Ohm's law - the voltage drop across an element is equal to the product of the resistance value of this element by the value of the current flowing through it.
Kirchhoff's first law - the sum of currents flowing into the node is equal to the sum of the currents flowing out of the node.
Kirchhoff's second law - in a closed loop, the algebraic sum of the voltages of the sources of electrical energy is equal to the algebraic sum of the voltage drops across the circuit elements. When traversing the contour in an arbitrarily chosen direction, the stress values \u200b\u200bare taken with a plus if the direction of the loop traversal and the direction of the stresses coincide and are taken with a minus if there is no such coincidence.
Equivalent conversion calculation
This method is used for not very complex passive electrical circuits, such circuits are quite common, and therefore this method is widely used. The main idea of \u200b\u200bthe method is that the electrical circuit is sequentially converted ("rolled") to one equivalent element, as shown in Fig. 1.13, and the input current is determined. Then a gradual return to the original scheme ("unrolling") is carried out with the sequential determination of currents and voltages.
Calculation sequence:
1. Arranged conditionally positive directions of currents and voltages.
2. Sections of the chain are equivalently transformed step by step. In this case, at each stage, currents and voltages are placed in the circuit newly obtained after the conversion in accordance with clause 1.
3. As a result of the equivalent conversion, the value of the equivalent resistance of the circuit is determined.
4. Determine the input current of the circuit using Ohm's law.
5. Step by step returning to the original circuit, all currents and voltages are found in series.
Let's consider this method using an example (Fig. 1.15). In the original circuit, we place conditionally positive directions of currents in the branches and voltages on the elements. It is easy to agree that under the influence of the source E with the indicated polarity, the direction of currents and voltages is as shown by the arrows. For convenience of further explanation of the method, we denote nodes a and b on the diagram. This can be omitted for normal calculation.
Then, combining all the series-connected elements, we complete the equivalent transformation of the circuit (Fig. 1.15, c):
In the last circuit (Fig.1.15, c) we find the current I 1:
Now we return to the previous scheme (Fig. 1.15, b). We see that the found current I 1 flows through R 1 , R 2,3 , R 4 and creates a voltage drop across them. Let's find these voltages:
.
Returning to the original circuit (Fig.1.15, a), we see that the found voltage U ab is applied to the elements R 2 and R 3 .
Hence, we can write that 
U 2
= U 3
= U a, b
The currents in these elements are found from completely obvious relationships:
So, the scheme is calculated.
calculation using Kirchhoff's laws
This method is the most versatile and is used to calculate any circuits. when calculating by this method, the currents in the branches are initially determined, and then the voltages on all elements. currents are found from equations obtained using Kirchhoff's laws. since each branch of the circuit has its own current, the number of initial equations must be equal to the number of branches of the circuit. the number of branches is usually denoted by n... some of these equations are written according to the first Kirchhoff's law, and some - according to the second Kirchhoff's law. all equations obtained must be independent. this means that there are no equations that can be obtained by permutations of terms in an existing equation or by arithmetic operations between the original equations. when drawing up equations, the concepts of independent and dependent nodes and contours are used. consider these concepts.
independent node a node is called, which includes at least one branch that is not included in other nodes. if the number of nodes is denoted by to, then the number of independent nodes is ( to-1). in the diagram (Fig. 1.16) of two nodes, only one is independent.
independent circuit is called a contour that differs from other contours by at least one branch that is not included in other contours. otherwise, such a contour is called addicted .
if the number of branches in the chain is n, then the number of independent contours is [ n– (to–1)].
in the circuit (Fig. 1.16) there are only three circuits, but only two independent circuits, and the third is dependent. you can select independent contours arbitrarily, that is, as independent contours, you can select some during the first calculation, and during the second (repeated) calculation - others that were previously dependent. the calculation results will be the same.
if, according to the first Kirchhoff law, we compose equations for ( to–1) independent nodes, and, according to the second Kirchhoff law, draw up equations for [ n– (to–1)] independent contours, then the total number of equations will be equal to:
(K–1) + [n – (K–1)] = n.
This means that the required number of equations is available for the calculation.
Calculation sequence:
1. Arrange conditionally - positive directions of currents and voltages.
2. Determine the number of unknown currents, which is equal to the number of branches ( n).
3. Select independent nodes and independent contours.
4
... Using the first Kirchhoff's law, we compose ( TO–1) equations for independent nodes.
