Analysis and calculation of electrical circuits. P1. Analysis and calculation of a direct current electrical circuit. Basic laws of DC circuits
Below write down the full group number (for example, 3ASU-2DB-202), last name and I. O. of the student, full code design option, for example, KR6-13 – code of the 13th version of tasks course work KR6.
At the bottom of the sheet (center) write down the name of the city and the current year.
2. On the next page, an “Abstract” of the completed work is presented (no more than 2/3 of the page) with brief description design diagrams of circuits, methods used (laws, rules, etc.) for analyzing circuit diagrams and the results obtained from completing tasks.
For example, an annotation for the completed first task.
"In task 1, a complex electrical circuit was calculated direct current with two voltage sources and six branches. When analyzing the circuit and calculating it, the following methods were used: the method of Kirchhoff’s laws, the method of nodal voltages (two nodes), the generalized Ohm’s law and the equivalent generator method. The correctness of the calculation results was confirmed by constructing a potential diagram of the second circuit and fulfilling the power balance condition."
Similarly, annotation of completed 2nd and 3rd tasks of the work is given.
3. On the third page, write down the topic of assignment 1 of the course work and below it (in brackets) the code for the calculated version of the assignment, for example, KR6.1-13. The electrical diagram of the circuit is drawn below (in compliance with GOST 2.721-74) and under it the initial data for calculating a given option is written out from Table 6.1, for example: E 1 = 10 V, E 2 = 35 V, R 1 = 15 Ohm, R 2 = ... etc.
4. Next, a step-by-step calculation of the circuit diagram is performed with the corresponding headings of each stage (step), with the drawing of the necessary design diagrams with conditionally positive directions of currents and voltages of the branches, with the recording of equations and formulas in general form, followed by the substitution of the numerical values of the physical quantities included in the formulas and recording intermediate calculation results (for searching possible errors calculated by the teacher). Calculation results should be rounded to no more than four to five significant figures, expressing floating point numbers if large or small.
Attention! When calculating values original data for calculating circuit diagrams (effective EMF values E, impedance values Z branches) it is recommended to round their values to whole numbers, for example Z= 13/3 » 4 ohms.
5. Diagrams and graphs are drawn on graph paper (or on sheets with a fine grid when working on a PC) according to GOST using uniform scales along the axes and indicating dimensions. Figures and diagrams must be numbered and labeled, for example, Fig. 2.5. Vector diagram of voltages and currents of an electrical circuit. The numbering of both figures and formulas is consistent across all three tasks!
7. It is recommended to submit reports for each assignment to the teacher for review on stapled sheets in A4 format, followed by binding them before defending the work.
8. Based on the results of calculations and graphic constructions conclusions are formulated for each task or at the end of the report - for the entire work. On last page The student puts his signature on the report and the date of completion of the work.
Attention!
1. Carelessly completed work will be returned to students for re-registration. The teacher also returns reports to individual students for revision with errors marked on the sheets or with a list of comments and recommendations for correcting errors on the title page.
2. After defending coursework, explanatory notes from students of groups with a mark and signature of the teacher (two teachers) on the title pages, also included in the appropriate statement and in the student record books, are handed over to the department for storage for two years.
Note: When compiling table 6.1. Variants of task 1, the Variant 2 program was used, developed by Associate Professor, Ph.D. Rumyantseva R.A. (RGGU, Moscow), and versions of task 6.2 and task 6.3. taken (with the consent of the authors) from the work of: Antonova O.A., Karelina N.N., Rumyantseva M.N. Calculation of electrical circuits (methodological instructions for course work on the course "Electrical engineering and electronics". - M.: MATI, 1997.
Exercise 1
ANALYSIS AND CALCULATION OF ELECTRICAL CIRCUIT
DIRECT CURRENT
For the option specified in table 6.1:
6.1.1. Write down the values of the parameters of the circuit elements and draw a design diagram of the circuit in accordance with GOST, indicating the conditionally positive directions of currents and voltages of the branches. Selecting a generalized circuit diagram (Fig. 1: A, b, V or G) is carried out as follows. If the option number assigned by the teacher to perform KR6 for the student N is divided by 4 without a remainder (and in option No. 1), then the diagram in Fig. 1 A; with a remainder of 1 (and in option No. 2), the scheme in Fig. 1 b; with a remainder of 2 (and in option No. 3) - diagram in Fig. 1 V; and finally, with a remainder of 3, the circuit in Fig. 1 G.
6.1.2. Conduct a topological analysis of the circuit diagram (determine the number of branches, nodes and independent circuits).
6.1.3. Compose the number of equations required to calculate the chain using Kirchhoff’s first and second laws.
6.1.4. Simplify the circuit diagram by replacing the passive triangle of the circuit with an equivalent star, calculating the resistance of its rays (branches).
6.1.7. Check the calculation of currents and voltages of all six branches of the original circuit by constructing a potential diagram on the scale of one of the circuits, in the branches of which at least one voltage source is included, and confirming that the power balance condition is met.
6.1.8. Check the correctness of the calculation of task 1 (together with the teacher) by comparing the data obtained with the data calculated using the Variant program installed on a computer in a specialized laboratory (class) of the department. Brief instructions for working with the program is displayed on the working field of the display along with the program interface.
6.1.9. Formulate conclusions based on the results of completed task 1.
Table 6.1
Options for assignment 1 course work KR6
| No. var | E 1, B | E 2,B | E 3, B | E 4 ,B | E 5 ,B | E 6,B | R 1 , Ohm | R 2 , Ohm | R 3 , Ohm | R 4 , Ohm | R 5 , Ohm | R 6 ohm | Branch for MEG | ||
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| No. var | E 1, B | E 2,B | E 3, B | E 4 ,B | E 5 ,B | E 6,B | R 1 , Ohm | R 2 , Ohm | R 3 , Ohm | R 4 , Ohm | R 5 , Ohm | R 6 ohm | Branch for MEG | ||
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Table 6.1(continuation)
| Var. No. | E 1, B | E 2,B | E 3, B | E 4 ,B | E 5 ,B | E 6,B | R 1 , Ohm | R 2 , Ohm | R 3 , Ohm | R 4 , Ohm | R 5 , Ohm | R 6 ohm | Branch for MEG |
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| A dash (--) in the table fields means the absence of this voltage source Ek in the circuit diagram |
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Department of Automation and Electrical Engineering
B3.B.11 Electrical engineering and electronics
Guidelines for practical exercises
by discipline Direction of training
260800 Product technology and catering organization
Training profile
Restaurant business organization technology
Graduate qualification (degree) bachelor
Ufa 2012UDK 378.147:621.3
Compiled by: senior teacher Gallyamova L.R.
senior teacher Filippova O.G.
Reviewer: Head of the Department of Electrical Machines and Electrical Equipment
Doctor of Technical Sciences, Professor Aipov R.S.
Responsible for the issue: Head of the Department of Automation and Electrical Engineering, Candidate of Technical Sciences, Associate Professor Galimardanov I.I.
2. Analysis of unbranched sinusoidal current circuits
and determination of parameters of equivalent circuits. Vector diagrams, triangles of voltages, resistances and powers
Bibliography
chain asynchronous motor three-phase
1. Analysis and calculation of linear DC electrical circuits
1.1 Theoretical information
An electrical circuit is a set of electrical devices that create a path for electric current, the electromagnetic processes in which are described by equations taking into account the concepts of electromotive force, electric current and electrical voltage.
