How does the current change in a parallel connection? Series and parallel connection of conductors

Electrical circuits, which we have to deal with in practice, usually consist of more than one receiver electric current, but from several different ones that can be connected to each other in different ways. Knowing the resistance of each and how they are connected, you can calculate the total resistance of the circuit.

Figure 78, a shows a circuit of a series connection of two electric lamps, and Figure 78, b - a diagram of such a connection. If you turn off one lamp, the circuit will open and the other lamp will go out.

Rice. 78. Sequential switching on of light bulbs and power sources

For example, a battery, a lamp, two ammeters and a key are connected in series in the circuit shown in Figure 62 (see § 38).

We already know that with a series connection, the current strength in any part of the circuit is the same, i.e.

What is the resistance of series-connected conductors?

By connecting conductors in series, we seem to increase the length of the conductor. Therefore, the resistance of the circuit becomes greater than the resistance of one conductor.

The total resistance of the circuit when connected in series is equal to the sum of the resistances of the individual conductors(or individual sections of the chain):

The voltage at the ends of individual sections of the circuit is calculated based on Ohm's law:

U 1 = IR 1, U 2 = IR 2.

From the above equalities it is clear that the voltage will be greater on the conductor with the greatest resistance, since the current strength is the same everywhere.

The total voltage in the circuit in a series connection, or the voltage at the poles of the current source, is equal to the sum of the voltages in individual sections of the circuit:

This equality follows from the law of conservation of energy. Electrical voltage on a section of a circuit is measured by the work of an electric current performed when passing through a section of the circuit electric charge in 1 Class. This work is done using energy electric field, and the energy spent on the entire section of the circuit is equal to the sum of the energies that are spent on the individual conductors that make up the section of this circuit.

All the above laws are valid for any number of series-connected conductors.

Example 1. Two conductors with resistance R 1 = 2 Ohms, R 2 = 3 Ohms are connected in series. The current in the circuit is I = 1 A. Determine the resistance of the circuit, the voltage on each conductor and the total voltage of the entire section of the circuit.

Let's write down the conditions of the problem and solve it.

Questions

1. Which connection of conductors is called serial? Draw it on the diagram.
2. What electrical quantity is the same for all conductors connected in series?
3. How to find the total resistance of a circuit, knowing the resistance of individual conductors, in a series connection?
4. How to find the voltage of a section of a circuit consisting of conductors connected in series, knowing the voltage on each?

Exercise

1. The circuit consists of two series-connected conductors whose resistance is 4 and 6 ohms. The current in the circuit is 0.2 A. Find the voltage on each conductor and the total voltage.
2. For electric trains, a voltage of 3000 V is used. How can lamps designed for a voltage of 50 V each be used to illuminate cars?
3. Two identical lamps, each rated at 220 V, are connected in series and connected to a network with a voltage of 220 V. What voltage will each lamp be under?
4. The electrical circuit consists of a current source - a battery that creates a voltage of 6 V in the circuit, a light bulb from a flashlight with a resistance of 13.5 Ohms, two spirals with a resistance of 3 and 2 Ohms, a key and connecting wires. All parts of the circuit are connected in series. Draw a diagram of the circuit. Determine the current strength in the circuit, the voltage at the ends of each of the current consumers.

1 What resistance R should be taken so that you can connect a lamp designed for voltage Vo = 120 V and current Iо = 4 A to a network with a voltage of V = 220 V?

2 Two arc lamps and resistance R are connected in series and connected to a network with voltage V=110V. Find the resistance R if each lamp is designed for voltage Vo = 40 V, and the current in the circuit is I = 12 A.

Resistance voltage

According to Ohm's law

3 To measure the voltage on a section of the circuit, two voltmeters are connected in series (Fig. 88). The first voltmeter gave a reading of V1 = 20 V, the second - V2 = 80 V. Find the resistance of the second voltmeter R2, if the resistance of the first voltmeter R1 = 5 kOhm.

The same current I flows through the voltmeters. Since the voltmeter shows the voltage across its own resistance, then

and the resistance of the second voltmeter

4 An iron wire rheostat, a milliammeter and a current source are connected in series. At temperature to = 0° C, the rheostat resistance is Ro = 200 Ohm. The resistance of the milliammeter is R = 20 Ohm, its reading is Iо = 30 mA. What current It will the milliammeter show if the rheostat is heated to a temperature of t = 50° C? Temperature coefficient of resistance of iron.

Serial and parallel connections of conductors. Additional resistances and shunts

5 A conductor with a resistance of R = 2000 Ohms consists of two parts connected in series: a carbon rod and a wire, both of which have temperature coefficients of resistance. What should the resistances of these parts be chosen so that the total resistance of the conductor R does not depend on temperature?

At temperature t, the total resistance of the series-connected parts of the conductor with resistances R1 and R2 will be

where R10 and R20 are the resistance of the carbon rod and wire at t0 = 0° C. The total resistance of the conductor does not depend on temperature if

In this case, at any temperature

From the last two equations we find

6 Create a wiring diagram for lighting a corridor with one light bulb, which allows you to turn the light on and off independently at any end of the corridor.

Wiring diagrams that allow you to turn on and off a light bulb at any end of the corridor are shown in Fig. 347. At the ends of the corridor, two switches P1 and P2 are installed, each of which has two positions. Depending on the location of the network terminals, option a) or b) may be more profitable in terms of saving wires.

7 V network with voltage V= 120 V included two light bulbs with the same resistance R = 200 Ohm. What current will flow through each light bulb when they are connected in parallel and in series?

I1 = V/R=0.6 A in parallel connection; I2=V/2R=0.3 A in series connection.

8 Rheostat with sliding contact, connected according to the circuit shown in Fig. 89, is a potentiometer (voltage divider). When the potentiometer slide is moved, the voltage Vx removed from it changes from zero to the voltage at the terminals of the current source V. Find the dependence of voltage Vx on the position of the slider. Construct a graph of this dependence for the case when the total resistance of the potentiometer Ro is many times less than the resistance of the voltmeter r.

Let the resistance of the potentiometer section ax be equal to rx for a given position of the engine (Fig. 89). Then the total resistance of this section and the voltmeter (they are connected in parallel) and the resistance of the rest of the potentiometer xb is Thus, the total resistance between points a and b will be

Current in the circuit I= V/R. Voltage in section ah

Since by condition R0<

those. voltage Vx is proportional to resistance rx. In turn, the resistance rx is proportional to the length of the section ax.

