Calculation of the total voltage of the circuit. DZ - Calculation of a complex DC circuit. Series connection of nonlinear elements
Fundamentals > Problems and Answers > Direct Electric Current
Methods for calculating DC circuits
The circuit consists of branches, has nodes and current sources. The formulas given below are suitable for calculating circuits containing both voltage sources and current sources. They are also valid for those special cases: when the circuit contains only voltage sources or only current sources.
Application of Kirchhoff's laws.Typically, all sources of emf and current sources and all resistances in a circuit are known. In this case, the number of unknown currents is set equal to. For each branch, the positive direction of the current is specified.
The number Y of mutually independent equations compiled according to Kirchhoff's first law is equal to the number of nodes minus one. The number of mutually independent equations compiled according to Kirchhoff’s second law,
When composing equations according to Kirchhoff’s second law, you should choose independent circuits that do not contain current sources. The total number of equations compiled according to the first and second Kirchhoff laws is equal to the number unknown currents.
Examples are given in the tasks of the section.
Loop current method (Maxwell).This method allows you to reduce the number of equations of the system to the number K, determined by formula (0.1.10). It is based on the fact that the current in any branch of the circuit can be represented as an algebraic sum of the loop currents flowing through this branch. When using this method, loop currents are selected and designated (at least one selected loop current must pass through any branch). It is known from theory that the total number of loop currents. It is recommended to chooseloop currents so that each of them passes through one current source (these loop currents can be considered to coincide with the corresponding currents of the current sourcesand they are usually given conditions of the problem), and the remainingselect loop currents passing through branches that do not contain current sources. To determine the last loop currents according to Kirchhoff’s second law for these loops, K equations are compiled in the following form:
Where - circuit’s own resistance n (the sum of the resistances of all branches included in the circuit n); - total circuit resistance n and l, and , if the directions of the loop currents in the common branch for the loops n and l coincide, then it is positive , otherwise negative; - algebraic sum of the EMF included in the branches forming the circuit n; - total resistance of the circuit branch n with a circuit containing a current source.
Examples are given in the tasks of the section.
Nodal stress method.This method allows you to reduce the number of equations of the system to a number Y equal to the number of nodes minus one
The essence of the method is that first, by solving the system of equations (0.1.13), the potentials of all nodes of the circuit are determined, and the currents of the branches connecting the nodes are found using Ohm’s law.
When composing equations using the nodal voltage method, the potential of any node is first assumed to be zero (it is called the basis potential). To determine the potentials of the remaining nodes, the following system of equations is compiled:
Here - the sum of the conductivities of the branches connected to node s;- the sum of the conductances of the branches directly connecting node s to node q;
- algebraic sum of the products of the emf of the branches adjacent to the node s , on their conductivity; in this case, those EMFs that act in the direction of node s are taken with the “+” sign, and with the “-” sign - in the direction from node s;- algebraic sum of currents of current sources connected to node s; in this case, those currents that are directed to the node are taken with the “+” sign s , and with the sign “-” - in the direction from node s.
It is recommended to use the nodal voltage method in cases where the number of equations is less than the number of equations compiled using the loop current method.
If in the circuit some nodes are connected by ideal emf sources, then the number Y of equations compiled using the nodal voltage method decreases:
Where - the number of branches containing only ideal emf sources.
Examples are given in the tasks of the section.
A special case is a two-node circuit. For circuits with two nodes (to be specific, nodes a and b ), nodal voltage
Where - the algebraic sum of the products of the EMF of the branches (EMFs are considered positive if they are directed to node a, and negative if from node a to node b ) on the conductivity of these branches;- currents of current sources (positive if they are directed to node a, and negative if directed from node a to node b) ; - sum conductivities of all branches connecting nodes a and b.
The principle of superposition.If in an electrical circuit the given values are the emf of the sources and the currents of the current sources, then the calculation of the currents based on the superposition principle is as follows. The current in any branch can be calculated as the algebraic sum of the currents caused in it by the EMF of each EMF source separately and the current passing through the same branch from the action of each current source. It should be borne in mind that when the currents caused by any one source of EMF or current are being calculated, the remaining sources of EMF in the circuit are replaced by short-circuited sections, and the branches with current sources of the remaining sources are turned off (the branches with current sources are opened).