The main elements of the electrical circuit (Figure 1.1) are sources and consumers of electrical energy.
Figure 1.1 Basic elements of an electrical circuit
DC generators and galvanic cells are widely used as sources of direct current electrical energy.
Sources of electrical energy are characterized by the emf E they develop and the internal resistance R0.
Consumers of electrical energy are resistors, electric motors, electrolysis baths, electric lamps, etc. In them, electrical energy is converted into mechanical, thermal, light, etc. In an electrical circuit, the positive direction of the emf E is taken to be the direction coinciding with force acting on a positive charge, i.e. from “-” source to “+” power source.
When calculating electrical circuits, real sources of electrical energy are replaced by equivalent circuits.
The equivalent circuit of the EMF source contains the EMF E and the internal resistance R0 of the source, which is much less than the resistance Rн of the electricity consumer (Rн >> R0). Often in calculations the internal resistance of the EMF source is equated to zero.
For a circuit section that does not contain an energy source (for example, for the circuit Figure 1.2, a), the relationship between current I and voltage U12 is determined by Ohm’s law for the circuit section:
where c1 and c2 are the potentials of points 1 and 2 of the circuit;
Y R is the sum of resistances in a section of the circuit;
R1 and R2 are the resistances of the circuit sections.
Figure 1.2 Electrical diagram section of the circuit: a - not containing an energy source; b - containing an energy source
For a section of a circuit containing an energy source (Figure 1.2, b), Ohm’s law is written as the expression
where E is the EMF of the energy source;
R = R1 + R2 is the arithmetic sum of the resistances of the circuit sections;
R0 is the internal resistance of the energy source.
The relationship between all types of power in an electrical circuit (power balance) is determined from the equation:
UR1 = UR2 + URp, (1.3)
where UR1 = UEI is the algebraic sum of the powers of energy sources;
UR2 - algebraic sum of consumer power (net power) (P2 = UI);
URp = УI2R0 - total power due to losses in source resistance.
Resistors, as well as the resistance of other electrical devices, are consumers of electrical energy. The power balance is determined by the law of conservation of energy, while in any closed electrical circuit the algebraic sum of the powers of energy sources is equal to the algebraic sum of powers consumed by consumers of electrical energy.
Coefficient useful action attitudes are determined by the attitude
When calculating unbranched and branched linear DC electrical circuits, various methods can be used, the choice of which depends on the type of electrical circuit.
When calculating complex electrical circuits, in many cases it is advisable to simplify them by folding, replacing individual sections of the circuit with series, parallel and mixed resistance connections with one equivalent resistance using the method of equivalent transformations (transfiguration method) of electrical circuits.
1.1.1 Method of equivalent transformations
Electric circuit with serial connection resistance (Figure 1.3, a) is replaced by a circuit with one equivalent resistance Rek (Figure 1.3, b), equal to the sum of all resistances of the circuit:
Rek = R1 + R2 +…+ Rn = , (1.5)
where R1, R2…Rn are the resistances of individual sections of the circuit.
Figure 1.3 Electrical circuit with series connection of resistances
In this case, the current I in the electrical circuit remains unchanged, all resistances are flown by the same current. The voltages (voltage drops) across the resistances when connected in series are distributed proportionally to the resistances of the individual sections:
U1/R1 = U2/R2 = … = Un/Rn.
When resistances are connected in parallel, all resistances are under the same voltage U (Figure 1.4). It is advisable to replace an electrical circuit consisting of parallel-connected resistances with a circuit with an equivalent resistance Rek, which is determined from the expression
where is the sum of the reciprocal values of the resistances of sections of parallel branches of the electrical circuit;
Rj is the resistance of the parallel section of the circuit;
n is the number of parallel branches of the chain.
Figure 1.4 Electrical circuit with parallel connection of resistances
The equivalent resistance of a section of a circuit consisting of identical resistances connected in parallel is equal to Rek = Rj/n. When two resistances R1 and R2 are connected in parallel, the equivalent resistance is determined as
and the currents are distributed inversely proportional to these resistances, while
U = R1I1 = R2I2 = … = RnIn.
With a mixed connection of resistances, i.e. in the presence of sections of an electrical circuit with series and parallel connection of resistances, the equivalent resistance of the circuit is determined in accordance with the expression
In many cases, it also turns out to be advisable to convert resistances connected by a triangle (Figure 1.5) to an equivalent star (Figure 1.5).
Figure 1.5 Electrical circuit with delta and star connection of resistances
In this case, the resistance of the rays of the equivalent star is determined by the formulas:
R1 = ; R2 = ; R3 = ,
where R1, R2, R3 are the resistances of the rays of the equivalent resistance star;
R12, R23, R31 - resistances of the sides of the equivalent resistance triangle. When replacing the resistance star with an equivalent resistance triangle, its resistance is calculated using the formulas:
R31 = R3 + R1 + R3R1/R2; R12 = R1 + R2 + R1R2/R3; R23 = R2 + R3 + R2R3/R1.
1.1.2 Method of applying Kirchhoff's laws
In any electrical circuit, in accordance with Kirchhoff’s first law, the algebraic sum of currents directed to a node is equal to zero:
where Ik is the current in the kth branch.
In accordance with Kirchhoff's second law, the algebraic sum of the EMF of power sources in any closed circuit of an electrical circuit is equal to the algebraic sum of the voltage drops on the elements of this circuit:
When calculating electrical circuits using the method of applying Kirchhoff's laws, conditional positive directions of currents in the branches are selected, then closed circuits are selected and the positive direction of bypassing the circuits is specified. In this case, for the convenience of calculations, it is recommended to choose the same bypass direction for all circuits (for example, clockwise).
To obtain independent equations, it is necessary that each new circuit includes at least one new branch (B) that was not included in the previous circuits.
The number of equations compiled according to Kirchhoff's first law is taken to be one less than the number of nodes Ny in the circuit: NI = Ny - 1. In this case, currents directed to the node are conventionally taken as positive, and those directed from the node as negative.
The remaining number of equations NII = NВ - Nу + 1 is compiled according to Kirchhoff's second law, where NВ is the number of branches.
When composing equations according to Kirchhoff’s second law, the emf of sources is assumed to be positive if their directions coincide with the selected direction of bypassing the circuit, regardless of the direction of the current in them. If there is a discrepancy, they are written down with a “-” sign. Voltage drops in branches in which the positive direction of the current coincides with the direction of bypass, regardless of the direction of the EMF in these branches - with a “+” sign. If they do not coincide with the direction of the bypass, the voltage drops are recorded with a “-” sign.
As a result of solving the resulting system of N equations, the actual values of the quantities being determined are found, taking into account their sign. In this case, quantities that have a negative sign actually have a direction opposite to the conventionally accepted one. The directions of quantities that have a positive sign coincide with the conventionally accepted direction.
1.2 Problems to be solved during the practical lesson
Determine the current in a direct current electrical circuit (Figure 1.5, a). EMF of the power source: E1 = 40 V, E2 = 20 V, internal resistances: R01 = 3 Ohm, R02 = 2 Ohm, potentials of points 1 and 2 circuits: c1 = 80 V, c2 = 60 V, resistor resistance R1 = 10 Ohm , R2 = 10 Ohm.