In Fig. 348, the solid line shows the dependence of Vx on rx, the dashed line shows the dependence of Vx on rx, when R0~r, that is, when in the expression for Vx the first term in the denominator cannot be neglected. This dependence is not linear, however, in this case, Vx varies from zero to the voltage at the terminals of the source V.

9 Find the resistance R of a bimetallic (iron-copper) wire of length l=100m. The diameter of the internal (iron) part of the wire is d=2 mm, the total diameter of the wire is D = 5 mm. Resistivities of iron and copper. For comparison, find the resistance of iron and copper wires Yazh and Rm of diameter D and length l.

Cross-sectional area of ​​the iron and copper parts of the wire

(Fig. 349). Their resistance

The resistance R of a bimetallic wire is found using the formula for parallel connection of conductors:

Resistance of iron and copper wires of diameter D and length l

10 Find the total resistance of the conductors connected to the circuit according to the diagram shown in Fig. 90, if resistance R1= = R2 = R5 = R6 = 1 Ohm, R3 = 10 Ohm, R4 = 8 Ohm.

11 The total resistance of two series-connected conductors is R = 5 Ohm, and of parallel connected conductors Ro = 1.2 Ohm. Find the resistance of each conductor.

When two conductors with resistances R1 and R2 are connected in series, their total resistance is

and in parallel connection

According to the well-known property of the reduced quadratic equation (Vieta’s theorem), the sum of the roots of this equation is equal to its second coefficient with the opposite sign, and the product of the roots is the free term, i.e. R1 and R2 must be the roots of the quadratic equation

Substituting the values ​​of Ro and R, we find R1 = 3 Ohm and R2 = 2 0m (or R1 = 2 Ohm and R2 = 3 Ohm).

12 The wires supplying current are connected to the wire ring at two points. In what ratio do the connection points divide the circumference of the ring if the total resistance of the resulting circuit is n = 4.5 times less than the resistance of the wire from which the ring is made?

The connection points of the supply wires divide the circumference of the ring in a ratio of 1:2, i.e., they are spaced 120 degrees apart along an arc.

13 In the circuit shown in Fig. 91, the ammeter shows the current I = 0.04 A, and the voltmeter shows the voltage V = 20 V. Find the resistance of the voltmeter R2 if the resistance of the conductor R1 = 1 kOhm.

14 Find the resistance R1 of the light bulb using the readings of a voltmeter (V=50 V) and an ammeter (I=0.5 A), connected according to the circuit shown in Fig. 92 if the voltmeter resistance R2 = 40 kOhm.

The current in the common circuit is I=I1+I2, where I1 and I2 are the currents flowing through the light bulb and the voltmeter. Because

Neglecting the current I2 = 1.25 mA compared to I = 0.5 A, we obtain from the approximate formula

the same light bulb resistance value: R1 = 100 Ohm.

15 Find the resistance of conductor R1 using the readings of an ammeter (I=5 A) and a voltmeter (V=100V), connected according to the circuit shown in Fig. 93 if the voltmeter resistance R2 = 2.5 kOhm. What will be the error in determining R1 if, assuming that , in the calculations we neglect the current flowing through the voltmeter?

Voltmeter reading

where I1 and I2 are the currents flowing through the resistance and the voltmeter. Total current

If we neglect the current I2 compared to I, then the required resistance

The error in determining R`1 will be

Considering that

let's find the relative error:

16 Two conductors with equal resistances R are connected in series to a current source with voltage V. What will be the difference in the readings of voltmeters with resistances R and 10R if they are connected alternately to the ends of one of the conductors?

Voltmeters with resistances R and 10R show voltages

therefore the difference in voltmeter readings

17 Two light bulbs are connected to a current source with a voltage of V= 12 V (Fig. 94). The resistance of the circuit sections is r1 = r2 = r3 = r4 = r = 1.5 Ohm. Bulb resistance R1 = R2 = R = 36 Ohm. Find the voltage on each light bulb.

18 In the diagram shown in Fig. 95, current source voltage V=200 V, and conductor resistance R1=60 Ohm, R2 = R3 = 30 Ohm. Find the voltage across resistance R1.

19 The electrical circuit consists of a current source with a voltage of V = 180V and a potentiometer with an impedance of R = 5 kOhm. Find the readings of the voltmeters connected to the potentiometer according to the circuit shown in Fig. 96. Voltmeter resistances R1 = 6 kOhm and R2 = 4 kOhm. The x slider is in the middle of the potentiometer.

20 Three resistors are connected according to the circuit shown in Fig. 97. If resistors are included in the circuit at points a and b, then the circuit resistance will be R = 20 Ohms, and if at points a and c, then the circuit resistance will be Ro = 15 Ohms. Find the resistance of resistors R1, R2, R3, if R1=2R2.

Equivalent switching circuits are shown in Fig. 350. Rheostat resistances

21 Into how many equal parts should a conductor having a resistance R = 36 Ohm be cut, the resistance of its parts connected in parallel was Ro - 1 Ohm?

The entire conductor has a resistance R = nr, where r is the resistance of each of n equal parts of the conductor. When n identical conductors are connected in parallel, their total resistance is R0 = r/n. Excluding r, we get

n can only be a positive integer greater than one. Therefore, solutions are possible only in cases where R/Ro = 4, 9, 16, 25, 36,... In our case

22 A cube-shaped frame is made of wire (Fig. 98), each edge of which has a resistance r. Find the resistance R of this frame if the current I in the common circuit goes from vertex A to vertex B.

In sections Aa and bB (Fig. 351), due to the equality of the resistances of the cube edges and their identical inclusion, the current I evenly branches into three branches and therefore is equal to I/3 in each of them. In sections ab, the current is equal to I/6, since at each point a the current again branches along two edges with equal resistances and all these edges are turned on equally.

The voltage between points A and B is the sum of the voltage in section Aa, the voltage in section ab and the voltage in section bB:

23 From a wire whose unit length has a resistance Rl, a frame is made in the shape of a circle of radius r, intersected by two mutually perpendicular diameters (Fig. 99). Find the resistance Rx of the frame if the current source is connected to points c and d.