Equivalent circuit transformations.In all cases of transformation, the replacement of some circuits with others that are equivalent to them should not lead to a change in currents or voltages in sections of the circuit that have not undergone transformation.
Replacement of series-connected resistances with one equivalent one. Resistances are connected in series if they flow around the same current (for example, resistancesconnected in series (see Fig. 0.1,3), also in series resistance).
n series connected resistances is equal to the sum of these resistances
With serial connection n voltage resistances across them are distributed in direct proportion to these resistances
In the special case of two series-connected resistances
where U - the total voltage acting on a section of the circuit containing two resistances(see Fig. 0.1.3).
Replacing parallel-connected resistances with one equivalent one. Resistors are connected in parallel if they are connected to the same pars of nodes, for example, resistance
(see Fig. 0.1.3).
Equivalent resistance of a circuit consisting of n parallel connected resistances (Fig. 0.1.4),
In the special case of parallel connection of two resistancesequivalent resistance
With parallel connection n resistances (Fig. 0.1.4, a) the currents in them are distributed inversely proportional to their resistances or directly proportional to their conductivities
Current in each of them is calculated through the current I in the unbranched part of the chain
In the special case of two parallel branches (Fig. 0.1.4, b)
Replacing a mixed resistance connection with one equivalent one. A mixed connection is a combination of series and parallel connections of resistances. For example, resistance (Fig. 0.1.4, b) are connected mixed. Their equivalent resistance
Formulas for converting a resistance triangle (Fig. 0.1.5, a) into an equivalent resistance star (Fig. 0.1.5, b), and vice versa, have the following form:
Equivalent source method(active two-terminal method, or open-circuit and short-circuit method). The use of the method is advisable for determining the current in any one branch of a complex electrical circuit. Let's consider two options: a) the equivalent EMF source method and b) the equivalent current source method.
With the equivalent EMF source methodto find the current I in an arbitrary branch ab, the resistance of which is R (Fig. 0.1.6, a,
the letter A means an active two-terminal network), you need to open this branch (Fig. 0.1.6,b), and replace the part of the circuit connected to this branch with an equivalent source with EMFand internal resistance(Fig. 0.1.6, c).
EMFof this source is equal to the voltage at the terminals of the open branch (open circuit voltage):
Calculation of circuits in idle mode (see Fig. 0.1.6, b) to determine carried out by any known method.
Internal resistanceequivalent EMF source is equal to the input resistance of the passive circuit relative to terminals a and b of the original circuit, from which all sources are excluded [EMF sources are replaced by short-circuited sections, and branches with current sources are disconnected (Fig. 0.1.6, d); the letter P indicates the passive nature of the circuit], with branch ab open. Resistance can be calculated directly from the diagram in Fig. 0.1.6, g.
The current in the desired branch of the circuit (Fig. 0.1.6, d), which has a resistance R, is determined according to Ohm’s law:
IN DC circuits Constant voltages operate, constant currents flow and only resistive elements (resistance) are present.
Ideal voltage source called a source, the voltage at the terminals of which, created by the internal electromotive force (EMF), does not depend on the current it generates in the load (Fig. 6.1a). In this case equality takes place. The current-voltage characteristic of an ideal voltage source is shown in Fig. 6.1b.
Ideal current source called a source that supplies a current to the load that does not depend on the voltage at the source terminals, Fig. 6.2a. Its current-voltage characteristic is shown in Fig. 6.2b.
IN resistance the relationship between voltage and current is determined by Ohm's law in the form
An example of an electrical circuit is shown in Fig. 6.3. It highlights branches, consisting of a series connection of several elements (source E and resistance) or one element (and) and nodes- points of connection of three or more branches, marked with bold dots. In the example considered, there are branches and nodes.
In addition, in the chain there are independent closed loops, not containing ideal current sources. Their number is equal. In the example in Fig. 6.3 their number, for example, contours with branches E and shown in Fig. 6.3 ovals with arrows indicating positive direction bypassing the circuit.
The relationship between currents and voltages in a circuit is determined by Kirchhoff's laws.
First Kirchhoff's law: the algebraic sum of currents converging at a node in an electrical circuit is equal to zero,
The currents flowing into the node have a plus sign, and the flowing currents have a minus sign.
Kirchhoff's second law: the algebraic sum of the voltages on the elements of a closed independent circuit is equal to the algebraic sum of the EMF of the ideal voltage sources connected in this circuit,
Voltages and EMF are taken with a plus sign if their positive directions coincide with the direction of the circuit bypass, otherwise a minus sign is used.