Answer: I = 1.6 A.
Figure 1.5 DC Electric Circuit
Determine the supply voltage U of the DC electrical circuit (Figure 1.5, b), as well as the load resistance Rн, if the voltage at the load terminals Un = 100 V, the current in the circuit I = 10 A, the resistance of each of the circuit wires Rп = 0.6 Ohm .
Answer: U = 112 V; Rн = 10 Ohm.
For the electrical circuit (Figure 1.1), determine the current I, the voltage at the consumer terminals U, the power of the power source P1, the power P2 of the external circuit, the efficiency of the installation, if the emf of the power source E = 10 V, its internal resistance R0 = 1 Ohm, load resistance Rн = 4 Ohm. Neglect the resistance of the supply wires.
Answer: I = 2 A; U = 8 V; P1 = 20 W; P2 = 16 W; z = 80%.
Determine the total resistance R0 and the distribution of currents in the DC electrical circuit (Figure 1.6). Resistor resistances: R1 = R2 = 1 Ohm, R3 = 6 Ohm, R4 = R5 = 1 Ohm, R6 = R7 = 6 Ohm, R8 = 10 Ohm, R9 = 5 Ohm, R10 = 10 Ohm. Power supply voltage U = 120 V.
Figure 1.6 Electrical circuit diagram for problem 1.2.4
For a direct current electrical circuit (Figure 1.7), determine the equivalent resistance Rek and the total current I in the circuit, as well as the voltage drop DU across resistors R1, R2, R8. Resistor resistances: R1 = 5 Ohm, R2 = 4 Ohm, R3 = 20 Ohm, R4 = 30 Ohm, R5 = 50 Ohm, R6 = 10 Ohm, R7 = 5 Ohm, R8 = 1.8 Ohm. EMF of the power source E = 50 V, neglect the internal resistance of the source.
Figure 1.7 Electrical circuit diagram for problem 1.2.5
For the conditions of problem 1.2.5, transform the star connection R3, R5, R6 into an equivalent triangle and calculate the resistance of its sides.
Figure 1.8 shows a bridge circuit for connecting resistors in a DC circuit with a power source voltage U = 120 V. Determine the magnitude and direction of current I5 in the diagonal of the bridge if the resistor resistances are: R1 = 25 Ohm, R2 = 5 Ohm, R3 = 20 Ohm, R4 = 10 Ohm, R5 = 5 Ohm.
Figure 1.8 Bridge circuit for connecting resistors
For a direct current electrical circuit (Figure 1.9), determine the currents I1 - I3 in the branches using Kirchhoff's laws. EMF E1 = 1.8 V, E2 = 1.2 V; resistor resistances: R1 = 0.2 Ohm, R2 = 0.3 Ohm, R3 = 0.8 Ohm, R01 = 0.6 Ohm, R02 = 0.4 Ohm.
Figure 1.9 Electrical circuit diagram for problem 1.2.8
Using Kirchhoff's laws, determine the currents I1 - I3 in the branches of the electrical circuit shown in Figure 1.10, a. EMF of power supplies: E1 = 100 V, E2 = 110 V; resistor resistances: R1 = 35 Ohm, R2 = 10 Ohm, R3 = 16 Ohm.
In a direct current electrical circuit (Figure 1.10, b) ammeter reading PA1: I5 = 5 A. Determine the currents in all branches of the circuit I1 I4 using Kirchhoff’s laws. Resistor resistances: R1 = 1 Ohm, R2 = 10 Ohm, R3 = 10 Ohm, R4 = 4 Ohm, R5 = 3 Ohm, R6 = 1 Ohm, R7 = 1 Ohm, R8 = 6 Ohm, R9 = 7 Ohm; EMF E1 = 162 V, E2 = 50 V, E3 = 30 V.
Figure 1.10 DC electrical circuits: a - to problem 1.2.9; b - to problem 1.2.10
In the direct current electrical circuit shown in Figure 1.11 a, determine the currents I1 I5 in the branches using the loop current method; voltage U12 and U34 between points 1-2 and 3-4 of the circuit. Create a power balance equation. EMF of the power source E = 30 V, current of the current source J = 20 mA, resistor resistances R1 = 1 kOhm, R2 = R3 = R4 = 2 kOhm, R5 = 3 kOhm.
In the DC electrical circuit shown in Figure 1.11 b, determine the currents in the branches using the loop current method. EMF of power supplies E 1 = 130 V, E2 = 40 V, E3 = 100 V; resistance R1 = 1 Ohm, R2 = 4.5 Ohm, R3 = 2 Ohm, R4 = 4 Ohm, R5 = 10 Ohm, R6 = 5 Ohm, R02 = 0.5 Ohm, R01 = R03 = 0 Ohm.
Figure 1.11 DC electrical circuits: a - to problem 1.2.11; b - to problem 1.2.12
2. Analysis of unbranched sinusoidal current circuits and determination of parameters of equivalent circuits. Vector diagrams, triangles of voltages, resistances and powers
2.1 Theoretical information
In an electrical circuit of a sinusoidal current with an active resistance R (Table 2.1), under the action of a sinusoidal voltage u = Umsinт, a sinusoidal current i = Imsinт arises, coinciding in phase with the voltage, since the initial phases of voltage U and current I are equal to zero (wu = 0, wi = 0). In this case, the phase shift angle between voltage and current μ = ūu - ūi = 0, which indicates that for this circuit the dependences of changes in voltage and current coincide with each other on a linear diagram over time.
The total resistance of the circuit is calculated using Ohm's law:
In an electrical circuit of a sinusoidal current containing a coil with inductance L (Table 2.1), under the influence of a voltage u = Um sin(мт + /2) varying according to a sinusoidal law, a sinusoidal current i = Imsincht appears, lagging in phase with the voltage by an angle /2.
In this case, the initial phase of the voltage wu = /2, and the initial phase of the current wi = 0. The phase shift angle between the voltage and current c = (wu - wi) = /2.
In an electrical circuit of a sinusoidal current with a capacitor with a capacitance C (Table 2.1), under the influence of voltage u = Umsin(ут - /2), a sinusoidal current i = Imsinт arises, advancing the voltage on the capacitor by an angle /2.
The initial phase angle of the current wi = 0, and the voltage wu = - /2. The phase shift angle between voltage U and current I c = (wu - wi) = - /2.
In an electrical circuit with a series connection of active resistance R and inductor L, the current lags behind the voltage by an angle μ › 0. In this case, the total resistance of the circuit is:
Circuit conductivity
where G = R/Z2 - active conductivity of the circuit;
BL = XL/Z2 - reactive inductive conductance chains.
Phase angle between voltage and current:
c = arctg XL/R = arctg BL/G. (2.4)
Similarly, you can obtain the corresponding calculation formulas for electrical circuits of sinusoidal current with various combinations of elements R, L and C, which are given in Table 2.1.
Circuit power with active, inductive and capacitive reactances (R, L and C):
where P = I2R - active power,
QL = I2XL - inductive component of reactive power,
QC = I2XC - capacitive component of reactive power.