If the current source is connected to points c and d, then the voltages in sections da and ab are equal, since the wire

homogeneous. Therefore, the potential difference between points a and b is zero. There is no current in this area. Therefore, the presence or absence of contact at the intersection point of conductors ab and cd is indifferent. Resistance Rx is thus the resistance of three conductors connected in parallel: cd with resistance 2rR1, cad and cbd with equal resistances prR1. From the relation

24 A wire of length L = 1 m is woven from three cores, each of which is a piece of bare wire with a resistance per unit length Rl = 0.02 Ohm/m. A voltage V = 0.01 V is created at the ends of the wire. By what value DI will the current in this wire change if a piece of length l = 20 cm is removed from one core?

25 The current source is initially connected to two adjacent vertices of a wire frame in the shape of a regular convex n-gon. Then the current source is connected to the vertices located one after the other. In this case, the current decreases by 1.5 times. Find the number of sides of an n-gon.

26 How should four conductors with resistances R1 = 10m, R2 = 2 0m, R3 = 3 ohms and R4 = 4 0m be connected to obtain a resistance R = 2.5 ohms?

Resistance R = 2.5 Ohm is achieved when the conductors are connected according to the sour cream connection circuit (Fig. 352).

27 Find the conductivity k of a circuit consisting of two consecutive groups of parallel-connected conductors. The conductivities of each conductor of the first and second groups are equal to k1 = 0.5 Sm and k2 = 0.25 Sm. The first group consists of four conductors, the second - of two.

28 The voltmeter is designed to measure voltages up to a maximum value of Vo = 30 V. In this case, a current I = 10 mA flows through the voltmeter. What additional resistance Rd needs to be connected to the voltmeter so that it can measure voltages up to V=150V?

To measure higher voltages with a voltmeter than those for which the scale is designed, it is necessary to connect an additional resistance Rd in series with the voltmeter (Fig. 353). The voltage across this resistance is Vd=V-Vo; therefore resistance Rd=(V-Vо)/I=12 kOhm.

29 The milliammeter needle deflects to the end of the scale if a current I = 0.01 A flows through the milliammeter. The resistance of the device is R = 5 0m. What additional resistance Rd must be connected to the device so that it can be used as a voltmeter with a voltage measurement limit of V = 300 V?

To measure voltages not exceeding V with the device, it is necessary to connect in series with it such an additional resistance Rd such that V = I(R + Rd), where I is the maximum current through the device; hence Rd = V/I-R30 kOhm.

30 A voltmeter connected in series with resistance R1 = 10 kOhm, when connected to a network with voltage V = 220 V, shows voltage V1 = 70 V, and connected in series with resistance R2, shows voltage V2 = 20 V. Find resistance R2.

31 A voltmeter with a resistance of R = 3 kOhm, connected to the city lighting network, showed a voltage of V = 125V. When the voltmeter was connected to the network through resistance Ro, its reading decreased to Vo = 115 V. Find this resistance.

The city lighting network is a current source with an internal resistance much lower than the resistance of the voltmeter R. Therefore, the voltage V = 125 V, which the voltmeter showed when directly connected to the network, is equal to the voltage of the current source. This means that it does not change when the voltmeter is connected to the network through the resistance Ro. Therefore, V=I(R + Ro), where I=Vо/R is the current flowing through the voltmeter; hence Ro = (V-Vо)R/Vо = 261 Ohm.

32 A voltmeter with a resistance R = 50 kOhm, connected to a current source together with an additional resistance Rd = 120 kOhm, shows a voltage Vo = 100 V. Find the voltage V of the current source.

The current flowing through the voltmeter and additional resistance is I=Vо/R. Current source voltage V=I(R+Rd)= (R+Rd)Vо/R = 340 V.

33 Find the reading of a voltmeter V with resistance R in the circuit shown in Fig. 100. The current before the branching is equal to I, the resistances of the conductors R1 and R2 are known.

34 There is a device with a division value i0=1 µA/division and the number of scale divisions N= 100. The resistance of the device is R = 50 Ohm. How can this device be adapted to measure currents up to a value of I = 10 mA or voltages up to a value of V = 1 V?

To measure higher currents than those for which the scale is designed, a shunt with resistance is connected in parallel with the device

to measure voltages, an additional resistance is switched on in series with the device - the current flowing through the device at the maximum deflection of the needle,

The voltage at its terminals in this case.

35 A milliammeter with a current measurement limit of I0 = 25 mA must be used as an ammeter with a current measurement limit of I = 5 A. What resistance Rsh should the shunt have? How many times does the sensitivity of the device decrease? Device resistance R=10 Ohm.

When a shunt is connected in parallel to the device (Fig. 354), the current I must be divided so that current Io flows through the milliammeter. In this case, current Ish flows through the shunt, i.e. I=Io + Ish. The voltages on the shunt and on the milliammeter are equal: IоR = IшRш; from here

Rш=IоR/(I-Iо)0.05 Ohm. The sensitivity of the device decreases, and the division price of the device increases by n=I/Iо=200 times.

36 An ammeter with a resistance R = 0.2 Ohm, short-circuited to a current source with a voltage V = 1.5 V, shows a current I = 5 A. What current I0 will the ammeter show if it is shunted with a resistance Rsh=0.1 Ohm?

37 When a galvanometer is shunted with resistances R1, R2 and R3, 90%, 99% and 99.9% of the current I of the common circuit is branched into them. Find these resistances if the galvanometer resistance R = 27 Ohms.

Since the shunts are connected to the galvanometer in parallel, the condition for equality of voltages on the galvanometer and on the shunts gives

38 A milliammeter with a number of scale divisions N=50 has a division value i0 = 0.5 mA/div and a resistance R = 200 Ohm. How can this device be adapted to measure currents up to a value of I = 1 A?

The greatest current flowing through the device is Iо = ioN. To measure currents significantly exceeding the current Iо, it is necessary to connect a shunt in parallel to the device, the resistance of which Rsh is significantly less than the resistance of the milliammeter R:

39 A shunt with resistance Rsh = 11.1 mOhm is connected to an ammeter with a resistance R = 0.1 Ohm. Find the current flowing through the ammeter if the current in the common circuit is I=27 A.

The current flowing through the shunt is Ish = I-Io. The voltage drops across the shunt and ammeter are equal: IшRш = IоR; hence Iо=IRsh/(R+Rsh) =2.7 A.

Content:

The flow of current in an electrical circuit is carried out through conductors, in the direction from the source to the consumers. Most of these circuits use copper wires and electrical receivers in a given quantity, having different resistances. Depending on the tasks performed, electrical circuits use serial and parallel connections of conductors. In some cases, both types of connections can be used, then this option will be called mixed. Each circuit has its own characteristics and differences, so they must be taken into account in advance when designing circuits, repairing and servicing electrical equipment.