5. With the help of Kirchhoff's second law, we compose [ n– (TO–1)] equations for independent contours. In this case, the voltages on the elements are expressed in terms of the currents flowing through them.
6. We solve the composed system of equations and determine the currents in the branches. When receiving negative values \u200b\u200bfor some currents, it is necessary to change their directions in the circuit to the opposite, which are true.
7. Determine the voltage drop across all circuit elements.
Let us consider the calculation sequence using the example of the circuit shown in Fig. 1.16. Given the direction of the source E, we arrange conditionally positive directions of currents and voltages. The circuit has three branches, so we need to write three equations. There are two nodes in the circuit, therefore, only one of them is independent. We choose node 1 as an independent node. For it we write the equation according to the first Kirchhoff's law:
I
1
= I 2
+ I 3 .
Next, you need to compose two equations according to the second Kirchhoff's law. There are only three circuits in the circuit, but only two independent ones. As independent contours, select a contour from the elements E–R 1 –R 2 and a contour of elements R 2 – R 3. Walking around these two contours in the clockwise direction, we write the following two equations:
E = I 1 ,R 1 + I 2 R 2 ,
0 = – I 2 R 2 + I 3 R 3 .
We solve the three obtained equations and determine the currents in the branches. Then, through the found currents, according to Ohm's law, we determine the voltage drops across all elements of the circuit.
loop current calculation
Complex schemes are characterized by the presence of a significant number of branches. If the previous method is applied, this leads to the need to solve a system of a significant number of equations.
The loop current method makes it possible to significantly reduce the number of initial equations. When calculating by the method of loop currents, the concepts of an independent loop and a dependent loop, which we already know, are used. In addition to them, the following concepts are also used in this method:
– own contour element - an element related to only one contour;
– common contour element - an element related to two or more circuit contours.
We denote, as before, through TO the number of nodes, and after n the number of branches in the chain. Then the number of independent circuit contours is determined by the already known formula [ n– (TO–1)].
The method is based on the assumption that each independent circuit has its own loop current (Fig. 1.17), and first find the loop currents in independent loops. The currents in the branches of the circuit are determined through the loop currents. In this case, it is assumed that in the proper elements of the circuit the currents coincide with the loop current of the given circuit, and in the common elements the current is equal to the algebraic sum of the loop currents of those circuits to which this element belongs.
Calculation sequence:
1. The number of branches is determined ( n) and the number of nodes ( TO) chain. Find the number of independent contours [ n– (TO–1)].
2. Select [ n– (TO–1)] independent contour.
3. Selected conditionally positive direction of loop currents in each of the independent loops (usually indicated by an arrow).
4. For each of the independent contours, an equation is drawn up according to the second Kirchhoff's law. In this case, the voltage drop across its own elements is defined as the product of the loop current by the resistance value, and on common elements - as the product of the algebraic sum of all loop currents flowing through this element by the value of its resistance. The loop is bypassed, as a rule, in the direction of its own loop current.
5. The system from [ n– (TO–1)] equations and the loop currents are found.
6. The currents in the branches of the circuit are found as follows:
- in own elements of the circuit, the current is equal to the circulating current;
- in the common elements of the circuit, the current is equal to the algebraic sum of the currents flowing through this element.
Let us consider in general the application of this method for calculating the circuit shown in Fig. 1.17.
This scheme has three branches and two nodes, therefore, it has only two independent contours. We select these contours and show in them the directions (arbitrarily) of the contour currents I k1 and I k2. We compose two equations according to the second Kirchhoff's law:
.
Having solved this system of equations, we find the loop currents I to 1 and I to 2. Then we determine the currents in the branches:
I 1 = I to 1, I 3 = I to 2, I 2 = I to 1 - I to 2.
CALCULATION BY THE OVERLAPPING METHOD
The method is used to calculate circuits containing several (two or more) sources of electrical energy. We emphasize that this method is applicable for calculating only linear circuits. The method is based on the assumption that the current in each branch of the circuit is equal to the algebraic sum of the currents generated by each source. Therefore, it is necessary to determine the currents generated by each source separately, and then sum them up taking into account the directions.
Calculation sequence:
1. Only one EMF source is left in the electrical circuit. Instead of the excluded EMF source, either a resistor is installed, the value of which is equal to the value of the internal resistance of the EMF source, or a jumper if the internal resistance of the source is zero.