For the one shown in Fig. 6.3 examples using Ohm’s law we obtain a subsystem of component equations
According to Kirchhoff's laws, the subsystem of topological equations of a chain has the form
Calculation based on Ohm's law
This method is convenient for calculating relatively simple circuits with one signal source. It involves calculating the resistance of sections of the circuit for which the value is known.
value of current (or voltage), followed by determination of the unknown voltage (or current). Let's consider an example of calculating a circuit, the diagram of which is shown in Fig. 6.4, with an ideal source current A and resistances Ohm, Ohm, Ohm. It is necessary to determine the currents of the branches and , as well as the voltages across the resistances , and .
The source current is known, then it is possible to calculate the resistance of the circuit relative to the terminals of the current source (parallel connection of resistance and series connection
Rice. 6.4 nal resistances and ),
The voltage at the current source (at the resistance) is equal to
Then you can find the branch currents
The results obtained can be verified using Kirchhoff's first law in the form. Substituting the calculated values, we obtain A, which coincides with the value of the source current.
Knowing the branch currents, it is not difficult to find the voltages across the resistances (the value has already been found)
According to Kirchhoff's second law. Adding up the results obtained, we are convinced of its implementation.
Circuit calculation using Kirchhoff equations
Let's calculate the currents and voltages in the circuit shown in Fig. 6.3 for and . The circuit is described by a system of equations (6.4) and (6.5), from which we obtain for the branch currents
From the first equation we express , and from the third
Then from the second equation we get
and therefore
From the equations of Ohm's law we write
For example, for the circuit in Fig. 6.3 in general we get
Substituting the previously obtained expressions for currents into the left side of equality (6.11), we obtain
which corresponds to the right side of expression (6.11).
Similar calculations can be done for the circuit in Fig. 6.4.
The power balance condition allows you to additionally control the correctness of the calculations.
In electrical engineering, it is generally accepted that a simple circuit is a circuit that reduces to a circuit with one source and one equivalent resistance. You can collapse a circuit using equivalent transformations of serial, parallel, and mixed connections. The exception is circuits containing more complex star and delta connections. Calculation of DC circuits produced using Ohm's and Kirchhoff's laws.
Example 1
Two resistors are connected to a 50 V DC voltage source, with internal resistance r = 0.5 Ohm. Resistor values R 1 = 20 and R2= 32 Ohm. Determine the current in the circuit and the voltage across the resistors.
Since the resistors are connected in series, the equivalent resistance will be equal to their sum. Knowing it, we will use Ohm's law for a complete circuit to find the current in the circuit.
Now knowing the current in the circuit, you can determine the voltage drop across each resistor.
There are several ways to check the correctness of the solution. For example, using Kirchhoff's law, which states that the sum of the emf in the circuit is equal to the sum of the voltages in it.
But using Kirchhoff's law it is convenient to check simple circuits that have one circuit. A more convenient way to check is power balance.
The circuit must maintain a power balance, that is, the energy given by the sources must be equal to the energy received by the receivers.
The source power is defined as the product of the emf and the current, and the power received by the receiver as the product of the voltage drop and the current.
The advantage of checking the power balance is that you do not need to create complex cumbersome equations based on Kirchhoff's laws; it is enough to know the EMF, voltages and currents in the circuit.
Example 2
Total current of a circuit containing two resistors connected in parallel R 1 =70 Ohm and R 2 =90 Ohm, equals 500 mA. Determine the currents in each of the resistors.
Two resistors connected in series are nothing more than a current divider. We can determine the currents flowing through each resistor using the divider formula, while we do not need to know the voltage in the circuit; we only need the total current and the resistance of the resistors.
Currents in resistors 
In this case, it is convenient to check the problem using Kirchhoff’s first law, according to which the sum of currents converging at a node is equal to zero.
If you do not remember the current divider formula, then you can solve the problem in another way. To do this, you need to find the voltage in the circuit, which will be common to both resistors, since the connection is parallel. In order to find it, you must first calculate the circuit resistance
And then the tension
Knowing the voltages, we will find the currents flowing through the resistors
As you can see, the currents turned out to be the same.
