Electromagnetic waves. Radio communication

Electromagnetic waves. Radio communication

1. How will the frequency of natural electromagnetic oscillations contour (see figure),

if the key K is moved from position 1 to position 2?
1) will decrease by 2 times
2) will increase by 2 times
3) decrease by 3 times
4) increase by 3 times

Answer:

Solution:
The frequency of natural electromagnetic oscillations of the circuit is related to the capacitance of the capacitor and the inductance of the coil by the ratio

Therefore, if the key is moved from position 1 to position 2, thereby reducing the inductance of the circuit by 4 times, the natural frequency of the circuit will increase by 2 times.

2. Which type of electromagnetic radiation (among the listed) has the highest frequency?
1) visible light
2) infrared radiation
3) radio waves
4) X-ray radiation

Answer:

3. Infrared radiation is emitted...
1) electrons during their directed movement in a conductor
2) atomic nuclei during their transformations
3) any charged particles
4) any heated bodies

Answer:

Solution:
Infrared radiation is electromagnetic radiation occupying the spectral region between the red end of visible light and microwave radiation. Such radiation is emitted by any heated body.

4. charged particle does not radiate electromagnetic waves in vacuum at...
1) uniform rectilinear motion in an inertial frame of reference (ISO)
2) uniform motion around the circle in ISO
3) oscillatory motion in ISO
4) any movement with acceleration in ISO

Answer:

Solution:
According to Maxwell's theory, a charged particle radiates electromagnetic waves in vacuum during any accelerated motion in the IFR. Consequently, a charged particle does not radiate in a vacuum only with uniformly rectilinear motion in an inertial frame of reference.

5. The circuit of the radio receiver is tuned to a wavelength of 30 m. How should the inductance of the coil of the oscillatory circuit of the receiver be changed so that, with a constant electric capacitance of the capacitor in the circuit, it is tuned to a wave length of 15 m?
1) increase by 2 times
2) increase by 4 times
3) reduce by 2 times
4) reduce by 4 times

Answer:

Solution:


Thus, to reduce the wavelength by 2 times with the same electric capacitance of the capacitor, it is necessary to reduce the inductance of the coil by 4 times.

6. The circuit of the radio receiver is tuned to a wavelength of 15 m. How should the inductance of the coil of the oscillatory circuit of the receiver be changed so that, with the capacitance of the capacitor unchanged, it is tuned to a wave length of 30 m?
1) increase by 2 times
2) increase by 4 times
3) reduce by 2 times
4) reduce by 4 times

Answer:

Solution:
The wavelength to which the radio receiver circuit is tuned is related to the inductance of the coil and the electrical capacitance of the capacitor of the oscillatory circuit by the ratio

Thus, to increase the wavelength by 2 times with the same electric capacitance of the capacitor, it is necessary to increase the inductance of the coil by 4 times.

7. As the frequency increases alternating current 4 times the inductive reactance of the coil...
1) will not change
2) will increase by 4 times
3) will decrease by 2 times
4) decrease by 4 times

Answer:

Solution:
The inductive reactance of a coil is proportional to the cyclic frequency of the alternating current flowing through it. Therefore, an increase in the frequency of the alternating current by a factor of 4 will lead to an increase inductive reactance also 4 times.

8. Which of the following types of electromagnetic radiation has the longest wavelength?
1) radio waves
2) visible light
3) infrared radiation
4) X-ray radiation

Answer:

9. The radio receiver circuit is tuned to a wavelength of 30 m. How should the electric capacitance of the capacitor in the receiver circuit be changed so that, with the inductance of the oscillating circuit coil unchanged, it is tuned to a wave length of 15 m?
1) increase by 2 times
2) increase by 4 times
3) reduce by 2 times
4) reduce by 4 times

Answer:

Solution:
The wavelength to which the radio receiver circuit is tuned is related to the inductance of the coil and the electrical capacitance of the capacitor of the oscillatory circuit by the ratio

Thus, to reduce the wavelength by 2 times with a constant inductance of the coil, it is necessary to reduce the capacitance of the capacitor by 4 times.

10. According to the section of the circuit with resistance R flowing alternating current. How will the alternating current power change in this section of the circuit if the effective value of the current strength on it is doubled, and its resistance is reduced by a factor of 2?
1) will not change
2) will increase by 2 times
3) decrease by 3 times
4) will increase by 4 times

Answer:

Solution:
AC power in a circuit section with resistance is proportional to the product of the square of the effective current value and the resistance value. Therefore, an increase in the effective value of the current by 2 times and a decrease in resistance by 2 times will lead to an increase in the current power in this section of the circuit by 2 times.