Series connection of conductors

In electrical engineering, the series and parallel connection of conductors in an electrical circuit is of great importance. Among them, a series connection scheme of conductors is often used, which assumes the same connection of consumers. In this case, inclusion in the circuit is performed one after another in order of priority. That is, the beginning of one consumer is connected to the end of another using wires, without any branches.

The properties of such an electrical circuit can be considered using the example of sections of a circuit with two loads. The current, voltage and resistance on each of them should be designated respectively as I1, U1, R1 and I2, U2, R2. As a result, relations were obtained that express the relationship between quantities as follows: I = I1 = I2, U = U1 + U2, R = R1 + R2. The data obtained are confirmed in practice by taking measurements with an ammeter and a voltmeter of the corresponding sections.

Thus, the series connection of conductors has the following individual features:

• The current strength in all parts of the circuit will be the same.
• The total voltage of the circuit is the sum of the voltages in each section.
• The total resistance includes the resistance of each individual conductor.

These ratios are suitable for any number of conductors connected in series. The total resistance value is always higher than the resistance of any individual conductor. This is due to an increase in their total length when connected in series, which also leads to an increase in resistance.

If you connect identical elements in series n, you get R = n x R1, where R is the total resistance, R1 is the resistance of one element, and n is the number of elements. Voltage U, on the contrary, is divided into equal parts, each of which is n times less than the total value. For example, if 10 lamps of the same power are connected in series to a network with a voltage of 220 volts, then the voltage in any of them will be: U1 = U/10 = 22 volts.

Conductors connected in series have a characteristic distinctive feature. If at least one of them fails during operation, the current flow stops in the entire circuit. The most striking example is when one burnt-out light bulb in a series circuit leads to failure of the entire system. To identify a burnt out light bulb, you will need to check the entire garland.

Parallel connection of conductors

In electrical networks, conductors can be connected in various ways: in series, in parallel and in combination. Among them, a parallel connection is an option when the conductors at the starting and ending points are connected to each other. Thus, the beginnings and ends of the loads are connected together, and the loads themselves are located parallel to each other. An electrical circuit may contain two, three or more conductors connected in parallel.

If we consider a series and parallel connection, the current strength in the latter option can be investigated using the following circuit. Take two incandescent lamps that have the same resistance and are connected in parallel. For control, each light bulb is connected to its own. In addition, another ammeter is used to monitor the total current in the circuit. The test circuit is supplemented by a power source and a key.

After closing the key, you need to monitor the readings of the measuring instruments. The ammeter on lamp No. 1 will show the current I1, and on lamp No. 2 the current I2. The general ammeter shows the current value equal to the sum of the currents of individual, parallel-connected circuits: I = I1 + I2. Unlike a series connection, if one of the bulbs burns out, the other will function normally. Therefore, parallel connection of devices is used in home electrical networks.

Using the same circuit, you can set the value of the equivalent resistance. For this purpose, a voltmeter is added to the electrical circuit. This allows you to measure the voltage in a parallel connection, while the current remains the same. There are also crossing points for the conductors connecting both lamps.

As a result of measurements, the total voltage for a parallel connection will be: U = U1 = U2. After this, you can calculate the equivalent resistance, which conditionally replaces all the elements in a given circuit. With a parallel connection, in accordance with Ohm's law I = U/R, the following formula is obtained: U/R = U1/R1 + U2/R2, in which R is the equivalent resistance, R1 and R2 are the resistances of both bulbs, U = U1 = U2 is the voltage value shown by the voltmeter.

One should also take into account the fact that the currents in each circuit add up to the total current strength of the entire circuit. In its final form, the formula reflecting the equivalent resistance will look like this: 1/R = 1/R1 + 1/R2. As the number of elements in such chains increases, the number of terms in the formula also increases. The difference in basic parameters distinguishes current sources from each other, allowing them to be used in various electrical circuits.

A parallel connection of conductors is characterized by a fairly low equivalent resistance value, so the current strength will be relatively high. This factor should be taken into account when a large number of electrical appliances are plugged into sockets. In this case, the current increases significantly, leading to overheating of cable lines and subsequent fires.

Laws of series and parallel connection of conductors

These laws concerning both types of conductor connections have been partially discussed earlier.

For a clearer understanding and perception in a practical sense, series and parallel connection of conductors, formulas should be considered in a certain sequence:

• A series connection assumes the same current in each conductor: I = I1 = I2.
• The parallel and series connection of conductors is explained in each case differently. For example, with a series connection, the voltages on all conductors will be equal to each other: U1 = IR1, U2 = IR2. In addition, with a series connection, the voltage is the sum of the voltages of each conductor: U = U1 + U2 = I(R1 + R2) = IR.
• The total resistance of a circuit in a series connection consists of the sum of the resistances of all individual conductors, regardless of their number.
• With a parallel connection, the voltage of the entire circuit is equal to the voltage on each of the conductors: U1 = U2 = U.
• The total current measured in the entire circuit is equal to the sum of the currents flowing through all conductors connected in parallel: I = I1 + I2.

In order to more effectively design electrical networks, you need to have a good knowledge of the series and parallel connection of conductors and its laws, finding the most rational practical application for them.

Mixed connection of conductors

Electrical networks typically use serial parallel and mixed connections of conductors designed for specific operating conditions. However, most often preference is given to the third option, which is a set of combinations consisting of various types of compounds.

In such mixed circuits, serial and parallel connections of conductors are actively used, the pros and cons of which must be taken into account when designing electrical networks. These connections consist not only of individual resistors, but also rather complex sections that include many elements.

The mixed connection is calculated according to the known properties of series and parallel connections. The calculation method consists of breaking the circuit down into simpler components, which are calculated separately and then summed up with each other.

A sequential connection is a connection of circuit elements in which the same current I occurs in all elements included in the circuit (Fig. 1.4).

Based on Kirchhoff’s second law (1.5), the total voltage U of the entire circuit is equal to the sum of the voltages in individual sections:

U = U 1 + U 2 + U 3 or IR eq = IR 1 + IR 2 + IR 3,

whence follows

R eq = R 1 + R 2 + R 3.