Example 3
In the electrical circuit shown in the diagram R 1 =50 Ohm, R 2 =180 Ohm, R 3 =220 Ohm. Find the power released by the resistor R 1, current through resistor R 2, voltage across resistor R 3 if it is known that the voltage at the circuit terminals is 100 V.
To calculate the DC power dissipated by resistor R 1, it is necessary to determine the current I 1, which is common to the entire circuit. Knowing the voltage at the terminals and the equivalent resistance of the circuit, you can find it.
Equivalent resistance and current in the circuit
Hence the power allocated to R 1
The essence of the calculations is, as a rule, to determine the currents in all branches and voltages on all elements (resistances) of the circuit using the known values of all circuit resistances and source parameters (emf or current).
For calculation electrical circuits dc different methods can be used. Among them the main ones are:
– a method based on the compilation of Kirchhoff equations;
– method of equivalent transformations;
– loop current method;
– application method;
– method of nodal potentials;
– equivalent source method;
The method, based on the compilation of Kirchhoff's equations, is universal and can be used for both single-circuit and multi-circuit circuits. In this case, the number of equations compiled according to Kirchhoff’s second law must be equal to the number of internal circuits of the circuit.
The number of equations compiled according to Kirchhoff's first law should be one less than the number of nodes in the circuit.
For example, for this scheme
2 equations are compiled according to Kirchhoff’s 1st law and 3 equations according to Kirchhoff’s 2nd law.
Let's consider other methods for calculating electrical circuits:
The equivalent transformation method is used to simplify circuit diagrams and calculations of electrical circuits. An equivalent conversion is understood as such a replacement of one circuit by another, in which the electrical quantities of the circuit as a whole do not change (voltage, current, power consumption remain unchanged).
Let's consider some types of equivalent circuit transformations.
A). series connection of elements
The total resistance of series-connected elements is equal to the sum of the resistances of these elements.
R E =Σ R j (3.12)
R E =R 1 +R 2 +R 3
b). parallel connection of elements.
Let's consider two parallel-connected elements R1 and R2. The voltages on these elements are equal, because they are connected to the same nodes a and b.
U R1 = U R2 = U AB
Applying Ohm's law we get
U R1 =I 1 R 1 ; U R2 =I 2 R 2
I 1 R 1 =I 2 R 2 or I 1 / I 2 =R 2 / R 1
Let's apply Kirchhoff's 1st law to node (a)
I – I 1 – I 2 =0 or I=I 1 +I 2
Let us express the currents I 1 and I 2 in terms of voltages and we get
I 1 = U R1 / R 1 ; I 2 = U R2 / R 2
I= U AB / R 1 + U AB / R 2 = U AB (1 / R 1 +1/R 2)
In accordance with Ohm's law, we have I=U AB / R E; where R E – equivalent resistance
Taking this into account, we can write
U AB / R E = U AB (1 / R 1 +1 / R 2),
1/R E =(1/R 1 +1/R 2)
Let us introduce the following notation: 1/R E = G E – equivalent conductivity
1/R 1 =G 1 – conductivity of the 1st element
1/R 2 =G 2 – conductivity of the 2nd element.
Let us write equation (6) in the form
G E =G 1 +G 2 (3.13)
From this expression it follows that the equivalent conductivity of parallel-connected elements is equal to the sum of the conductivities of these elements.
Based on (3.13), we obtain the equivalent resistance
R E = R 1 R 2 / (R 1 + R 2) (3.14)
V). Converting a resistance triangle into an equivalent star and the reverse conversion.
The connection of three elements of the chain R 1, R 2, R 3, which has the form of a three-ray star with a common point (node), is called a “star” connection, and the connection of these same elements, in which they form the sides of a closed triangle, is called a “triangle” connection.
Fig.3.14. Fig.3.15.
connection - star () connection - delta ()
The transformation of a resistance triangle into an equivalent star is carried out according to the following rule and relationships:
The resistance of the beam of an equivalent star is equal to the product of the resistances of the two adjacent sides of the triangle divided by the sum of all three resistances of the triangle.
The transformation of a resistance star into an equivalent triangle is carried out according to the following rule and relationships:
The resistance of the side of an equivalent triangle is equal to the sum of the resistances of the two adjacent rays of the star plus the product of these two resistances divided by the resistance of the third ray:
G). Converting a current source into an equivalent EMF source If the circuit has one or more current sources, then often for convenience of calculations it is necessary to replace the current sources with EMF sources
Let the current source have parameters I K and G HV.