In an unbranched electrical circuit of a sinusoidal current with inductance L, capacitance C and active resistance, under certain conditions, voltage resonance can occur (a special state of an electrical circuit in which its inductive reactance XL is equal to the capacitive reactance XC of the circuit). Thus, voltage resonance occurs when the circuit reactances are equal, i.e. at XL = XС.
Circuit resistance at resonance Z = R, i.e. The total resistance of the circuit at voltage resonance has a minimum value equal to the active resistance of the circuit.
Phase angle between voltage and current at voltage resonance
ц = су - сi = arctg = 0,
in this case, the current and voltage are in phase. The power factor of the circuit has a maximum value: cos c = R/Z = 1 and the current in the circuit also reaches a maximum value I = U/Z = U/R.
Reactive power of the circuit at voltage resonance:
Q = QL - QC = I2XL - I2XС = 0.
The active power of the circuit at resonance acquires the greatest value, equal to the total power: P = UI cos c = S.
When constructing a vector diagram for an electrical circuit with resistances connected in series, the initial value is the current, since in this case the current value in all sections of the circuit is the same.
The current is plotted on the appropriate scale (mi = n A/cm), then relative to the current on the accepted scale (mu = n V/cm) the voltage drops DU at the corresponding resistances are plotted in the sequence of their location in the circuit and voltage (Figure 2.1).
Figure 2.1 Construction of a vector diagram
2.2 Example of solving a typical problem
Determine the readings of instruments in the AC electrical circuit (Figure 2.2). Power supply voltage U = 100 V, active and reactance resistances are R = 3 Ohms, XL = 4 Ohms, XC = 8 Ohms. Construct a vector diagram of current and voltage.
Figure 2.2 AC Electrical Circuit
Electrical circuit impedance:
Coil impedance:
Ammeter reading PA1 (circuit current):
Uк = I?Zк = 20 ? 5 = 100 V.
UC = I?ХС = 20 ? 8 = 160 V.
Wattmeter reading PW1:
Р = I2?R = 202? 3 = 1200 W = 1.2 kW.
The vector diagram is shown in Figure 2.3.
Figure 2.3 Vector diagram
2.3 Problems to be solved during the practical lesson
For a single-phase unbranched AC electrical circuit, determine the voltage drop UL across the inductive reactance XL, the voltage U applied in the circuit, the active P, reactive Q and total power S and the power factor cos of the circuit, if the active and reactive resistance R = XL = 3 Ohm, and the voltage drop across the active element UR = 60 V.
Answer: UL = 60 V; U = 84.8 V; P = 1.2 kW;
Q = 1.2 kVAr; S = 1.697 kVA; cos= 0.71.
A coil with active resistance R = 10 Ohm and inductance L = 133 mH and a capacitor with capacitance C = 159 μF are connected in series to the AC network. Determine the current I in the circuit and the voltage on the coil UC and capacitor UC at a supply voltage of U = 120 V, construct a vector diagram of currents and voltages.
Answer: I = 5A; UK = 215 V; UC = 100 V..
Determine the current in an unbranched AC electrical circuit containing active and reactive resistance: R = 1 Ohm; XC = 5 Ohm; XL = 80 Ohm, as well as the frequency f0 at which voltage resonance occurs, current I0, voltage on the capacitor UC and inductance UL at resonance, if the supply voltage U = 300 V at a frequency f = 50 Hz.
Answer: I =3.4 A; f0 = 12.5 Hz; I0 = 300 A; UC = UL = 6000 V.
Calculate at what capacitance of the capacitor in the circuit in Figure 2.2 there will be voltage resonance if R = 30 Ohm; XL = 40 Ohm.
Answer: C = 78 µF.
3. Calculation of three-phase circuits for various methods of connecting receivers. Circuit analysis for symmetrical and asymmetrical operating modes
3.1 Theoretical information
A three-phase power supply system for electrical circuits is a combination of three sinusoidal EMFs or voltages, identical in frequency and amplitude, shifted in phase relative to each other by an angle of 2/3, i.e. 120є (Figure 3.1).
Figure 3.1 Vector diagram
In symmetrical power supplies, the EMF values are equal. Neglecting the internal resistance of the source, we can take the corresponding EMF of the source equal to the voltages acting at its terminals EA = UA, EB = UB, EC = UC.
An electrical circuit in which a three-phase system of emf or voltage operates is called three-phase. Exist various ways connection of phases of three-phase power sources and three-phase electricity consumers. The most common are star and delta connections.
When connecting the phases of a three-phase electricity consumer with a “star” (Figure 3.2), the ends of the phase windings x, y and z are combined into a common neutral point N, and the beginnings of phases A, B, C are connected to the corresponding linear wires.
Figure 3.2 Connection diagram of the receiver phase windings “star”
The voltages UA, UB, UC acting between the beginnings and ends of the consumer phases are its phase voltages. The voltages UАВ, УВС, УСА, acting between the beginnings of the consumer phases are linear voltages (Figure 3.2). Linear currents Il in the supply lines (IA, IB, IC) are also phase currents Iph flowing through the consumer phases. Therefore, in the presence of a symmetrical three-phase system, when connecting the consumer phases with a “star”, the following relationships are valid:
Il = Iph, (3.1)
Ul = Uph. (3.2)
The active P, reactive Q and total S power of the electricity consumer with a symmetrical load (ZA = ZB = ZC = Zph) and star-connected phases are determined as the sum of the corresponding phase powers.
R = RA + RV + RS = 3 Rf;
Рф = Uф Iф cos ф;
Р = 3Uф Iф cos cph = 3 RфUл Iл cos cph;
Q = QA + QB + QC = 3 Qph;
Q = 3Uф Iф sin ф = 3 ХфUл Iл sin ф;
The connection in which the beginning of the subsequent winding of the electricity consumer phase is connected to the end of the previous phase (in this case, the beginnings of all phases are connected to the corresponding linear wires) is called a “triangle”.
When connected by a “triangle” (Figure 3.3), the phase voltages are equal to the linear voltages
Ul = Uph. (3.3)
Figure 3.3 Connection diagram of the receiver phase windings in a triangle
With a symmetrical power system
UАВ = UВС = UА = Uф = Uл.
The relationship between linear and phase currents when connecting a consumer with a triangle and a symmetrical load
Il = Iph. (3.4)
With a symmetrical electricity consumer with a “triangle” phase connection, the total S, active P and reactive Q powers of the individual phases of the consumer are determined using the formulas obtained for the “star” phase connection.
Three groups of lighting lamps with a power of P = 100 W each with a rated voltage Unom = 220 V are connected in a star configuration with a neutral wire (Figure 3.4, a). In this case, nA = 6 lamps are connected in parallel in phase A, nA = 6 lamps in phase B, nB = 4 lamps in phase C, nC = 2 lamps in phase C. Linear symmetrical voltage of the power source Ul = 380 V. Determine the phase resistance Zph and phase currents Iph of the electricity consumer, construct a vector diagram of currents and voltages, determine the current IN in the neutral wire.
Figure 3.4 Three-phase power system: a - star connection diagram; b - vector diagram
Active resistance of consumer phases:
RВ = = 120 Ohm;
RC = = 242 Ohm,
here Uf = = 220 V.