Thus, when connecting circuit elements in series, the total equivalent resistance of the circuit is equal to the arithmetic sum of the resistances of the individual sections. Consequently, a circuit with any number of series-connected resistances can be replaced by a simple circuit with one equivalent resistance R eq (Fig. 1.5). After this, the calculation of the circuit is reduced to determining the current I of the entire circuit according to Ohm’s law

and using the above formulas, calculate the voltage drop U 1 , U 2 , U 3 in the corresponding sections of the electrical circuit (Fig. 1.4).

The disadvantage of sequential connection of elements is that if at least one element fails, the operation of all other elements of the circuit stops.

Electric circuit with parallel connection of elements

A parallel connection is a connection in which all consumers of electrical energy included in the circuit are under the same voltage (Fig. 1.6).

In this case, they are connected to two circuit nodes a and b, and based on Kirchhoff’s first law, we can write that the total current I of the entire circuit is equal to the algebraic sum of the currents of the individual branches:

I = I 1 + I 2 + I 3, i.e.

whence it follows that

.

In the case when two resistances R 1 and R 2 are connected in parallel, they are replaced by one equivalent resistance

.

From relation (1.6), it follows that the equivalent conductivity of the circuit is equal to the arithmetic sum of the conductivities of the individual branches:

g eq = g 1 + g 2 + g 3.

As the number of parallel-connected consumers increases, the conductivity of the circuit g eq increases, and vice versa, the total resistance R eq decreases.

Voltages in an electrical circuit with resistances connected in parallel (Fig. 1.6)

U = IR eq = I 1 R 1 = I 2 R 2 = I 3 R 3.

It follows that

those. The current in the circuit is distributed between parallel branches in inverse proportion to their resistance.

According to a parallel-connected circuit, consumers of any power, designed for the same voltage, operate in nominal mode. Moreover, turning on or off one or more consumers does not affect the operation of the others. Therefore, this circuit is the main circuit for connecting consumers to a source of electrical energy.

Electric circuit with a mixed connection of elements

A mixed connection is a connection in which the circuit contains groups of parallel and series-connected resistances.

For the circuit shown in Fig. 1.7, the calculation of equivalent resistance begins from the end of the circuit. To simplify the calculations, we assume that all resistances in this circuit are the same: R 1 =R 2 =R 3 =R 4 =R 5 =R. Resistances R 4 and R 5 are connected in parallel, then the resistance of the circuit section cd is equal to:

.

In this case, the original circuit (Fig. 1.7) can be represented in the following form (Fig. 1.8):

In the diagram (Fig. 1.8), resistance R 3 and R cd are connected in series, and then the resistance of the circuit section ad is equal to:

.

Then the diagram (Fig. 1.8) can be presented in an abbreviated version (Fig. 1.9):

In the diagram (Fig. 1.9) the resistance R 2 and R ad are connected in parallel, then the resistance of the circuit section ab is equal to

.

The circuit (Fig. 1.9) can be represented in a simplified version (Fig. 1.10), where resistances R 1 and R ab are connected in series.

Then the equivalent resistance of the original circuit (Fig. 1.7) will be equal to:

Rice. 1.10

Rice. 1.11

As a result of the transformations, the original circuit (Fig. 1.7) is presented in the form of a circuit (Fig. 1.11) with one resistance R eq. Calculation of currents and voltages for all elements of the circuit can be made according to Ohm's and Kirchhoff's laws.

LINEAR CIRCUITS OF SINGLE-PHASE SINEUSOIDAL CURRENT.

Obtaining sinusoidal EMF. . Basic characteristics of sinusoidal current

The main advantage of sinusoidal currents is that they allow the most economical production, transmission, distribution and use of electrical energy. The feasibility of their use is due to the fact that the efficiency of generators, electric motors, transformers and power lines in this case is the highest.

To obtain sinusoidally varying currents in linear circuits, it is necessary that e. d.s. also changed according to a sinusoidal law. Let's consider the process of occurrence of sinusoidal EMF. The simplest sinusoidal EMF generator can be a rectangular coil (frame), uniformly rotating in a uniform magnetic field with angular velocity ω (Fig. 2.1, b).

Magnetic flux passing through the coil as the coil rotates abcd induces (induces) in it based on the law of electromagnetic induction EMF e . The load is connected to the generator using brushes 1 , pressed against two slip rings 2 , which in turn are connected to the coil. Coil induced value abcd e. d.s. at each moment of time is proportional to the magnetic induction IN, the size of the active part of the coil l = ab + dc and the normal component of the speed of its movement relative to the field vn:

e = Blvn (2.1)

Where IN And l- constant values, a vn- a variable depending on the angle α. Expressing the speed v n through the linear speed of the coil v, we get

e = Blv·sinα (2.2)

In expression (2.2) the product Blv= const. Therefore, e. d.s. induced in a coil rotating in a magnetic field is a sinusoidal function of the angle α .

If the angle α = π/2, then the product Blv in formula (2.2) there is a maximum (amplitude) value of the induced e. d.s. E m = Blv. Therefore, expression (2.2) can be written in the form

e = Emsinα (2.3)

Because α is the angle of rotation in time t, then, expressing it in terms of angular velocity ω , we can write α = ωt, and rewrite formula (2.3) in the form

e = Emsinωt (2.4)

Where e- instantaneous value e. d.s. in a reel; α = ωt- phase characterizing the value of e. d.s. at a given moment in time.

It should be noted that instant e. d.s. over an infinitesimal period of time can be considered a constant value, therefore for instantaneous values ​​of e. d.s. e, voltage And and currents i the laws of direct current are valid.

Sinusoidal quantities can be represented graphically by sinusoids and rotating vectors. When depicting them as sinusoids, instantaneous values ​​of quantities are plotted on the ordinate on a certain scale, and time is plotted on the abscissa. If a sinusoidal quantity is represented by rotating vectors, then the length of the vector on the scale reflects the amplitude of the sinusoid, the angle formed with the positive direction of the abscissa axis at the initial time is equal to the initial phase, and the rotation speed of the vector is equal to the angular frequency. Instantaneous values ​​of sinusoidal quantities are projections of the rotating vector onto the ordinate axis. It should be noted that the positive direction of rotation of the radius vector is considered to be the direction of rotation counterclockwise. In Fig. 2.2 graphs of instantaneous e values ​​are plotted. d.s. e And e".