Fig.3.16. Fig.3.17.
Then the parameters of the equivalent EMF source can be determined from the relations
E E =I K / G VN; R VN.E =1 / G VN (3.17)
When replacing an EMF source with an equivalent current source, the following relations must be used
I K E =E / R VN; G VN, E =1 / R VN (3.18)
Loop current method.
This method is used, as a rule, when calculating multi-circuit circuits, when the number of equations compiled according to Kirchhoff’s 1st and 2nd laws is six or more.
To calculate using the loop current method in a complex circuit diagram, internal loops are determined and numbered. In each of the circuits, the direction of the circuit current is arbitrarily selected, i.e. current that closes only in this circuit.
Then, for each circuit, an equation is drawn up according to Kirchhoff’s 2nd law. Moreover, if any resistance simultaneously belongs to two adjacent circuits, then the voltage on it is defined as the algebraic sum of the voltages created by each of the two circuit currents.
If the number of contours is n, then there will be n equations. By solving these equations (using the method of substitution or determinants), the loop currents are found. Then, using equations written according to Kirchhoff’s 1st law, the currents are found in each of the branches of the circuit.
Let's write down the contour equations for this circuit.
For 1st circuit:
I 1 R 1 +(I 1 +I 2)R 5 +(I I +I III)R 4 =E 1 -E 4
For 2nd circuit
(I I +I II)R 5 + I II R 2 +(I II -I III)R 6 =E 2
For 3rd circuit
(I I +I III)R 4 +(I III -I II)R 6 +I III R 3 =E 3 -E 4
Carrying out the transformations, we write the system of equations in the form
(R 1 +R 5 +R 4)I I +R 5 I II +R 4 I III =E 1 -E 4
R 5 I I +(R 2 +R 5 +R 6) I II -R 6 I III =E 2
R 4 I I -R 6 I II +(R 3 +R 4 +R 6) I III =E 3 -E 4
Deciding this system equations, we determine the unknowns I 1, I 2, I 3. Branch currents are determined using the equations
I 1 = I I ; I 2 = I II; I 3 = I III; I 4 = I I + I III; I 5 = I I + I II; I 6 = I II – I III
Overlay method.
This method is based on the superposition principle and is used for circuits with multiple power sources. According to this method, when calculating a circuit containing several emf sources. , in turn all emfs except one are set equal to zero. The currents in the circuit created by this one EMF are calculated. The calculation is made separately for each EMF contained in the circuit. The actual values of currents in individual branches of the circuit are determined as the algebraic sum of currents created by the independent action of individual emfs.
Fig.3.20. Fig.3.21.
In Fig. 3.19 is the original circuit, and in Fig. 3.20 and Fig. 3.21 the circuits are replaced with one source in each.
The currents I 1 ’, I 2 ’, I 3 ’ and I 1 ”, I 2 ”, I 3 ” are calculated.
The currents in the branches of the original circuit are determined using the formulas;
I 1 =I 1 ’ -I 1 ”; I 2 = I 2 “-I 2 ’; I 3 =I 3 ' +I 3 "
Nodal potential method
The method of nodal potentials allows you to reduce the number of jointly solved equations to Y – 1, where Y is the number of nodes of the equivalent circuit. The method is based on the application of Kirchhoff's first law and is as follows:
1. We take one node of the circuit diagram as the basic one with zero potential. This assumption does not change the values of currents in the branches, since - the current in each branch depends only on the potential differences of the nodes, and not on the actual potential values;
2. For the remaining Y - 1 nodes, we compose equations according to Kirchhoff’s first law, expressing the branch currents through the potentials of the nodes.
In this case, on the left side of the equations, the coefficient at the potential of the node under consideration is positive and equal to the sum of the conductivities of the branches converging to it.
The coefficients at the potentials of nodes connected by branches to the node under consideration are negative and equal to the conductivities of the corresponding branches. The right side of the equations contains the algebraic sum of the currents of the branches with current sources and the short-circuit currents of the branches with EMF sources converging to the node under consideration, and the terms are taken with a plus (minus) sign if the current of the current source and the EMF are directed towards the node in question (from the node).
3. By solving the compiled system of equations, we determine the potentials of the U-1 nodes relative to the base one, and then the currents of the branches according to the generalized Ohm’s law.
Let us consider the application of the method using the example of calculating a circuit according to Fig. 3.22.