Phase currents:
IB = = 1.82 A;
The current in the neutral wire is determined graphically. Figure 3.4, b) shows a vector diagram of voltages and currents, from which we find the current in the neutral wire:
3.3 Problems to be solved during the practical lesson
A three-phase symmetrical consumer of electrical energy with phase resistance ZА = ZВ = ZС = Zф = R = 10 Ohm is connected by a star and connected to a three-phase network with a symmetrical voltage Ul = 220 V (Figure 3.5, a). Determine the ammeter reading when line wire B breaks and the total power of a three-phase symmetrical consumer. Construct a vector diagram of voltages and currents with a symmetrical load and with a break in line wire B.
Answer: IA = 12.7 A; P = 4839 W.
A three-phase consumer of electrical energy with active and reactive phase resistances: R1 = 10 Ohm, R2 = R3 = 5 Ohm and ХL = XC = 5 Ohm, connected by a triangle (Figure 3.5, b) and connected to a three-phase network with linear voltage Ul = 100 V with symmetrical supply. Determine the ammeter reading when line wire C breaks; determine phase and linear currents, as well as active, reactive and apparent power of each phase and the entire electrical circuit. Construct a vector diagram of currents and voltages.
Answer: IA = 20 A (at break); IAB = 10 A, IBC = ICA = 14.2 A;
IA = 24 A, IB = 15 A, IC = 24 A; RAV = 10 kW, RVS = RSA = 1 kW, R = 3 kW;
QAB = 0 VAr, QВС = - 1 kVAr, QCA = 1 kVAr, Q = 0;
SАВ = 1 kVA, SВС = SСА = 1.42 kVA, S = 4.85 kVA.
Figure 3.5 Electrical circuit diagram: a - to problem 3.3.1; b - to problem 3.3.2
In the electrical circuit of a three-phase symmetrical consumer of electrical energy, connected by a triangle, the reading of the ammeter connected to line A IA = Il = 22 A, the resistance of the resistors RАВ = RВС = RСА = 6 Ohms, the capacitors ХАВ = ХВС = ХСА = 8 Ohms. Determine line voltage, active, reactive and apparent power. Construct a vector diagram.
Answer: Ul = 127 V, P = 2.9 kW, Q = 3.88 kVAr, S = 4.85 kVA.
An electricity consumer connected by a “star” with active and reactive (inductive) phase resistances: RA = RВ = RC = Rф = 30 Ohm, ХА = ХВ = ХС = Хф = 4 Ohm is included in a three-phase symmetrical network with a linear voltage Ul = 220 V Determine phase and line currents and active power of the consumer. Construct a vector diagram of voltages and currents.
Answer: If = Il = 4.2 A; P = 1.6 kW.
For the conditions of problem 4.3.1, determine the phase voltages and currents, the active power Pk of the consumer during a short circuit of phase B, and construct a vector diagram for this case.
4. Calculation of the mechanical characteristics of an asynchronous motor
4.1 Theoretical information
An asynchronous machine is an electric machine in which a rotating magnetic field is excited during operation, but the rotor rotates asynchronously, that is, with an angular velocity different from the angular velocity of the field.
A three-phase asynchronous machine consists of two main parts: a stationary stator and a rotating rotor.
Like any electrical machine, an asynchronous machine can operate as a motor or generator.
Asynchronous machines mainly differ in the design of the rotor. The rotor consists of a steel shaft, a magnetic circuit made of sheets of electrical steel with stamped grooves. The rotor winding can be short-circuited or phase.
The most widespread are asynchronous motors with a squirrel-cage rotor. They are the simplest in design, easy to use and economical.
Asynchronous motors are the main converters of electrical energy into mechanical energy and form the basis for the drive of most mechanisms used in all areas of human activity. The operation of asynchronous motors does not have a negative impact on the environment. The space occupied by these machines is small.
The rated power of the LV motor is the mechanical power on the shaft in the operating mode for which it is intended by the manufacturer. A number of nominal capacities are established by GOST 12139.
The synchronous rotation speed nc is established by GOST 10683-73 and at a network frequency of 50 Hz has the following values: 500, 600, 750, 1000, 1500 and 3000 rpm.
The energy efficiency indicators of an asynchronous motor are:
Efficiency factor (efficiency), representing the ratio of useful power on the shaft to the active power consumed by the engine from the network
Power factor cosс, representing the ratio of active power consumed to the total power consumed from the network;
Slip characterizes the difference between the rated n1 and synchronous nc motor speed
The values of efficiency, cosс and slip depend on the load of the machine and are given in the catalogues. The mechanical characteristic represents the dependence of the engine torque on its rotational speed at constant voltage and frequency of the supply network. Starting properties are characterized by the values of starting torque, maximum (critical) torque, starting current or their multiples. Rated current can be determined from the formula for rated engine power
The starting current is determined by the catalog data of the starting current multiplicity.
The rated torque of the engine is determined by the formula
The nominal rotor speed pN is determined by the formula
The starting torque is determined from the catalog data.
The maximum torque is determined from catalog data.
The power consumed by the motor from the network at rated load is greater than the rated power by the amount of losses in the motor, which is taken into account by the efficiency value.
Total power loss in the engine at rated load
The mechanical characteristic of an asynchronous motor is calculated using the formula
where sKP is the critical slip at which the engine develops the maximum (critical) torque MMAX;
s - current slip (independently take 8-10 values from 0 to 1, including sKP and sН).
The shaft rotation speed is determined by slip
5. Electrical measurements and instruments
5.1 Theoretical information
The objects of electrical measurements are all electrical and magnetic quantities: current, voltage, power, energy, magnetic flux, etc. Electrical measuring devices are also widely used to measure non-electrical quantities (temperature, pressure, etc.). There are electrical measuring instruments for direct evaluation and comparison devices. The instrument scales indicate the type of current, the instrument system, its name, the operating position of the scale, the accuracy class, and the insulation test voltage.
Based on the principle of operation, there are magnetoelectric, electromagnetic, electrodynamic, ferrodynamic, as well as thermal, induction, electrochemical and other electrical measuring instruments. Electrical measurements can also be made using digital meters. Digital measuring instruments (DMTs) are multi-range, universal instruments designed for measuring various electrical quantities: alternating and direct current and voltage, capacitance, inductance, timing parameters of the signal (frequency, period, pulse duration) and recording the signal shape , its spectrum, etc.
In digital measuring instruments, the input measured analog (continuous) quantity is automatically converted into the corresponding discrete quantity, followed by presentation of the measurement result in digital form.
According to the principle of operation and design, digital devices are divided into electromechanical and electronic. Electromechanical devices have high accuracy, but low measurement speed. Electronic devices use modern electronics base.
One of the most important characteristics of electrical measuring instruments is accuracy. The results of measurements of electrical quantities inevitably differ from their true value due to the presence of corresponding errors (random, systematic, misses).
Depending on the method of numerical expression, errors are distinguished between absolute and relative, and in relation to indicating instruments - also reduced.
