Make a system of equations on the second Circhoff law for contour currents;
5. Solve the system of equations, find contour currents and determine valid branches.
Example 3. We will solve the problem (example 2) by the method of contour currents, the source data are the same.
1. Only two independent circuits are possible in the task: choose the contours AUD.BUT and Sun.D.IN, and take the direction of contour currents in them I.11 and I.22 clockwise (Fig. 2.3).
2. Valid branches I.1 , I.2, I.3 And their directions are also shown on (Fig. 2.3).
3. Communication of valid and contour currents:
I.1 = I.11 ; I.2 = I.22 — I.11 ; I.3 = I.22
4. Make a system of equations for contour currents according to the second law of Kirchhoff:
E1 - E2 \u003d I11 · (R1 + R5 + R2) - I22 · R2
E2 \u003d i22 · (R2 + R4 + R3) - I11 · R2;
5-15 \u003d 11 · I.11 -five· I.22
15 \u003d 11 · I.22 -five· I.11 .
Deciding the system of equations, we get:
I.11 = -0,365
I.22 \u003d 1,197, then
I.1 = -0,365; I.2 = 1,562; I.3 = 1,197
As we see the actual values \u200b\u200bof the branches of the branches coincide with the obtained values \u200b\u200bin Example 2.
2.4 Node voltage method (two node method).
Often there are schemes containing only two nodes; In fig. 2.4 depicts one of these schemes.
Figure 2.4. By calculating the electrical circuits by two nodes.
The most rational method for calculating currents in them is Method of two nodes.
Under Method of two nodes Understand the method for calculating the electrical circuits, in which the current voltage (with its help then the currents of the branches are determined) take the voltage between the two nodes BUT and IN schemes - U.AU.
Voltage U.AU It can be found from Formula:
U.AU=
In the numerator of the formula, the sign "+", for the branch containing the EMF, is taken if the direction of the EMF of this branch is directed towards increasing the potential, and the "-" sign. In our case, if the potential of the node and take the above potential of the node in (the potential of the node in taking equal to zero), E1G.1 , takes with the "+" sign, and E2 ·G.2 with the sign "-":
U.AU=
Where G. - conductivity of branches.
Determining the nodal voltage, you can calculate currents in each branch of the electrical circuit:
I.TO\u003d (U.AU) G.TO.
If the current has a negative value, then its actual direction is the opposite of the indicated on the diagram.
In this formula, for the first branch, since the current I.1 Coincides with the direction E1, then its value is taken with a plus sign, and U.AU With a minus sign, since the current is sent to towards the current. In the second branch and E2. and U.AU They are directed towards the current and take a minus sign.
Example 4.. For the scheme Fig. 2.4 If E1 \u003d 120V, E2 \u003d 5Ω, R1 \u003d 2, R2 \u003d 1, R3 \u003d 4Ω, R4 \u003d 10).
UAV \u003d (120 · 0.5-50 · 1) / (0.5 + 1 + 0.25 + 0.1) \u003d 5.4 V
I1 \u003d (E1-UAV) · G1 \u003d (120-5.4) · 0.5 \u003d 57,3A;
I2 \u003d (- E2-UAV) · G2 \u003d (-50-5.4) · 1 \u003d -55,4A;
I3 \u003d (O-UAV) · G3 \u003d -5.4 · 0.25 \u003d -1,35a;
I4 \u003d (O-UAV) · G4 \u003d -5.4 · 0.1 \u003d -0.54a.
2.5. Nonlinear DC circuits and their calculation.
So far, we have considered the electrical chains whose parameters (resistances and conductivity) were considered independent of the value and the direction of the current passing or the voltage applied to them.
In practical conditions, most encountered elements have parameters depend on current or voltage, the volt-amps characteristic of such elements is non-linear (Fig. 2.5), such elements are called Nonlinear. Nonlinear elements are widely used in different areas Techniques (automation, computing equipment and others).
Fig. 2.5. Volt-ampere characteristics of nonlinear elements:
1 - semiconductor element;
2 - thermal resistance
Nonlinear elements allow you to implement processes that are impossible in linear circuits. For example, stabilize the voltage, strengthen the current and others.
Nonlinear elements are managed and unmanageable. Unmanaged nonlinear elements work without the influence of control exposure (semiconductor diodes, thermal resistance and others). Controlled elements are under the influence of control exposure (thyristors, transistors and others). Uncontrollable nonlinear elements have one volt-ampere characteristic; Managed - family characteristics.
The calculation of the electrical circuits of DC is most often produced by graphic methods that are applicable in any form of volt-ampere characteristics.
Sequential connection of nonlinear elements.
In fig. 2.6 shows a diagram of a sequential connection of two nonlinear elements, and in Fig. 2.7 their voltamper characteristics - I.(U.1 ) and I.(U.2 )
|
|
Fig. 2.6 Sequential Connection Scheme
Nonlinear elements.