To solve by the method of nodal potentials we take 
.
System of nodal equations: number of equations N = N y – N B -1,
where: N y = 4 – number of nodes,
N B = 1 – number of degenerate branches (branches with 1st source of emf),
those. for this chain: N = 4-1-1=2.
We compose equations according to Kirchhoff’s first law for (2) and (3) nodes;
I2 – I4 – I5 – J5=0; I4 + I6 –J3 =0;
Let us represent the currents of the branches according to Ohm’s law through the potentials of the nodes:
I2 = (φ2 − φ1) / R2 ; I4 = (φ2 +E4 − φ3) / R4
I5 = (φ2 − φ4) / R5 ; I6 = (φ3 – E6 − φ4) / R6;
Where, 
Substituting these expressions into the node current equations, we obtain a system;
Where 
,
By solving a system of equations using the numerical method of substitution or determinants, we find the values of the potentials of the nodes, and from them the values of voltages and currents in the branches.
Equivalent source method (active two-terminal network)
A two-terminal circuit is a circuit that is connected to the external part through two terminals - poles. There are active and passive two-terminal networks.
An active two-terminal network contains sources of electrical energy, while a passive one does not contain them. Legend two-terminal circuits in a rectangle with the letter A for active and P for passive (Fig. 3.23.)
To calculate circuits with two-terminal networks, the latter are represented by equivalent circuits. The equivalent circuit of a linear two-terminal network is determined by its current-voltage or external characteristic V (I). The current-voltage characteristic of a passive two-terminal network is straight. Therefore, its equivalent circuit is represented by a resistive element with resistance:
rin = U/I (3.19)
where: U is the voltage between the terminals, I is the current and rin is the input resistance.
The current-voltage characteristic of an active two-terminal network (Fig. 3.23, b) can be constructed from two points corresponding to idle modes, i.e. at r n = °°, U = U x, I = 0, and short circuit, i.e. i.e. when g n =0, U = 0, I =Iк. This characteristic and its equation have the form:
U = U x – g eq I = 0 (3.20)
g eq = U x / Ik (3.21)
where: g eq – equivalent or output resistance of a two-terminal network, coincident
are given with the same characteristic and equation of the electric energy source, represented by the equivalent circuits in Fig. 3.23. 
So, an active two-terminal network seems to be an equivalent source with EMF - Eek = U x and internal resistance - g eq = g out (Fig. 3.23, a) An example of an active two-terminal network.- galvanic cell. When the current changes within 0 If a receiver with a load resistance Mr is connected to an active two-terminal network, then its current is determined using the equivalent source method: I = E eq / (g n + g eq) = U x / (g n + g out) (3.21) As an example, consider calculating the current I in the circuit in Fig. 3.24, using the equivalent source method. To calculate the open-circuit voltage U x between terminals a and b of the active two-terminal network, we open the branch with the resistive element g n (Fig. 3.24, b). Using the superposition method and taking into account the symmetry of the circuit, we find: U x =J g / 2 + E / 2 By replacing the sources of electrical energy (in this example, sources of emf and current) of an active two-terminal network with resistive elements with resistances equal to the internal resistances of the corresponding sources (in this example, zero resistance for the emf source and infinitely large resistance for the current source), we obtain the output resistance (resistance measured at the terminals a and b) g out = g/2 (Fig. 3.24, c). According to (3.21), the desired current is: I = (J r / 2 + E / 2) / (r n + r / 2). Determining the conditions for transmitting maximum energy to the receiver In communication devices, electronics, automation, etc., it is often desirable to transfer the greatest energy from the source to the receiver (actuator), and the transmission efficiency is of secondary importance due to the smallness of the energy. Let's consider the general case of powering the receiver from an active two-terminal network, in Fig. 3.25 the latter is represented by an equivalent source with EMF E eq and internal resistance g eq. Let's determine the power Рн, PE and the efficiency of energy transmission: Рн = U n I = (E eq – g eq I) I ; PE = E eq I = (g n – g eq I) I 2 η= Рн / PE 100% = (1 – g eq I / E eq) 100% With two limiting resistance values r n = 0 and r n = °°, the power of the receiver is zero, since in the first case the voltage between the terminals of the receiver is zero, and in the second case the current in the circuit is zero. Consequently, some specific value r corresponds to the highest possible (given e eq and g ek) value of the receiver power. To determine this resistance value, we equate to zero the first derivative of the power pn with respect to gn and get: (g eq – g n) 2 – 2 g n g eq -2 g n 2 = 0 whence it follows that, provided g n = g eq (3.21) The receiver power will be maximum: Рн max = g n (E 2 eq / 2 g n) 2 = E 2 eq / 4 g n I (3.22) Equality (1.38) is called the condition for maximum receiver power, i.e. transfer