2. The currents in all branches created by this EMF source are determined.
3. The next EMF source is left in the circuit, and the rest are treated in the same way as described in item 1.
4. The currents in the circuit created by the second EMF source are determined.
5. Do the same with the remaining sources.
6. True currents in the branches of a circuit are defined as the algebraic sum of currents in these branches, created by each of the sources.
Let's calculate the circuit shown in fig. 1.18, overlay method. We will assume that the internal resistances of the EMF sources are equal to zero.
At the beginning we leave the source E 1, and the source E 2 is removed and a jumper is placed in its place (Fig. 1.18, b). In the resulting circuit, we find currents by the equivalent conversion method:
Then we only leave the source E 2, and instead of E 1 a jumper is placed (Fig. 1.18, c). In the resulting circuit, we determine the currents in the branches also by the equivalent conversion method:
We find the real currents in the original circuit (Fig. 1.18, a) by algebraic summation of the found currents.
Current I 1 is equal to the current difference I 11 and current I 12:
I 1 = I 11 – I 12 .
The current I 2 is equal to the sum of the currents I 21 and I 22, because they coincide in direction:
I 2 = I 21 + I 22 .
Current I 3 is equal to the current difference I 32 and current I 31:
I 3 = I 32 – I 31 .
2.2. Parallel connection of elements
electrical circuits
In fig. 2.2 shows an electrical circuit with parallel connection of resistances.
Figure: 2.2
The currents in the parallel branches are determined by the formulas:
where 
- conductivity of the 1st, 2nd and nth branches.
In accordance with the first Kirchhoff's law, the current in the unbranched part of the circuit is equal to the sum of the currents in the parallel branches.
The equivalent conductivity of an electric circuit consisting of n parallel-connected elements is equal to the sum of the conductivity of parallel-connected elements.
The equivalent resistance of a circuit is the reciprocal of the equivalent conductance
Let the electrical circuit contains three resistors connected in parallel.
Equivalent conductivity
The equivalent resistance of a circuit consisting of n identical elements is n times less than the resistance R of one element
Let's take a circuit consisting of two parallel-connected resistors (Fig. 2.3). The values \u200b\u200bof the resistances and the current in the unbranched part of the circuit are known. It is necessary to determine the currents in the parallel branches.
Figure: 2.3 Equivalent circuit conductance
,
and the equivalent resistance
Input voltage
Currents in parallel branches
Similarly
The current in the parallel branch is equal to the current in the unbranched part of the circuit multiplied by the resistance of the opposite, foreign parallel branch and divided by the sum of the resistances of the foreign and its own parallel branches.
2.3 Converting the resistance triangle
to an equivalent star
There are circuits in which there are no resistances connected in series or in parallel, for example, the bridge circuit shown in Fig. 2.4. It is impossible to determine the equivalent resistance of this circuit relative to the branch with the EMF source using the methods described above. If the triangle of resistances R1-R2-R3, connected between nodes 1-2-3, is replaced by a three-beam star of resistances, the rays of which diverge from point 0 to the same nodes 1-2-3, the equivalent resistance of the resulting circuit is easily determined.
Figure: 2.4 The resistance of the beam of an equivalent resistance star is equal to the product of the resistances of the adjacent sides of the triangle divided by the sum of the resistances of all sides of the triangle.
In accordance with the specified rule, the resistances of the rays of the star are determined by the formulas:
The equivalent connection of the resulting circuit is determined by the formula
Resistors R0 and Rλ1 are connected in series, and branches with resistances Rλ1 + R4 and Rλ3 + R5 are connected in parallel.
2.4 Resistance Star Transformation
into an equivalent triangle
Sometimes it is useful to convert the resistance star to an equivalent delta to simplify the circuit.
Consider the circuit in Fig. 2.5. Replace the star of resistances R1-R2-R3 with an equivalent triangle of resistances RΔ1-RΔ2-RΔ3, connected between nodes 1-2-3.