Fig. 2.7 Voltample characteristics of nonlinear elements.
Build a volt-ampere characteristic I.(U.), expressing current dependence I. In the chain from the voltage attached to it U.. Since the current of both sections of the chain are the same, and the amount of voltages on the elements is equal to the applied (Fig. 2.6) U.= U.1 + U.2 then to build a characteristic I.(U.) It is enough to sum up the abscissions of the specified curves I.(U.1 ) and I.(U.2 ) For certain current values. Using the characteristics (Fig. 2.6), you can solve the various task for this chain. Let, for example, a value applied to the current U. And it is required to determine the current in the chain and the distribution of stresses at its sections. Then on the characteristics I.(U.) We celebrate the point BUT corresponding to the applied voltage U. and spend from it horizontal crossing curves I.(U.1 ) and I.(U.2 ) before intersection with the owner of the ordinate (point D.) which shows the amount of current in the chain, and the segments IND. and FROMD. The magnitude of the voltage on the chain elements. And on the contrary, according to a specified current, you can determine the voltage of both the general and on the elements.
Parallel compounds of nonlinear elements.
With a parallel connection of two nonlinear elements (Fig. 2.8) with specified volt-ampere characteristics in the form of curves I.1 (U.) and I.2 (U.) (Fig. 2.9) Voltage U. It is general, and current I in the unbranched part of the chain is equal to the amount of currents of the branches:
I. = I.1 + I.2
Fig. 2.8 Scheme of parallel connection of nonlinear elements.
Therefore, to obtain a general characteristic I (U), it is sufficient for arbitrary values \u200b\u200bof the voltage U in Fig. 2.9 To sum up the ordinates of the characteristics of individual elements.
Fig. 2.9 Volt-ampere characteristics of nonlinear elements.
05.12.2014
Lesson 25 (9 Class)
Subject. Calculation of simple electrical chains
Solving any task for calculating the electrical circuit should be started from the choice of the method that calculates will be made. As a rule, one and the same task can be solved by several methods. The result in any case will be the same, and the complexity of calculations can differ significantly. To correctly select the calculation method, you must first determine which class this electric chain includes: to simple electrical circuits or complex.
TO simple Electrical circuits that contain either one source of electrical energy or several electrical circuits are in one branch. Below are two schemes of simple electrical circuits. The first scheme contains one voltage source, in which case the electrical circuit is unambiguously related to simple circuits. The second contains two sources already, but they are in one branch, therefore it is also a simple electrical chain.
The calculation of simple electrical chains is usually produced in such a sequence:
1. First simplify the scheme sequentially converting all passive elements of the circuit in one equivalent resistor. To do this, it is necessary to identify the sections of the circuit on which the resistors are connected in series or in parallel, and according to the known formulas to replace them with equivalent resistors (resistances). The chain is gradually simplified and lead to the presence in the chain of one equivalent resistor.
2. Next, this procedure is carried out with active elements of the electrical circuit (if their number of more than one source). By analogy with the previous point, we simplify the scheme until we receive in the diagram one equivalent voltage source.
3. As a result, we give any simple electrical circuit To the following form: 
Now it is possible to apply the Ohma law - the relation (1.22) and actually determine the value of the current flowing through the source of electrical energy.
combined Homework
1. F.Y. Boxinov, N.M. Kiryukhin, E.A. Kiryukhina. Physics, grade 9, "Ranok", Kharkov, 2009. § 13-14 (p. 71-84) repeat.
2. Exercise 13 (task 2, 5), exercise 14 (task 3, 5, 6) decide.
3. Remove in the working notebook problem 1, 3, 4 (see the following page).
aI with a balance
PI DC. Examples of solved tasks
Introduction
The problem of tasks is an integral part of physics learning, since in the process of solving problems, the formation and enrichment of physical concepts occurs, the physical thinking of students is developing and their skills to apply knowledge in practice are being improved.
In the course of solving problems, the following didactic objectives can be completed and successfully implemented:
- Nomination of the problem and creating a problem situation;
- Generalization of new information;
- Formation of practical skills and skills;
- Checking the depth and strength of knowledge;
- Consolidation, generalization and repetition of the material;
- Implementation of the principle of polytechnism;
- Development of creative abilities of students.
Along with this, when solving tasks, schoolchildren are raised by the hardworking, inquisitiveness of the mind, cutter, independence in judgments, interest in teaching, will and character, perseverance in achieving the goal. To implement these purposes, it is especially convenient to use non-traditional tasks.
Tasks for calculating the electrical circuits of DC
According to the school program for consideration of this topic, very little time is given, so students more or less successfully seize the methods of solving problems this type. But often such types of tasks are found. olympiad tasksBut they are based on the school course.