If the number of pairs of magnet poles p ≠ 1, then in one revolution of the coil (see Fig. 2.1) occurs p full cycles of change e. d.s. If the angular frequency of the coil (rotor) n revolutions per minute, then the period will decrease by pn once. Then the frequency e. d.s., i.e. the number of periods per second,

f = Pn / 60

From Fig. 2.2 it is clear that ωТ = 2π, where

ω = 2π / T = 2πf (2.5)

Size ω , proportional to the frequency f and equal to the angular velocity of rotation of the radius vector, is called the angular frequency. Angular frequency is expressed in radians per second (rad/s) or 1/s.

Graphically depicted in Fig. 2.2 e. d.s. e And e" can be described by expressions

e = Emsinωt; e" = E"msin(ωt + ψe") .

Here ωt And ωt + ψe"- phases characterizing the values ​​of e. d.s. e And e" at a given point in time; ψ e"- the initial phase that determines the value of e. d.s. e" at t = 0. For e. d.s. e the initial phase is zero ( ψ e = 0 ). Corner ψ always counted from the zero value of the sinusoidal value when it passes from negative to positive values ​​to the origin (t = 0). In this case, the positive initial phase ψ (Fig. 2.2) are laid to the left of the origin (towards negative values ωt), and the negative phase - to the right.

If two or more sinusoidal quantities that change with the same frequency do not have the same sinusoidal origins in time, then they are shifted relative to each other in phase, i.e., they are out of phase.

Angle difference φ , equal to the difference in the initial phases, is called the phase shift angle. Phase shift between sinusoidal quantities of the same name, for example between two e. d.s. or two currents, denote α . The phase shift angle between the current and voltage sinusoids or their maximum vectors is denoted by the letter φ (Fig. 2.3).

When for sinusoidal quantities the phase difference is equal to ±π , then they are opposite in phase, but if the phase difference is equal ±π/2, then they are said to be in quadrature. If the initial phases are the same for sinusoidal quantities of the same frequency, this means that they are in phase.

Sinusoidal voltage and current, the graphs of which are presented in Fig. 2.3 are described as follows:

u = Umsin(ω t+ψ u) ; i = Imsin(ω t+ψ i) , (2.6)

and the phase angle between current and voltage (see Fig. 2.3) in this case φ = ψ u - ψ i.

Equations (2.6) can be written differently:

u = Umsin(ωt + ψi + φ) ; i = Imsin(ωt + ψu - φ) ,

because ψ u = ψ i + φ And ψ i = ψ u - φ .

From these expressions it follows that the voltage leads the current in phase by an angle φ (or the current is out of phase with the voltage by an angle φ ).

Forms of representation of sinusoidal electrical quantities.

Any sinusoidally varying electrical quantity (current, voltage, emf) can be presented in analytical, graphical and complex forms.

1). Analytical presentation form

I = I m sin( ω·t + ψ i), u = U m sin( ω·t + ψ u), e = E m sin( ω·t + ψ e),

Where I, u, e– instantaneous value of sinusoidal current, voltage, EMF, i.e. values ​​at the considered moment in time;

I m , U m , E m– amplitudes of sinusoidal current, voltage, EMF;

(ω·t + ψ ) – phase angle, phase; ω = 2·π/ T– angular frequency, characterizing the rate of phase change;

ψ i, ψ u, ψ e – the initial phases of current, voltage, EMF are counted from the point of transition of the sinusoidal function through zero to a positive value before the start of time counting ( t= 0). The initial phase can have both positive and negative meanings.

Graphs of instantaneous current and voltage values ​​are shown in Fig. 2.3

The initial phase of the voltage is shifted to the left from the origin and is positive ψ u > 0, the initial phase of the current is shifted to the right from the origin and is negative ψ i< 0. Алгебраическая величина, равная разности начальных фаз двух синусоид, называется сдвигом фаз φ . Phase shift between voltage and current

φ = ψ u – ψ i = ψ u – (- ψ i) = ψ u+ ψ i.

The use of an analytical form for calculating circuits is cumbersome and inconvenient.

In practice, one has to deal not with instantaneous values ​​of sinusoidal quantities, but with actual ones. All calculations are carried out for effective values, the passport data of various electrical devices indicates effective values ​​(current, voltage), most electrical measuring instruments show effective values. Effective current is the equivalent of direct current, which generates the same amount of heat in the resistor at the same time as alternating current. The effective value is related to the amplitude simple relation

2). Vector the form of representation of a sinusoidal electrical quantity is a vector rotating in a Cartesian coordinate system with a beginning at point 0, the length of which is equal to the amplitude of the sinusoidal quantity, the angle relative to the x axis is its initial phase, and the rotation frequency is ω = 2πf. The projection of a given vector onto the y-axis at any time determines the instantaneous value of the quantity under consideration.

Rice. 2.4

A set of vectors depicting sinusoidal functions is called a vector diagram, Fig. 2.4

3). Complex The presentation of sinusoidal electrical quantities combines the clarity of vector diagrams with accurate analytical calculations of circuits.

Rice. 2.5

We depict current and voltage as vectors on the complex plane, Fig. 2.5 The abscissa axis is called the axis of real numbers and is designated +1 , the ordinate axis is called the axis of imaginary numbers and is denoted +j. (In some textbooks, the real number axis is denoted Re, and the axis of imaginary ones is Im). Let's consider the vectors U And I at a point in time t= 0. Each of these vectors corresponds to a complex number, which can be represented in three forms:

A). Algebraic

U = U’+ jU"

I = I’ – jI",

Where U", U", I", I" – projections of vectors on the axes of real and imaginary numbers.

b). Indicative

Where U, I– modules (lengths) of vectors; e– the base of the natural logarithm; rotation factors, since multiplication by them corresponds to rotation of the vectors relative to the positive direction of the real axis by an angle equal to the initial phase.

V). Trigonometric

U = U·(cos ψ u+ j sin ψ u)

I = I·(cos ψ i – j sin ψ i).

When solving problems, they mainly use the algebraic form (for addition and subtraction operations) and the exponential form (for multiplication and division operations). The connection between them is established by Euler's formula

e jψ = cos ψ + j sin ψ .

Unbranched electrical circuits

Fundamentals > Problems and Answers > Direct Electric Current

Serial and parallel connections of current sources
Kirchhoff's rule

1 Find the potential difference between points a and b in the diagram shown in Fig. 118. E. d.s. current sources e 1 = 1 V and e 2 =1.3 V, resistor resistance R 1 = 10 ohms and R 2 = 5 ohms.
Solution:
Since e 2 > e 1 then the current I will flow in the direction shown in Fig. 118, while the potential difference between points a and b

2 Two elements with e. d.s. e 1 = 1.5 V and e 2 r1 = 0.6 Ohm and r 2 = 0.4 Ohm are connected according to the circuit shown in Fig. 119. What potential difference between points a and b will the voltmeter show if the resistance of the voltmeter is large compared to the internal resistances of the elements?