2.5. Resistance star transformation
into an equivalent triangle
The resistance of the side of the equivalent triangle of resistances is equal to the sum of the resistances of two adjacent rays of the star plus the product of these same resistances, divided by the resistance of the remaining (opposing) ray. The resistances of the sides of a triangle are determined by the formulas:
The equivalent resistance of the converted circuit is
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Transcript
1 NI DOBROZHANOVA, VN TRUBNIKOVA Calculation of direct current electric circuits by the method of equivalent transformations PRACTICE ON THEORETICAL BASIS OF ELECTRICAL ENGINEERING Recommended for publication by the Editorial and Publishing Council of the State Educational Institution of Higher Professional Education "Orenburg State University" Orenburg 00
2 BBK.ya D UDC..0. (0.) Reviewer candidate of technical sciences, associate professor N.Yu. Ushakova D Dobrozhanova N.I., Trubnikova V.N. Calculation of DC electric circuits by the method of equivalent transformations: Workshop on the theoretical foundations of electrical engineering. Orenburg: GOU OSU, p. The workshop is intended for independent training of students in the section "DC circuits". Contains examples of calculating circuits by the method of equivalent transformations, as well as tasks for independent solution. BBK.ya Dobrozhanova N.I., Trubnikova V.N., 00 GOU OSU, 00
3 Introduction The main laws that determine the electrical state of any electrical circuit are Kirchhoff's laws. On the basis of these laws, a number of practical methods for calculating direct current circuits have been developed, which make it possible to reduce calculations when calculating complex circuits. To significantly simplify the calculations, and in some cases, to reduce the complexity of the calculation, perhaps using equivalent circuit transformations. Convert parallel and series connections of elements, star connection to an equivalent triangle and vice versa. The current source is replaced with an equivalent EMF source. By the method of equivalent transformations, it is theoretically possible to calculate any circuit, and at the same time use simple computational means. Or determine the current in any one branch, without calculating the currents of other sections of the circuit. In this workshop on the theoretical foundations of electrical engineering, examples of calculating linear DC electric circuits using equivalent transformations of typical circuits for connecting energy sources and consumers are considered, calculation formulas are given, as well as tasks for independent solution. The workshop is intended for deep self-study and self-control of mastering the TOE course.
4 Calculation of linear DC electric circuits by the method of equivalent transformations. Examples of solutions g Problem .. For a circuit (figure), determine the equivalent resistance relative to the input terminals g, if known: 0, Ohm, Ohm, 0 Ohm, Ohm, Ohm, Ohm, 0 Ohm, 0 0 Ohm. f d c Solution: We start transforming the circuit from the branch farthest from the source, i.e. terminals g Pattern: Ohm; 0 0 Ohm; 0 0 Ohm; Ohm; Ohm; Ohm; e 0.0, Ohm. Task .. For the circuit (Figure a), determine the input resistance if known: 0 Ohm) b) Figure
5 Solution: The original circuit can be drawn in relation to the input terminals (Figure b), from which it can be seen that all resistances are connected in parallel. Since the values \u200b\u200bof the resistances are equal, then to determine the value of the equivalent resistance, you can use the formula: e, n where the value of resistance, Ohm; n the number of parallel connected resistors. 0 e 0 Ohm. Task .. Find the equivalent resistance of the circuit (Figure a), which is formed by dividing a nichrome wire with a resistance of 0, Ohm into five equal parts and soldering copper jumpers at the obtained points -, -, -. Disregard the resistances of jumpers and transition contacts. a a a) b) Figure Solution: With a wire resistance of 0, Ohm and provided that all five parts are equal, the resistance of each separate section of the wire is: 0, 0, 0 Ohm. Let's designate each section of the wire and depict the original circuit with an equivalent equivalent circuit (Figure b). The figure shows that the circuit is a series connection of two parallel-connected resistance groups. Then the value of the equivalent resistance will be determined: 0, 0 0, 0 0, e 0, 0 Ohm. Task .. Determine the equivalent resistance relative to the terminals, if 0 Ohm (Figure a).
6 We transform the connection "triangle" f d c into an equivalent "star", determine the values \u200b\u200bof the converted resistances (Fig. B): f, Ohm By the condition of the problem, the values \u200b\u200bof all resistances are equal, which means :, Ohm. fdcffeecd) b) Figure On the transformed circuit, a parallel connection of branches between nodes e was obtained, then the equivalent resistance is: e (c) (d) () () cdc (, 0) (, 0) (, 0) (, 0) d , Ohm. And then the equivalent resistance of the original circuit is a series connection of resistances: 0.0 ohms. f e On the example of this circuit, consider the transformation "star" - "triangle". We transform the connection "star" with resistances into an equivalent "triangle" with resistances, and d (Figure a): f fd Ohm; Ohm; 0 f fd
7 0 0 d Ohm. 0 Then we transform the parallel connections of the branches with resistances fd and (Figure b): d fd fd "Ohm; 0 0 fd d 00 0 d" Ohm. d 0 0 fff fd dd) b) Figure Resistance value f "is determined by converting the parallel connection and (" "): f" fff fd (fd "d") ("") fd dd 0 0 f, fd, d (0 0 ) (0 0) 0 0 Ohm. Then the equivalent resistance is the sum of the resistances and ": f eq f" 000 Ohm. Task .. In a given circuit (Figure a), determine the input resistances of the branches, c d and f, if it is known that: Ohm, Ohm, Ohm, Ohm, Ohm, Ohm, Ohm, Ohm. Solution: To determine the input resistance of the branches, exclude all EMF sources from the circuit. In this case, points c and d, as well as f are short-circuited, i.e. internal resistances of voltage sources are equal to zero.