11. In a certain area of ​​space bounded by planes AB and CD, a uniform magnetic field is created.

Metal square frame moves with constant speed directed along the frame plane and perpendicular to the field induction lines. Which of the graphs correctly shows the time dependence of the induction EMF in the frame, if at the initial time the frame begins to cross the plane MN(see figure), and at the moment of time touches the front side of the CD line?

Answer:

Solution:
Because by the time t0 the frame touches the line CD with its front side, it covers the distance l in the time t 0 /2. According to the law of electromagnetic induction, the module of the EMF of induction that occurs in the frame is directly proportional to the rate of change of the magnetic flux through the circuit:

Until the frame enters the magnetic field ( t), the magnetic flux through it does not change and is equal to zero, which means that the induction EMF is also equal to zero. In the time interval

when the frame enters the area magnetic field, the magnetic flux through the frame grows at a constant rate:

Therefore, the induction emf has a constant value. In the time interval

when the frame leaves the magnetic field region, the magnetic flux through the frame decreases at a constant rate:

Thus, during this period of time, the induction EMF is again constant, modulo has the same value, but the sign is opposite, since, according to Lenz's rule, the inductive current now has the opposite direction. Finally, after the rear side of the frame crosses the CD line, the magnetic flux through it again stops changing. The induction emf is zero. The correct dependence of the induction emf on time is shown in graph 4.

12. The figure shows a graph of the current strength versus time in an oscillatory circuit with a capacitor and a coil connected in series, the inductance of which is 0.2 H.

The maximum value of the energy of the electric field of the capacitor is ...

Answer:µJ.

Solution:
It can be seen from the graph that the amplitude of the current oscillation in the oscillatory circuit is equal to i 0=5 mA. For the oscillatory circuit, the law of conservation of energy is fulfilled: the sum of the energy of the electric field of the capacitor and the energy of the magnetic field of the coil is a constant value. The maximum value of the energy of the electric field of the capacitor coincides with the maximum energy of the magnetic field of the coil, which means that

Date of: ______________ year Signature: __________

Class: Grade 11

Item: physics

Topic: Test on the topic "Oscillations and waves"

The purpose of the lesson: test students' knowledge and skills on the topic Vibrations and waves »

Educational: foster a culture of physical labor; care when explaining new material.

Educational: test students' knowledge and skills

Developing: promote the development of mental activity.

Requirements for knowledge and skills:

Students should know :

- basic concepts, formulas, definitions, units of measurement

Students must be able to :

- apply formulas in solving problems, formulate definitions of a physical quantity

Lesson type:combined lesson

Software : textbook, workbook, blackboard, reference material and additional material offered by the teacher.

Plan:

IOrg. moment

II

III Test

IVReflection

VSummarizing

VIHomework

During the classes:

IOrg. moment

IIOral survey on the past topic(formulas, formulations, laws.)

III. The work is given in two versions with solutions.

For a differentiated approach to the control of knowledge, skills and abilities of students, tasks are systematized according to the level of complexity: each task contains three questions of the first, second and third levels, respectively, arising from one another. The grade for each task is determined by the achievement of the corresponding level by the students.

Evaluation criteria.

Grade "2" - solved 2 tasks of level "a" and less

Grade "3" - solved 2 tasks of level "b" and less

Grade "4" - solved 2 tasks of level "c" or less

Score "5" - solved 2 tasks of level "c" and more

1 option

1.

A)

b)

V)

2.

A)

b)

V)

3.

A)

b)

V)

Option 2

1.

A)

b)

V)

2-

A) What is the current strength in the circuit?

b)

V)

3.

A)

b)

V)

Answers and solution.


1 option

1. The oscillatory circuit of the radio receiver consists of a capacitor with a capacity of 1000 pF and a coil with an inductance of 50 μH.

A) What is the period of natural oscillations in the circuit?

b) What wavelength is this radio tuned to?

V) By how much and how should the capacitance of the capacitor be changed to tune the radio receiver to a wavelength of 300 m?

2. A coil with an inductance of 50 mH is included in the AC network with a voltage of 220 V.

A) What is the frequency of alternating current if the current in the circuit is 1.75 A? ( active resistance coils are ignored.)

b) Determine the capacitance of the capacitor that must be included in this circuit so that resonance occurs in the circuit.