Solution:
Since e 2 > e 1 , then the current I will flow in the direction shown in Fig. 119. We neglect the current through the voltmeter due to
the fact that its resistance is high compared to the internal resistances of the elements. The voltage drop across the internal resistances of the elements must be equal to the difference e. d.s. elements, since they are included towards each other:
from here

Potential difference between points a and b (voltmeter reading)

3 Two elements with e. d.s. e 1 =1.4B and e 2 = 1.1 V and internal resistances r =0.3 Ohm and r 2 = 0.2 Ohm are closed by opposite poles (Fig. 120). Find the voltage at the terminals of the elements. Under what conditions is the potential difference between points a and b is equal to zero?

Solution:

4 Two current sources with the same e. d.s. e = 2 V and internal resistances r1 =0.4 Ohm and r 2 = 0.2 Ohm connected in series. At what external circuit resistance R will the voltage at the terminals of one of the sources be equal to zero?

Solution:
Circuit current

(Fig. 361). Voltages at the terminals of current sources

Solving the first two equations under the condition V1=0, we obtain

The condition V2=0 is not feasible, since the joint solution of the first and third equations leads to the value R<0.

5 Find internal resistance r1 the first element in the circuit shown in Fig. 121 if the voltage at its terminals is zero. Resistor values R 1 = ZOm, R 2 = 6 0m, internal resistance of the second element r 2 = 0.4 Ohm, e. d.s. elements are the same.

Solution:
Current in the common circuit



According to the conditions of the problem, the voltage at the terminals of the first element

from here

6 At what ratio between the resistances of resistors R 1 , R2, R3 and internal resistances of the elements r1, r2 (Fig. 122) voltagewill it be zero at the terminals of one of the elements? E.m.f. elements are the same.

Solution:

7 Two generators with the same e. d.s. e = 6 V and internal resistances r1 =0.5 Ohm and r2 = 0.38 Ohm are included according to the circuit shown in Fig. 123. Resistor resistances R 1 = 2 ohms, R2 = 4 ohms, R3 = 7 Ohm. Find voltage V 1 and V2 at the generator terminals.

Solution:
Current in the common circuit

where is the external resistance of the circuit

Voltage at the terminals of the first and second generator
voltage at the terminals of the second generator

8 Three elements with e. d.s. e 1 = 2.2 V, e 2 = 1.1 V and e 3 = 0.9 V and internal resistance r 1 = 0.2 Ohm, r 2 = 0.4 Ohm and r h = 0.5 Ohm are connected in series. External circuit resistance R= 1 Ohm. Find the voltage at the terminals of each element.

Solution:
According to Ohm's law for a complete circuit, the current

The voltage at the terminals of each element is equal to the difference e. d.s. and voltage drop across the internal resistance of the element:


The voltage at the terminals of the battery of cells is equal to the voltage drop across the external resistance of the circuit:

The voltage at the terminals of the third element turned out to be negative, since the current is determined by all the circuit resistances and the total emf, and the voltage drop across the internal resistance r3 is greater than the emf. e 3 .

9 A battery of four elements connected in series with e. d.s. e = 1.25 V and internal resistance r = 0.1 Ohm powers two parallel connected conductors with resistances R1 = 50 Ohm and R 2 = 200 Ohm. Find the voltage at the battery terminals.

Solution:

10 How many identical batteries with e. d.s. e = 1 .25V and internal resistance r = 0.004 Ohm must be taken to create a battery that would produce a voltage V= at the terminals 11 5 V at current I = 25 A?

Solution:
Battery terminal voltage

Hence,

11 Battery of n = 40 batteries connected in series with e. d.s. e = 2.5 V and internal resistance r = 0.2 Ohm is charged from a network with a voltage of V = 121 V. Find the charging current if a conductor with resistance is introduced in series into the circuit R = 2 Ohm.

Solution:

12 Two elements with e. d.s. e 1 = 1.25 V and e 2 = 1.5 V and identical internal resistances r = 0.4 Ohm connected in parallel (Fig. 124). Resistor value R = 10 Ohm. Find the currents flowing through the resistor and each element.

Solution:
The voltage drop across the resistor if currents flow in the directions shown in Fig. 124,

Considering that I=I1+I2, we find

Note that I1<0. Это значит, что направление тока противоположно указанному на рис. 124.
13 Two elements with e. d.s. e 1 =6 V and e 2 = 5 V and internal resistances r1 = 1 ohm and r2 = 20m connected according to the diagram shown in Fig. 125. Find the current flowing through a resistor with resistance R = 10 Ohm.

Solution:
By choosing the directions of the currents shown in Fig. 362, let's compose the Kirchhoff equations. For node b we have I1+I2-I=0; for abef circuit (clockwise circuit)

and for the bcde circuit (counterclockwise bypass)

From these equations we find

14 Three identical elements with e. d.s. e = 1.6 V and internal resistance r =0.8 Ohm are included in the circuit according to the diagram shown in Fig. 126. Milliammeter shows current I =100 mA. Resistor values R 1 = 10 Ohm and R2 = 15 0m, resistor resistance R unknown. What voltage V does the voltmeter show? The resistance of a voltmeter is very high, the resistance of a milliammeter is negligible.

Solution:
Internal element resistance

Resistance of parallel connected resistors

General e. d.s. elements e 0 =2 e According to Ohm's law for a complete circuit

15 Resistor values ​​R 1 and R 2 and e. d.s. e 1 and e 2 current sources in the circuit shown in Fig. 127 are known. At what e.m.f. e 3 the third source does not current flow through resistor R3?

Solution:
Let us select the directions of currents I1, I2 and I3 through resistors R1, R2 and R3, shown in Fig. 363. Then I3=I1+I2. The potential difference between points a and b will be equal to

If

Excluding I1 we find

16 A circuit of three identical elements connected in series with an emf. e and internal resistance r short-circuited (Fig. 128). Whichwill the voltage be shown by a voltmeter connected to the terminals of one of the elements?