8 b) cd f a) a a f e e d c c d f Picture The branch is torn, and since resistance, then the input resistance of the branch is equal to the equivalent resistance of the circuit relative to the points and (Figure b): 0 "Ohm;" "Ohm;" "" "" "Ohm. The input resistances of the branches and are determined in a similar way. Moreover, when calculating the resistances, it was taken into account that the connection shortly points and excludes resistances from the circuit, in the first case, and, in the second case.cd f cd Ohm f Ohm. Problem .. Twelve wire segments of the same length, the resistance of each segment is equal to Ohm, soldered in such a way that they occupy the positions of the edges of the cube (Figure a). To two vertices lying
9 on one diagonal of the cube, two more of the same segments are soldered. Determine the equivalent resistance between the free ends of the last two segments. Solution: A star with rays -, -, - is transformed into an equivalent triangle, the resistance of the sides of which is determined (Figure b): Ohm; Ohm; Ohm. a a) a b) Figure Triangles -; -, - we transform into equivalent stars, the resistance of the rays of which will be the following (Figure a): - Ohm; Ohm; Ohm; - 0 Ohm;
10 0 Ohm; 0 ohm; - Ohm; Ohm; Ohm. In the diagram (Figure a), the sections connected in series - and -0; - and -; - and -0; - and - we replace with equivalent resistances, respectively (Figure b): 0 0 Ohm; Ohm; 0 0 Ohm; Ohm. Then, in the resulting circuit (Figure b), a star with rays -, -0, and - is transformed into an equivalent triangle with side resistances (Figure a): "Ohm; 0" Ohm; "Ohm. Next, the star with rays -, -0, - we transform into an equivalent triangle connection with the resistances of the sides (Figure b):" Ohm; 0 "Ohm;
11 0 "0 0 Ohm. 0 a a) -0-0 a b) Figure In the diagram (Figure b), parallel sections are replaced by equivalent ones (Figure 0a), the resistance of which is:" 0 "" 0 0 Ohm; "" "0 0" "" Ohm; "" "" 0 "" 0 0 Ohm. "" "a a) a b) Figure
12 In the circuit (Figure 0a), we transform the triangle -0- into an equivalent star with rays -, -0, - (Figure 0b): 0 Ohm; Ohm; Ohm a a a) b) Figure 0 Then, transforming the parallel connection of the sections between the nodes and, the diagram of Figure 0b will take the form of a series connection of the sections -, -, - and -: (0 0) () () () Ohm. () () () () 0 0 in Ohm. 0 Task .. Using the transformation method, determine the parameters of the equivalent circuit (Figure a), if 0 V, 0 V, J A, 0 Ohm. Solution: Replace the parallel-connected branches with a current source J and resistance with an equivalent branch with an EMF source (Figure b): J 00 V. Then we transform two parallel active branches (Figure c): 0 0 Ohm; "0 V; 0 0
13 eq Ohm; "000 V. eq, J a) b) c) Figure Let's solve the problem differently. Let's use the formula for the transformation of parallel branches: J V; Ohm; 0 0 eq 000 V. Problem .. In the circuit (figure) determine the currents by the method of equivalent transformations and draw up a power balance, if known: Ohm, 0 Ohm, 0 Ohm, 0 V. Solution: Equivalent resistance for parallel connected resistances: Figure 0 0 Ohm. 0 0 Equivalent resistance of the entire circuit: e Ohm. Current in the unbranched part of the circuit: e 0 A. Voltage across parallel resistors: 0 V. Currents in parallel branches:
14 0 0 A; 0 0 A. Power balance: 000 W; P source P consumption W. Task .. In the circuit (Figure a), determine the ammeter readings, if known: Ohm, 0 Ohm, 0 Ohm, 0 Ohm; 0 Ohm, 0 Ohm, V. The resistance of the ammeter can be considered equal to zero. A E e A E A) b) Figure Solution: If the resistance is replaced by one equivalent e, then the original circuit can be represented in a simplified form (Figure b). The value of the equivalent resistance: e Ohm By converting the parallel connection of the resistances e and the circuit (Figure b), we get a closed loop, for which, according to the second Kirchhoff's law, we can write the equation: e, e from where the current: A. e e