To such, non-standard tasks for the calculation of the electrical circuits of DC include tasks whose schemes are:
2) symmetrical;
3) consist of complex mixed connections of elements.
In the general case, any chain can be calculated using the laws of Kirchhoff. However, these laws are not included in the school curriculum. In addition, it is not possible to solve a system from a large number of equations with many unknowns, not many students and this path is not best way waste time. Therefore, you need to be able to use methods that allow you to quickly find resistance and containers.
Method of equivalent schemes
The equivalent scheme method is that the initial scheme must be represented as sequential sections, on each of which the connection of the diagram elements is either sequentially or in parallel. For such a presentation, the scheme must be simplified. Under the simplification of the scheme, we understand the connection or disconnection of any nodes of the scheme, removing or adding resistors, capacitors, ensuring that the new scheme from sequentially and parallel to the connected elements is equivalent to the original.
The equivalent scheme is such a scheme that when the identical voltage is applied to the original and transformed circuit, the current in both chains will be the same in the respective areas. In this case, all calculations are made with a transformed circuit.
To draw an equivalent circuit for a chain with a complex mixed compound of resistors, you can use multiple receptions. We will restrict ourselves to considering the details of only one of them - the method of equipotential nodes.
This method is that in symmetric diagrams, points with equal potentials are found. These nodes are connected to each other, and if some section of the scheme was included between these points, then it is discarded, since due to equality of potentials at the ends of the current it does not flow, and this section does not affect the general resistance of the circuit.
Thus, the replacement of several nodes of equal potentials leads to a simpler equivalent scheme. But sometimes it is more expedient to reverse the replacement of one node
several nodes with equal potentials, which does not disrupt electrical conditions in the rest.
Consider examples of solving problems with these method.
Z A D A C A №1
Decision:
By virtue of the symmetry of the branches of the circuit point C and D are equipotential. Therefore, the resistor between them can exclude. Equipotential points C and D connect to one node. We get a very simple equivalent scheme:
The resistance of which is equal:
Rav \u003d Rac + RCD \u003d R * R / R * R + R * R / R + R \u003d R.
Z A D A C N № 2
Decision:
At points F and F` The potentials are equal, then resistance between them can be discarded. The equivalent scheme looks like this:
Resistance areas dnb; f`c`d` D`, n`, b`; FCD are equal to each other and equal to R1:
1 / R1 \u003d 1 / 2R + 1 / R \u003d 3/2R
With this in mind, a new equivalent scheme is obtained:
Its resistance and resistance of the source chain Rava is:
1 / Rav \u003d 1 / R + R1 + R1 + 1 / R + R1 + R1 \u003d 6 / 7R
Z A D A C A number 3.
Decision:
Points C and D have equal potentials. Except resistance between them. We get an equivalent scheme:
The desired resistance is Rava equal:
1 / Rav \u003d 1/2R + 1 / 2R + 1 / R \u003d 2 / R
Z A D A C number 4.
Decision:
As can be seen from the circuit, the nodes 1,2,3 have equal potentials. Connect them to the node 1. Nodes 4,5,6 have also equal to the potentially connected to the node 2. We obtain such an equivalent scheme:
Resistance on the A-1 section, R 1-equal to the resistance on the section 2-B, R3 and equal:
Resistance in section 1-2 is equal: R2 \u003d R / 6.
Now the equivalent scheme is obtained:
General resistance Rava is equal:
RV \u003d R1 + R2 + R3 \u003d (5/6) * R.
C and D A h and number 5.
Decision:
Points C and F-equivalent. Connect them into one node. Then the equivalent scheme will have the following form:
Resistance on the AC section:
Resistance on the FN plot:
Resistance on the DB plot:
It turns out an equivalent scheme:
The desired general resistance is:
Task number 6.
Decision:
We will replace a common node about three nodes with equal potentials O, O 1, O 2. We will receive an equivalent system:
Resistance on the ABCD plot:
Resistance on the area A`b`d`
Resistance to the site ACV
We get an equivalent scheme:
The desired total resistance of the R AB chain is:
R AB \u003d (8/10) * R.
Task number 7.
Decision:
"We divide" a node about two equipotential angle about 1 and 2. Now the scheme can be represented as parallel connection two identical chains. Therefore, it is enough to consider one of them in detail:
Resistance to this scheme R 1 is:
Then the resistance of the entire chain will be equal to:
Z A D A h and number 8
Decision:
Nodes 1 and 2 are equipotential, so connect them to one node I. Nodes 3 and 4 are also equipotential - connections to another node II. Equivalent scheme has the form:
Resistance to the section A - i is equal to resistance at the site B - II and equal:
The resistance of the I-5-6- II region is:
The resistance of the I- II site is equal.