Solution:
Let's consider the same circuit without a voltmeter (Fig. 364). From Ohm's law for a complete circuit we find

From Ohm's law for the section of the chain between points a and b we obtain

Connecting a voltmeter to points where the potential difference is zero cannot change anything in the circuit. Therefore, the voltmeter will show a voltage of zero.
17 Current source with emf. e 0 included in the circuit, the parameters of which are given in Fig. 129. Find the emf. e current source and direction of its connection to pins a and b , in which no current flows through the resistor with resistance R2.

Solution:
Let's connect the current source to terminals a and b and select the current directions shown in Fig. 365. For node e we have I=I0+I2. By traversing the contours aefb and ecdf clockwise we get
Using the condition I2 = 0, we find

The minus sign shows that the poles of the current source in Fig. 365 needs to be swapped.
18 Two elements with the same emf. e connected in series. External circuit resistance R = 5 Ohm. Ratio of voltage at the terminals of the first element to the voltage at the terminals of the second elementequals 2/3. Find internal resistance of elements r1 and r 2, if r 1=2 r 2.

Solution:

19 Two identical elements with emf. e = 1.5 V and internal resistance r = 0.2 Ohm shorted toresistor whose resistance is one case R1 = 0.2 Ohm, in another - R 2 = 20 Ohm. As needed connect the elements (series or parallel) in the first and second cases in order to obtain the maximum current in the circuit?

Solution:
When two elements are connected in parallel, the internal resistance and emf. are equal to r/2 and e when connected in series they are 2r and 2 e . Currents flow through resistor R
This shows that I2>I1 if R/2+r r. Therefore, the current is higher in series connection.
20 Two elements with emf. e 1 = 4V and e 2 = 2V and internal resistances r1 = 0.25 Ohm and r 2 = 0.75 ohms included in the circuit shown inrice. 130. Resistor resistances R1 = 1 Ohm and R2 = 3 Ohm, capacitance C = 2 µF.Find the charge on the capacitor.

Solution:

21 To a battery of two parallel-connected elements with e.m.f. e 1 and e 2 and internal resistances r1 and r 2 a resistor with resistance R is connected. Find the current I , flowing through resistor R, and currents I1 and I 2 in the first and second elements. At whatconditions, the currents in individual circuits can be equalzero or change its direction to the opposite?

Solution:
Let us choose the directions of currents shown in Fig. 366. For node b we have I-I1-I2=0. By traversing the abef and bcde contours clockwise we get

From these equations we find

Current I=0 when the polarity of one of the elements is changed and, in addition, the condition is met

Current I1=0 at

and current I2 = 0 at

Currents I1 and I2 have the directions shown in Fig. 366, if

They change their direction when

22 Battery of n identical batteries,connected in one case in series, in the other in parallel, is connected to a resistor with resistance R. Under what conditions is the current flowing throughwill the resistor be the same in both cases?

Solution:
When n(R-r) = R-r. If R=r, then the number of elements is arbitrary; if R№ r, the problem has no solution ( n =1).
23 Battery of n = 4 identical elements with internal resistance r =2 ohms connected in one casein series, in the other - in parallel, closes to a resistor with resistance R =10Ohm. How many times does the voltmeter reading in one case differ from the voltmeter reading in another case? The resistance of the voltmeter is high compared to R and r.

Solution:

where V1 is the voltmeter reading when the elements are connected in series, V2 is when the elements are connected in parallel.
24 How will the current flowing through a resistor with resistance R = 2 Ohm change if n =10 identical elements connected in series with this resistor, should they be connected in parallel with it? E.m.f. element e = 2 V, its internal resistance r = 0.2 Ohm.

Solution:

25 The battery is made up of N=600 identicalelements so that n groups are connected in seriesand each of them contains m elements connected in parallel. E.m.f. each element e = 2 V, its internal resistance r = 0.4 Ohm. At what values n and m battery, being shorted to externalresistance R = 0.6 Ohm, will be transferred to an external circuitmaximum power? Find the current flowingthrough resistance R.

Solution:
The total number of elements is N=nm (Fig. 367). External circuit current

where r/m - internal resistance of a group of t parallel-connected elements, and n r/ m - internal resistance n groups connected in series. The maximum power (see problem 848) is given to the external circuit when the resistance R is equal to the internal resistance of the battery of cells n r/ m, i.e.
In this case, points I = 46 A flow through resistance R.

26 Battery capacity=80 A H h. Find the battery capacity from n=3 such batteries connected in series and parallel.

Solution:
When connected in series, the same current flows through all the cells in a battery, so they will all discharge within the same amount of time. Therefore, the battery capacity will be equal to the capacity of each battery:
In parallel connection n batteries, 1/n part of the total current flows through each of them; therefore, with the same discharge current in the common circuit, the batteries will be discharged in n times longer than one battery, i.e. the battery capacity is n times greater than the capacity of a separate battery:

Note, however, that the energy

given by the battery to the circuit, both in series and in parallel connection n batteries in n times the energy supplied by one battery. This happens because when connected in series, e.g. d.s. batteries in n times more e. d.s. one battery, and with a parallel connection the emf. battery remains the same as for each battery, but Q increases by n times.
27 Find the battery capacity of the batteries connected according to the diagram shown in Fig. 131. Capacity of each battery Qo =64 A H h .

Solution:
Each group of five batteries connected in series has a capacity

Three groups connected in parallel give the total battery capacity

28 The bridge for measuring resistance is balanced so that no current flows through the galvanometer (Fig. 132). Current in the right branch I =0.2 A. Find the voltage V at the terminals of the current source. Resistor resistances R1 = 2 Ohm, R2 = 4 Ohm, R3 = 1 Ohm.

Solution:

29 Find the currents flowing in each branch of the circuit shown in Fig. 133. E.m.f. current sources e 1 = 6.5 V and e 2 = 3.9 V. Resistor resistances R1=R2=R3=R4=R5=R6=R=10 Ohm.

Solution:
We compose the Kirchhoff equations in accordance with the directions of the currents indicated in Fig. 133: I1 + I2 - I3 = 0 for node b;
I3 - I4 - I5 =0 for node h; I5 - I1 - I6 = 0 for node f: in this case

For the abfg circuit (clockwise traversal),

For circuit bcdh (counterclockwise bypass) and

for circuit hdef (clockwise bypass
arrow). Solving this system of equations, taking into account that all resistances are the same and equal to R = 10 Ohms, we obtain

Negative values ​​of currents I2, I4 and I6 show that for a given emf. sources and resistor resistances, these currents flow in directions opposite to those indicated in Fig. 133.