15 The voltage at the terminals of the parallel branches is expressed from the equation according to Ohm's law for the passive branch, obtained by transforming e and: e. e Then the ammeter will show the current: 0 A A. 0 0 e Task..0 Using the method of equivalent conversions, determine all currents in the circuit (Figure a), if 0 V, 0 V, 0 V, 0 Ohm. Solution: First, we transform the original circuit to one circuit, and determine the current in the unbranched part. To do this, we determine the values \u200b\u200bof the equivalent resistances and equivalent EMF (Figure b): 0 0 Ohm; Ohm; IN; C. 0 0 Let's compose the equations according to the second Kirchhoff's law for a given contour: - - a) b) Figure (),
16 then A. Determine the voltages at the terminals of the parallel branches - and - according to Ohm's law: 0 0 V 0 0 V Determine the currents of the branches: A; AND; 0 0 A; 0 0 A. Tasks .. Determine the currents of the branches of the circuit (Figure a), if Ohm, J A, Ohm. Solution: We transform the "triangle" of resistances into an equivalent "star" (Figure b) and determine the values \u200b\u200bof the obtained resistances: Ohm; Ohm; Ohm. Let's transform the parallel connection of branches between nodes and. () () () () () (), Ohm.
17 The current in the circuit obtained as a result of the transformations is considered equal to the current of the current source J, and then the voltage is: J V., J J a) b) Figure And now you can determine the currents and: A; AND; Returning to the original scheme, we determine the voltage from the equation according to the second Kirchhoff's law: V. 0 Then the current in the branch with resistance will be determined: 0, A. The values \u200b\u200bof the remaining unknown currents can be determined from the equations according to the first Kirchhoff's law for nodes and: 0 J 0, A ; -, A. 0 J Problem .. Using the method of equivalent transformations, find the current 0 (Figure a), if 0 0 V, 0 V, Ohm, 0 Ohm. Solution: To transform the active "star", introduce additional nodes, and. The resulting passive "star" is transformed into a passive "triangle" (Figure b), the resistances of which are equal: Ohm;
18 Ohm; Ohm) b) Figure We will transfer the EMF sources through additional nodes (Figure a) and determine the parameters of the equivalent EMF sources a) b) Figure Obviously, with the same EMF values \u200b\u200band their multidirectionality, the values \u200b\u200bof the equivalent EMF sources are equal to zero. The resulting passive "triangle" is transformed with a "triangle" (Figure b): Ohm; Ohm;
19 ohm. We replace the connection of the obtained resistances with one equivalent: () () ec Ohm. () For the resulting circuit, we write the equation according to the second Kirchhoff's law, from which we express the current 0: 0 0 A. 0 eq eq. Problem .. Using the method of equivalent circuit transformations (Figure a), determine the current 0 if 0 0 V, 0 V, 0 V, Ohm, Ohm J) b) Figure Solution: In the active branch of the "triangle" of resistances - - transform the EMF source into an equivalent current source (Figure b): 0 J A. The resulting passive "triangle" of resistances is converted into a "star". The values \u200b\u200bof the resistances obtained, due to the equality of the values \u200b\u200bof the initial resistances, will be equal to: Ohm.
20 Then we replace the branch with the current source between the nodes with two connected in parallel with the resistances and, and convert it into EMF sources (Figure a): J 0 V; J 0 V. We transform the parallel branches between the nodes and (Figure b): eq () () () () Ohm; () () () () 0 () 0 () eq 0 V eq eq) b) Figure For the resulting circuit, we write the equation according to the second Kirchhoff's law: whence we express the current 0: 0 (eq) 0 eq 0 0 () 0 eq ek A.
21. Tasks for an independent solution Task .. For a circuit (Figure 0), determine the input resistance (equivalent) relative to the input terminals, if known: 0 Ohm, 0 Ohm. Task .. For a circuit (Figure), find the input resistance, if known: Ohm, 0 ohm, ohm, ohm, ohm, ohm, ohm. c c d Figure 0 Figure Task .. Determine the equivalent circuit resistance (figure) between terminals B and D, if Ohm, Ohm. Task .. Determine the currents and voltages in individual sections of the circuit (figure), if the voltage at the input is 0 V, and the resistances of the circuit sections: 0, Ohm, 0 Ohm, Ohm DACB Figure Figure Task .. Find the current in the resistance (figure), if : 00 V, Ohm, 0 Ohm, 0 Ohm, Ohm. Task .. Determine the value of resistance (figure), if Ohm, readings of ammeters A A, A A.