of maximum energy. In Fig. Figure 3.26 shows the dependences of Рн, PE, U n and η on current I. TOPIC 4: LINEAR AC ELECTRICAL CIRCUITS An electric current that periodically changes in direction and amplitude is called a variable. Moreover, if the alternating current changes according to a sinusoidal law, it is called sinusoidal, and if not, it is called non-sinusoidal. An electrical circuit with such a current is called an alternating (sinusoidal or non-sinusoidal) current circuit. AC electrical devices are widely used in various areas of the national economy, in the generation, transmission and transformation of electrical energy, in electric drives, household appliances, industrial electronics, radio engineering, etc. The predominant distribution of electrical devices of alternating sinusoidal current is due to a number of reasons. Modern energy is based on the transfer of energy over long distances using electric current. A prerequisite for such transmission is the possibility of simple current conversion with low energy losses. Such a transformation is feasible only in alternating current electrical devices - transformers. Due to the enormous advantages of transformation, modern electric power industry primarily uses sinusoidal current. A great incentive for the design and development of electrical devices with sinusoidal current is the possibility of obtaining high-power electrical energy sources. Modern turbogenerators of thermal power plants have a power of 100-1500 MW per unit, and generators of hydroelectric power stations also have greater power. The simplest and cheapest electric motors include asynchronous sinusoidal alternating current motors, which have no moving electrical contacts. For electric power plants (in particular, for all power plants) in Russia and in most countries of the world, the standard frequency is 50 Hz (in the USA - 60 Hz). The reason for this choice is simple: lowering the frequency is unacceptable, since already at a current frequency of 40 Hz incandescent lamps blink noticeably to the eye; An increase in frequency is undesirable, since the induced emf increases in proportion to the frequency, which negatively affects the transmission of energy through wires and the operation of many electrical devices. These considerations, however, do not limit the use of alternating current of other frequencies to solve various technical and scientific problems. For example, the frequency of alternating sinusoidal current in electric furnaces for smelting refractory metals is up to 500 Hz. In radio electronics, high-frequency (megahertz) devices are used, so at such frequencies the radiation of electromagnetic waves increases. Depending on the number of phases, AC electrical circuits are divided into single-phase and three-phase. The solution to any problem of calculating an electrical circuit should begin with the choice of the method by which the calculations will be made. As a rule, one and the same problem can be solved by several methods. The result will be the same in any case, but the complexity of the calculations may differ significantly. To correctly select a calculation method, you must first determine which class this electrical circuit belongs to: simple electrical circuits or complex ones. TO simple include electrical circuits that contain either one source of electrical energy or several located in the same branch of the electrical circuit. Below are two diagrams of simple electrical circuits. The first circuit contains one voltage source, in which case the electrical circuit clearly belongs to simple circuits. The second already contains two sources, but they are in the same branch, therefore it is also a simple electrical circuit. Simple electrical circuits are usually calculated in the following sequence: The described technique is applicable to the calculation of any simple electrical circuits; typical examples are given in example No. 4 and example No. 5. Sometimes calculations using this method can be quite voluminous and time-consuming. Therefore, after finding a solution, it would be useful to check the correctness of manual calculations using specialized programs or drawing up a power balance. The calculation of a simple electrical circuit in combination with drawing up a power balance is given in example No. 6. Complex electrical circuits TO complex electrical circuits include circuits containing several sources of electrical energy included in different branches. The figure below shows examples of such circuits. To fully calculate complex electrical circuits, the following methods are usually used: Features of the application of each method for calculating complex electrical circuits are described in more detail in the corresponding subsections.
For complex electrical circuits, the calculation method for simple electrical circuits is not applicable. Simplification of the circuits is impossible, because It is impossible to select in the diagram a section of a circuit with a serial or parallel connection of elements of the same type. Sometimes, transforming a circuit with its subsequent calculation is still possible, but this is rather an exception to the general rule.