The absolute error of a measuring device is the difference between the measured Ai and the actual Ad values of the measured quantity:
YES = Ai - Hell. (4.1)
The absolute error does not give an idea of the accuracy of the measurement, which is assessed by the relative measurement error, which is the ratio of the absolute measurement error to the actual value of the measured value, expressed as a fraction or percentage of its actual value
To assess the accuracy of the indicating measuring instruments themselves, the reduced error is used, i.e. the ratio of the absolute error of the DA reading to the nominal value Anom, expressed as a percentage, corresponding to the largest reading of the device:
Electrical measuring instruments are divided into eight accuracy classes: 0.05; 0.1; 0.2; 0.5; 1.0; 1.5; 2.5; 4 indicated on the scales. Accuracy classes of instruments are determined by the given error.
When measuring sufficiently large currents, when the measuring device is not designed for such currents, shunts are connected parallel to the device circuit, which are a resistance of a known value, having a relatively low resistance Rsh, through which most of the measured current is passed. The distribution of currents between the device and the shunt IA and Ish is inversely proportional to the resistances of the corresponding branches.
in this case, the measured current I = IA + Ish, then
To simplify calculations, the shunt coefficient is taken equal to Ksh = 10; 100 and 1000. When measuring sufficiently large voltages, an additional resistance Rd is connected in series with the device, to which most of the measured voltage is supplied.
Measuring shunts and additional resistance are used only in DC electrical circuits. AC electrical circuits use current transformers (for measuring very high currents) and voltage transformers (for measuring high voltages).
5.2 Example of solving a typical problem
To measure the voltage in an electrical circuit, a voltmeter of accuracy class 1.0 with a measurement limit Unom = 300 V is used. The voltmeter reading Ui = 100 V. Determine the absolute DU and relative measurement errors and the actual value of the measured voltage.
Since the true (actual) value of the measured quantity is unknown, to determine the absolute error we use the accuracy class of the device (the reduced error of the device is equal to its accuracy class, i.e. r = 1%):
Relative error
Consequently, the measured voltage value Ui = 100 V may differ from its actual value by no more than 3%.
5.3 Problems to be solved during the practical lesson
Determine the absolute DI and relative error of current measurement with an ammeter with a nominal current limit value Inom = 5 A and an accuracy class of 0.5. If its reading (measured value) Ii = 2.5 A.
Answer: DI = 0.025 A, d = 1%.
The limit value of the current measured by a milliammeter is I = 4?10-3 A, the resistance of which is RA = 5 Ohms. Determine the resistance Rsh of the shunt used to expand the current measurement limit to I = 15A.
Answer: Rsh = 1.33 mOhm.
Electrical measuring set K-505 is equipped with a voltmeter with a scale having NV = 150 divisions, and an ammeter with a scale having NA = 100 divisions. Determine the value of the instrument scale division, the readings of the voltmeter, the arrow of which indicates = 100 divisions, as well as the readings of the ammeter, the arrow of which indicates = 50 divisions, for the limits of measurement of currents and voltages, the nominal values of which are presented in table 54.1
Table 4.1 Device parameters
For the electrical circuit (Figure 54.1), determine the currents in the branches and the reading of the voltmeter РV1, which has an internal resistance Rв = 300 Ohm. Resistor resistances: R1 = 50 Ohm, R2 = 100 Ohm, R2 = 150 Ohm, R4 = 200 Ohm. EMF of power supplies: E1 = 22 V, E2 = 22 V.
Answer: I1 = 0.026 A, I2 = 0.026 A, I3 = 0.052 A, Uv = 15.6 V.
Figure 5.1 Electrical circuit diagram
The K-505 electrical measuring kit is equipped with a wattmeter designed for the current and voltage limits given in Table 5.2; the wattmeter scale has N = 150 divisions. Determine the division price of the CW wattmeter for all voltage and current limits corresponding to its readings. During the measurement, the wattmeter needle deviated by Nґ = 100 divisions in all cases.
Table 5.2 Instrument parameters
An ammeter designed for a maximum direct current Inom = 20 A is included in the direct current electrical circuit to measure current. Ammeter reading I = 10 A, actual current Id = 10.2 A. Determine the absolute DI, relative d and reduced g measurement error.
Answer: DI = 0.2 A; d = 2%; r = 1%.
An electrical circuit with a voltage of U = 220 V includes a voltmeter with an additional resistance Rd = 4000 Ohms, the resistance of the voltmeter RB = 2000 Ohms. Determine the voltmeter readings.
Answer: UB = 73.33 V.
An ammeter type M-61 with a measurement limit Inom = 5 A is characterized by a voltage drop across the terminals ДУА = 75?10-3 V = 75 mV. Determine the resistance of the ammeter RA and the power it consumes RA.
An additional resistance Rd = 12 kOhm is connected to a voltmeter with an internal resistance of 8 kOhm. If there is additional resistance, you can use this voltmeter to measure voltage up to 500 V. Determine what voltage can be measured with this device without additional resistance.
Answer: U = 200 V.
The meter panel says “220 V, 5 A, 1 kWh = 500 revolutions.” Determine the relative error of the meter if the following values were obtained during verification: U = 220 V, I = 3 A, the disk made 63 revolutions in 10 minutes. Provide a circuit diagram for switching on the meter.
Answer: d = 14.5%.
The meter panel says “1 kWh = 2500 disk revolutions.” Determine the power consumption if the meter disk makes 20 revolutions in 40 seconds.
Answer: P = 720 W.
The resistance of a magnetoelectric ammeter without a shunt is RA = 1 Ohm. The device has 100 divisions, the division price is 0.001 A/division. Determine the measurement limit of the device when connecting a shunt with a resistance RШ = 52.6?10-3 Ohm and the division value.
Answer: 2 A; 0.02 A/div.
The upper limit of microammeter measurement is 100 μA, internal resistance is 15 Ohms. What should the shunt resistance be equal for the upper limit of measurement to increase by 10 times?
Answer: 1.66 Ohm.
For an electromagnetic voltmeter with a total deflection current of 3 mA and an internal resistance of 30 kOhm, determine the upper limit of measurement and the resistance of the additional resistor required to extend the upper limit of measurement to 600 V.
Answer: 90 V; 170 kOhm.
Bibliography
1. Kasatkin, A.S. Electrical engineering [Text]: textbook for students. non-electrical specialist. universities / A.S. Kasatkin, M.V. Nemtsov. - 6th ed., revised. - M.: Higher school, 2000. - 544 p.: ill.
2. Theoretical basis electrical engineering [Text]: textbook / A.N. Gorbunov [etc.]. - M.: UMC "TRIADA", 2003. - 304 p.: ill.
3. Nemtsov, M.V. Electrical engineering [Text]: textbook / M.V. Nemtsov, I.I. Svetlakova. - Rostov-n/D: Phoenix, 2004. - 567 p.: ill.
4. Rekus, G.G. Fundamentals of electrical engineering and industrial electronics in examples and problems with solutions [Text]: textbook. a manual for university students studying non-electrotechnical specialties. direction preparation of diploma specialist. in the field of engineering and technology: admitted by the Ministry of Education and Science of the Russian Federation / G.G. Recus. - M.: Higher school, 2008. - 343 p.: ill.