22 А А Figure Figure Task .. Using the conversion method, determine the parameters of the equivalent circuit eq, eq, if 0 V, 0 V, 0 V, 0 Ohm, 0 Ohm (picture). Task .. Find the voltage at the terminals of the current source J 0 A (figure), if: Ohm, Ohm. eq eq J Figure Figure Task .. Using the transformation of the circuit, find the current and voltage, if: 0 V, 0 V, 0 V, 0 Ohm (figure). Task..0 Using the method of equivalent transformations, determine the current (figure) if: 0 V, 0 V, 0 V, J A, 0 Ohm, Ohm, Ohm. J Figure Figure
23 Problem .. In the circuit (Figure 0) EMF of the power source V, the resistances of the branches are :, Ohm ;, Ohm ;, Ohm; Ohm; Ohm. Determine the currents in all branches of the circuit in two ways: a) transformation of the resistance star - - into an equivalent triangle; b) transformation of one of the resistance triangles into an equivalent star. Task .. The circuit (figure) is connected to a network with a constant voltage of 0 V. EMF and internal resistances of the sources are as follows: 00 V, 0 V, 0 0, Ohm, 0 0, Ohm. Resistance values \u200b\u200bin branches :, Ohm, Ohm, 0, Ohm. Determine the voltmeter reading, currents in all branches and draw up a power balance. _ V Figure 0 Figure Problem .. In the circuit (figure) the EMF of the power supplies are 0 V, 0 V, and the resistance of the branches is Ohm; Ohm ;, Ohm, Ohm. Determine the current in the branch with resistance by the method of equivalent transformations. Problem .. In the circuit (figure) the values \u200b\u200bof 00 V and resistances of the Ohms branches are known. Determine the readings of the wattmeter W for four cases: a) the keys K, K, K are open; b) the key K is closed, K and K are open; c) the keys K, K are closed, K is open; d) the keys K, K, K are closed. K W K Picture Picture K
24 Problem .. In the circuit (figure) the current values \u200b\u200bof the current source J ma with internal conductivity g0 0 S and the conductivity of two parallel connected consumers g 0 S and g 0 are known. See Determine currents 0, parameters of the equivalent voltage source. Task .. Determine the voltages ed, ec, cd and currents in the branches of the circuit (figure), if 0 A, Ohm, Ohm, Ohm, Ohm. J 0 e c g g g 0 c d Picture Picture f
25 List of used sources Bessonov L.A. Theoretical foundations of electrical engineering. Electrical circuits: Textbook. for universities / L.A. Bessonov. 0th ed. M .: Gardariki, 000.s .: ill. Goldin O.E. and others. Programmed study of the theoretical foundations of electrical engineering: textbook. /O.E.Goldin, A.E. Kaplyansky, L.S.Polotovksky. M: High school ,. from: ill. Collection of tasks and exercises on the theoretical foundations of electrical engineering: a textbook for universities. / Ed. P.A. Ionkin. M .: Energoizdat ,. from: ill. Collection of problems on the theoretical foundations of electrical engineering: a textbook for universities. / Ed. L.A. Bessonova. ed., revised. and add. M .: Vysshaya shkola, 0. p .: silt Collection of problems on the theoretical foundations of electrical engineering: Textbook. manual for universities / Ed. L.A. Bessonova. ed., revised. and add. M .: Higher school ,. from: ill. Repyev Yu.G., Semenko L.P., Poddubny G.V. Theoretical foundations of electrical engineering. Circuit theory. Krasnodar: Krasnodar Polytechnic Institute, 0. p. Ogorelkov, B.I. Methodical instructions to the RGZ on TOZ. Analysis of steady-state processes in DC electric circuits / A. N. Ushakov, N. Yu. Ushakova, B. I. Ogorelkov. Orenburg: ORPTI ,. from. Methods for calculating electric circuits of direct current: Methodical instructions / B. I. Ogorelkov, A. N. Ushakov, N. Yu. Ushakova. Orenburg: ORPTI, 0.- p.
MINISTRY OF EDUCATION OF THE RUSSIAN FEDERATION State educational institution of higher professional education "Orenburg State University" Department of Theoretical and General Electrical Engineering
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The analysis of any electrical circuit begins with the construction of its model, which is described by an equivalent circuit.
In electrical circuits, the following simplest connections of passive elements are distinguished: serial, parallel, connection in the form of a triangle and in the form of a three-pointed star. Before starting the circuit analysis, it is advisable to carry out preliminary equivalent circuit transformations.The essence of such transformations is to replace some part of the circuit with another, electrically equivalent to it, but with a more convenient structure for calculation. More often than others, two types of such transformations are used: replacement of series and parallel connected elements with one equivalent; transformation of a three-pointed star into a triangle and vice versa.
The equivalent resistance of series-connected elements is equal to the arithmetic sum of their resistances:
. (1.26)
The equivalent conductance of parallel connected resistive elements is equal to the arithmetic sum of their conductances:
. (1.27)
When transforming a triangle (Figure 1.14) into a star (Figure 1.15) at the given resistances of the sides of the triangle RAB, RBV, RBA, the equivalent resistances of the rays of the star RA, RB, RB are determined.
Figure: 1.14. Circuit diagram - triangle
Figure: 1.15. Circuit diagram - star
The equivalent resistances of the rays of the star are:
When converting a star into an equivalent triangle for given RA, RB, RB, the equivalent resistances are determined as follows.
Electrical circuits count simple, if they contain only serial or only parallel connection of elements.
The section of the circuit containing both parallel and series connection of elements is called complicated or mixed connection area elements.
Transformations of electrical circuits are considered equivalent if during their implementation the voltages and currents in the sections of interest to us do not change.
When converting complex electrical circuits, a sequential method is used, that is, sections of the circuit are sequentially transformed that have a simple connection of elements.
4.3.1. Equivalent transformation of the circuit when connecting elements in series
Consider a complex equivalent circuit of an electric circuit consisting of a series connection of individual elements (Fig. 4.6). This circuit is a circuit in which a current common to all elements flows through all the elements. Equivalently, we transform the circuit to one element, but so that the voltage and current at the circuit terminals retain their values. This is possible when the resistance of the original circuit and the equivalent circuit are the same. Based on Ohm's law and Kirchhoff's second law in complex form, the equation of electrical equilibrium can be written
The voltage and current for both circuits are the same when
Conclusion. With an equivalent transformation, with a series connection of elements, their complex resistances add up.
1) Equivalent resistance conversion
Consider the electrical circuit diagram shown in Figure 4.7. Equivalently we transform the resistances R 1 and R 2 to one resistance R eq.
Considering that Z R \u003d R, and the ratio obtained above, we get R eq \u003d R 1 + R 2.
2) Equivalent conversion of capacities.
R 
consider the electrical circuit diagram shown in Figure 4.8. Equivalently we transform capacities С 1 and С 2 to one equivalent capacitance С eq.
Considering that Z С \u003d 1 / (jωC), and the ratio obtained above, we obtain
.
3) Equivalent inductance conversion
R 
consider the electrical circuit diagram shown in Figure 4.9. Equivalently transform the inductance L 1 and L 2 to one equivalent inductance L eq.
Considering that Z L \u003d jωL, and the ratio obtained above, we get L equiv \u003d L 1 + L 2.
4.3.2. Equivalent transformation of the circuit with parallel connection of elements
Consider a complex equivalent circuit of an electrical circuit consisting of a parallel connection of individual elements (Fig. 4.10). This chain contains two nodes, between which all elements are connected. Common to all elements is the stress on them. Equivalently, we transform the circuit to one element, but so that the voltage and current at the circuit terminals retain their values. This is possible when the resistance of the original circuit and the equivalent circuit are the same. On the basis of Ohm's law and the first Kirchhoff's law in complex form, the equation of electrical equilibrium can be written
I \u003d I 1 + I 2 + ... + I n, or (U / Z eq) \u003d (U / Z 1) + (U / Z 2) +… (U / Z n).
Hence it follows that
(1/Z eq) \u003d (1 / Z 1) + (1/Z 2) + … +(1/Z n), or Z eq \u003d 1 / [(1 / Z 1) + (1/Z 2) + … +(1/Z n)].
Taking into account, (1 / Z) = Y Is the complex conductivity of an element, it can be written that
Y eq \u003d Y 1 + Y 2 + … + Y n.
Conclusion. With an equivalent transformation, with parallel connection of elements, their complex conductivities are added.