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Introduction........................................................ ...................................... 4
1 Section 1. Calculation of a complex DC electrical circuit 5
1.1 Calculation of currents according to Kirchhoff’s laws.................................... 5
1.2 Replacing the resistance triangle with an equivalent star.................................................... ........................................................ ........ 6
1.3 Calculation using the “Loop Currents” method.................................................. 8
1.4 Power balance of the electrical circuit.................................... 9
1.5 Calculation of potentials of points in an electrical circuit.................................. 10
2 Section 2. Calculation and analysis of the AC electrical circuit 12
2.1 Calculation of currents using the complex method.................................... 12
2.2 Determination of the active power of a wattmeter.................................. 14
2.3 Balance of active and reactive power.................................... 14
2.4 Vector diagram of currents.................................................... 14
3 Section 3. Calculation of a three-phase electrical circuit.................................. 15
3.1 Calculation of phase and line currents.................................................. 15
3.2 Power of a three-phase electrical circuit.................................... 16
3.3 Vector diagram of currents and voltages.................................. 17
4 Section 4. Calculation of a three-phase asynchronous motor....... 18
Conclusion................................................. ................................ 23
List of references......................................................... 24
Introduction
Electrical engineering as a science is a field of knowledge that deals with electrical and magnetic phenomena and their practical use. Electronics, radio engineering, electric drives and other related sciences began to develop on the basis of electrical engineering.
Electrical energy is used in all areas of human activity. Production plants in enterprises are mainly electrically driven, i.e. are driven by electric motors. Electrical instruments and devices are widely used to measure electrical and non-electrical quantities.
The continuously expanding use of various electrical and electronic devices necessitates the knowledge of specialists in all fields of science, technology and the production of basic concepts about electrical and electromagnetic phenomena and their practical application.
Students' knowledge of this discipline will ensure their fruitful work in the future as engineers given the current state of power supply of enterprises.
As a result of the acquired knowledge, an engineer of non-electrical specialties must be able to skillfully operate electrical and electronic equipment and electric drives used in modern production conditions, and know the way and methods of saving energy.
SECTION 1. CALCULATION OF A COMPLEX ELECTRICAL DC CIRCUIT
The circuit parameters are given in Table 1.
Table 1 – Electrical circuit diagram parameters.
|
EMF of power source 1 (E 1) |
|
|
EMF of power source 2 (E 2) |
|
|
EMF of power source 3 (E 3) |
|
|
Internal resistance of power supply (R 01) |
|
|
Internal resistance of power supply (R 02) |
|
|
Internal resistance of power supply (R 03) |
|
|
Resistor 1 value (R 1) |
|
|
Resistor 2 (R 2) |
|
|
Resistor 3 value (R 3) |
|
|
Resistor 4 value (R 4) |
|
|
Resistor value 5 (R 5) |
|
|
Resistor value 6 (R 6) |
1.1 Calculation of currents according to Kirchhoff's laws
We show on the diagram the direction of currents in the branches (Fig. 1).
According to Kirchhoff’s first law for direct current circuits, the algebraic sum of currents in any node of an electrical circuit is zero, i.e. the sum of currents directed from the node is equal to the sum of currents directed to the node.
We compose equations according to Kirchhoff’s first law for nodes, the number of which is equal to (n–1), where n is the number of nodes in the circuit:
A) +I 1 + I 3 – I 2 = 0; (1.1)
B) I 4 + I 6 – I 3 = 0; (1.2)
D) I 5 – I 1 – I 4 = 0. (1.3)
According to Kirchhoff's second law for direct current circuits in any closed circuit, the algebraic sum of the voltages on the resistive elements is equal to the algebraic sum of the emf.
We compose equations according to Kirchhoff’s second law for each circuit:
I) I 3 ∙ (R 3 + R 03) – I 1 ∙ (R 1 + R 01) + I 4 ∙ R 4 = E 3 – E 1; (1.4)
II) I 1 ∙ (R 1 + R 01) + I 2 ∙ (R 2 + R 02) + I 5 ∙ R 5 = E 1 + E 2; (1.5)
III) I 6 ∙ R 6 – I 4 ∙ R 4 – I 5 ∙ R 5 = 0. (1.6)
We solve all the resulting equations together as a system, substituting all known values:
=>
(1.7)
Having solved the matrix, we obtain unknown values of currents in the branches:
I 1 = – 0.615 A;
If the current in the branch turns out to be negative, it means that its direction is opposite to that chosen in the diagram.
1.2 Replacing the resistance triangle with an equivalent star
Let us transform the “triangle” bcd, corresponding to the electrical circuit diagram, into an equivalent “star” (Fig. 2). The initial triangle is formed by resistances R 4, R 5, R 6. During the transformation, the condition of circuit equivalence is necessarily preserved, i.e. the currents in the wires passing to the converted circuit and the voltages between the nodes do not change their values.
When converting a “triangle” into a “star” we use the following calculation formulas:
Ohm. (1.10)
As a result of the transformation, the original circuit is simplified (Fig. 3).
In the converted circuit there are only three branches and, accordingly, three currents I 1, I 2, I 3. To calculate these currents, it is enough to have a system of three equations compiled according to Kirchhoff’s laws:
(1.11)
When drawing up equations, the direction of current and circuit bypass is selected in the same way as in a three-circuit circuit.
We compose and solve the system:
(1.12)
Having solved the matrix, we obtain the unknown values of the currents I 1, I 2, I 3:
I 1 = –0.615 A;
By substituting the obtained current values into the equations compiled for the three-circuit circuit, we determine the remaining currents I 4, I 5, I 6:
1.3 Calculation using the “Loop Currents” method
We arbitrarily set the direction of the loop currents in the cells of the original circuit. It is more convenient to indicate all currents in one direction - clockwise
INTRODUCTION
The topic of this course work: “Calculation and analysis of electrical circuits.”
The course project includes 5 sections:
1) Calculation of DC electrical circuits.
2) Calculation of non-linear DC circuits.
3) Solution of single-phase linear AC electrical circuits.
4) Calculation of three-phase linear electrical circuits of alternating current.
5) Study of transient processes in electrical circuits.
Each assignment involves making diagrams.
The goal of the course project is to study various methods for calculating electrical circuits and based on these calculations to build various types diagrams.
The course project uses the following designations: R-active resistance, Ohm; L - inductance, H; C - capacitance, Ф; XL, XC - reactance (capacitive and inductive), Ohm; I - current, A; U - voltage, V; E - electromotive force, V; shi, shi - voltage and current shift angles, degrees; P - active power, W; Q - reactive power, Var; S - total power, VA; ts - potential, V; NE is a nonlinear element.
CALCULATION OF LINEAR DC ELECTRICAL CIRCUITS
For the electrical circuit (Fig. 1), do the following:
1) Create a system of equations based on Kirchhoff’s laws to determine currents in all branches of the circuit;
2) Determine the currents in all branches of the circuit using the loop current method;
3) Determine the currents in all branches of the circuit based on the method of nodal potentials;
4) Draw up a power balance;
5) Present the results of current calculations for points 2 and 3 in table form and compare;
6) Construct a potential diagram for any closed loop that includes an emf.
E1=30 V; R4=42 Ohm;
E2=40 V; R5=25 Ohm;
R1=16 Ohm; R6=52 Ohm;
R2=63 Ohm; r01=3 Ohm;
R3=34 Ohm; r02=2 Ohm;
R1"=R1+r01=16+3=19 Ohm;
R2"=R2+r02=63+2=65 Ohm.
Let's choose the direction of the currents.
Let's choose the direction of traversing the contours.
Let's create a system of equations according to Kirchhoff's law:
E1=I1R1"+I5R5-I4R4
E2=I2R2"+I5R5+I6R6
E2=I4R4+I3R3+I2R2"
Figure 1. DC circuit diagram
Calculation of electrical circuits using the loop current method.
Let's arrange the currents
Let's choose the direction of the loop currents according to the EMF
Let's create equations for loop currents:
Ik1 Х(R1"+R4+R5)-Ik2ЧR4+Ik3R5"=E1
Ik2 Х(R3+R+R2")-Ik1ЧR4+Ik3Ч=E2
Ik3 H(R6+R2"+R5)+Ik1×R5+Ik2×R2"=E2
Let's substitute the numerical values of EMF and resistance into the equation:
Ik1 CH86-Ik2CH42-+Ik3CH25=30
Ik1 CH42+Ik2CH141+Ik3CH65=40
Ik1 H(25)+Ik2H65+Ik3H142=40
Let's solve the system using the matrix method (Cramer's method):
D1= =5.273Х105
D2= =4.255Х105
D3= =-3.877Х105
We calculate Ik:
Let us express the circuit currents in terms of contour currents:
I2 =Ik2+Ik3=0.482+(-44)=0.438 A
I4 =-Ik1+Ik2=0.482-0.591=-0.109A
I5 =Ik1 + Ik3=0.591+(-0.044)=0.547A
Let's draw up a power balance for a given circuit:
Fig.=E1I1+E2I2=(30Х91)+(40Х38)=35.25 W
Rpr.=I12R1"+I22R2"+I32R3+I42R4+I52R5+I62R6=(91)2H16+(38)2H 63 + (82)2H H34+(-09)2H42+(47)2H25+(44)H52=41.53 watts .
1 Calculation of electrical circuits using the nodal potential method
2 Let's arrange the currents
3 Place the nodes
4 Let’s create an equation for the potentials:
ts1=(1?R3+1?R4+1?R1")-ts2Ч(1/R3)-ts3-(1/R4)=E1?R1"
ts2Ч(1/R3+1?R6+1?R2")-ts1Ч(1/R3)-ts3(1/R2") =(-E2 ?R2")
ts3Ch(1/R5+1?R4+1?R2")-ts2H(1/R2")-ts1H(1/R4)=E2?R2"
Let's substitute the numerical values of EMF and resistance:
ts1H0.104-ts2H0.029-ts3H0.023=1.57
Ts1H0.029+ts2H0.063-ts3H0.015=(-0.61)
Ts1H0.023-ts2H0.015+ts3H0.078=0.31
5 Let’s solve the system using the matrix method (Cramer’s method):
1= = (-7.803Х10-3)
2= = (-0.457Х10-3)
3= = 3.336Х10-3
6 Calculate c:
q2= = (-21Х103)
7 Finding the currents:
I1= (ts4- ts1+E)1?R1"=0.482A
I2= (ts2- ts3+E2) ?R2"=0.49A
I3= (ts1- ts2) ?R3=(-0.64)A
I4= (ts3- ts1) ?R4=(-0.28)A
I5= (ts3- ts4) ?R5= 0.35A
I6= (ts4- ts2) ?R6=(-0.023)A
8 The results of current calculations using two methods are presented in the form of a free table
Table 1 - Results of current calculations using two methods
Let's construct a potential diagram for any closed loop including EMF.
Figure 3 - DC circuit circuit
E1=30 V; R4=42 Ohm;
E2=40 V; R5=25 Ohm;
R1=16 Ohm; R6=52 Ohm;
R2=63 Ohm; r01=3 Ohm;
R3=34 Ohm; r02=2 Ohm;
R1"=R1+r01=16+3=19 Ohm;
R2"=R2+r02=63+2=65 Ohm.
We calculate the potentials of all points of the circuit when moving from element to element, knowing the magnitude and direction of the branch currents and EMF, as well as the resistance values.
If the current coincides in direction with the bypass, it means -, if it coincides with the EMF, it means +.
ts2=ts1-I2R2"= 0 - 0.438 H 65 = - 28.47B
ts3=ts2+E2= - 28.47+40=11.53B
ts4=ts3-I4R4 = 11.58-(-4.57)=16.15B
ts4=ts4-I3R3 = 16.15-16.32=-0.17B
We build a potential diagram, plot the circuit resistance along the abscissa axis, and the potentials of the points along the ordinate axis, taking into account their signs.
An electrical circuit is a set of electrical devices that create a path for electric current, the electromagnetic processes in which are described by equations taking into account the concepts of electromotive force, electric current and electric voltage.
The main elements of the electrical circuit (Figure 1.1) are sources and consumers of electrical energy.
Figure 1.1 Basic elements of an electrical circuit
DC generators and galvanic cells are widely used as sources of direct current electrical energy.
Sources of electrical energy are characterized by the emf E they develop and the internal resistance R0.
Consumers of electrical energy are resistors, electric motors, electrolysis baths, electric lamps, etc. In them, electrical energy is converted into mechanical, thermal, light, etc. In an electrical circuit, the positive direction of the emf E is taken to be the direction coinciding with force acting on a positive charge, i.e. from “-” source to “+” power source.
When calculating electrical circuits, real sources of electrical energy are replaced by equivalent circuits.
The equivalent circuit of the EMF source contains the EMF E and the internal resistance R0 of the source, which is much less than the resistance Rн of the electricity consumer (Rн >> R0). Often in calculations the internal resistance of the EMF source is equated to zero.
For a circuit section that does not contain an energy source (for example, for the circuit Figure 1.2, a), the relationship between current I and voltage U12 is determined by Ohm’s law for the circuit section:
where c1 and c2 are the potentials of points 1 and 2 of the circuit;
Y R is the sum of resistances in a section of the circuit;
R1 and R2 are the resistances of the circuit sections.
Figure 1.2 Electrical diagram of a circuit section: a - not containing an energy source; b - containing an energy source
For a section of a circuit containing an energy source (Figure 1.2, b), Ohm’s law is written as the expression
where E is the EMF of the energy source;
R = R1 + R2 is the arithmetic sum of the resistances of the circuit sections;
R0 is the internal resistance of the energy source.
The relationship between all types of power in an electrical circuit (power balance) is determined from the equation:
UR1 = UR2 + URp, (1.3)
where UR1 = UEI is the algebraic sum of the powers of energy sources;
UR2 - algebraic sum of consumer power (net power) (P2 = UI);
URp = УI2R0 - total power due to losses in source resistance.
Resistors, as well as the resistance of other electrical devices, are consumers of electrical energy. The power balance is determined by the law of conservation of energy, while in any closed electrical circuit the algebraic sum of the powers of energy sources is equal to the algebraic sum of powers consumed by consumers of electrical energy.
The efficiency of the installation is determined by the ratio
When calculating unbranched and branched linear DC electrical circuits, various methods can be used, the choice of which depends on the type of electrical circuit.
When calculating complex electrical circuits, in many cases it is advisable to simplify them by folding, replacing individual sections of the circuit with series, parallel and mixed resistance connections with one equivalent resistance using the method of equivalent transformations (transfiguration method) of electrical circuits.