V) Determine the resonant frequency in the circuit if the same capacitor is connected in series with the existing capacitor.

3. The primary winding of the step-down transformer contains 10,000 turns and is connected to an alternating current network with a voltage of 380 V.

A) What is the voltage in the secondary winding if it consists of 1000 turns?

b) The resistance of the secondary winding of the transformer is 1 ohm, the current strength in it is 3 A. What is the voltage on the load connected to the secondary winding of the transformer?

V) What is the efficiency of a transformer?

Option 2

1. Open oscillatory circuit emits radio waves with a wavelength of 300 m.

A) Determine the frequency of the emitted waves.

b) Determine the inductance of the circuit if its capacitance is 5000 pF.

V) By how much and how should the inductance of the circuit be changed so that radio waves of twice the wavelength are emitted?

2- A 4 μF capacitor is included in the AC network with a frequency of 50 Hz and a voltage of 220 V.

A) What is the current strength in the circuit?

b) Determine the inductance of the coil that must be included in this circuit so that resonance occurs in the circuit.

V) What will be equal to resonant frequency in a circuit if the same capacitor is connected in parallel with the existing capacitor?

3. The voltage on the primary winding of the transformer is 6 V, and on the secondary winding 120 V.

A) What is the current strength in the secondary winding if the current strength in the primary winding is 4 A?

b) Determine the voltage at the output of the transformer if its efficiency is 95%.

V) What is the resistance of the secondary winding of the transformer?

1 option

1. The oscillatory circuit of the radio receiver consists of a capacitor with a capacity of 1000 pF and a coil with an inductance of 50 μH.

A) What is the period of natural oscillations in the circuit?

b) What wavelength is this radio tuned to?

V) By how much and how should the capacitance of the capacitor be changed to tune the radio receiver to a wavelength of 300 m?

2. A coil with an inductance of 50 mH is included in the AC network with a voltage of 220 V.

A) What is the frequency of alternating current if the current in the circuit is 1.75 A? (Ignore the active resistance of the coil.)

b) Determine the capacitance of the capacitor that must be included in this circuit so that resonance occurs in the circuit.

V) Determine the resonant frequency in the circuit if the same capacitor is connected in series with the existing capacitor.

3. The primary winding of the step-down transformer contains 10,000 turns and is connected to an alternating current network with a voltage of 380 V.

A) What is the voltage in the secondary winding if it consists of 1000 turns?

b) The resistance of the secondary winding of the transformer is 1 ohm, the current strength in it is 3 A. What is the voltage on the load connected to the secondary winding of the transformer?

V) What is the efficiency of a transformer?

Option 2

1. An open oscillatory circuit emits radio waves with a wavelength of 300 m.

A) Determine the frequency of the emitted waves.

b) Determine the inductance of the circuit if its capacitance is 5000 pF.

V) By how much and how should the inductance of the circuit be changed so that radio waves of twice the wavelength are emitted?

2- A 4 μF capacitor is included in the AC network with a frequency of 50 Hz and a voltage of 220 V.

A) What is the current strength in the circuit?

b) Determine the inductance of the coil that must be included in this circuit so that resonance occurs in the circuit.

V) What will be the resonant frequency in the circuit if the same capacitor is connected in parallel with the existing capacitor?

3. The voltage on the primary winding of the transformer is 6 V, and on the secondary winding 120 V.

A) What is the current strength in the secondary winding if the current strength in the primary winding is 4 A?

b) Determine the voltage at the output of the transformer if its efficiency is 95%.

V) What is the resistance of the secondary winding of the transformer?

Setting up a transistor receiver, in principle, differs little from setting up a tube receiver. After making sure that the bass amplifier is corrected and the tubes or transistors of the receiver work in normal modes, start setting up the contours. The tuning starts with the detector stage, then goes to the IF amplifier, local oscillator and input circuits.

It is best to tune the contours with a generator high frequency. If it is not there, then you can tune by ear, according to the received radio stations. In this case, only an avometer of any type (TT-1, VK7-1) and another receiver may be required, the intermediate frequency of which is equal to the intermediate frequency of the tuned receiver, but sometimes they are tuned without any instruments. The autometer during adjustment serves as an indicator of the output signal.

When setting up the IF amplifier circuits in a tube receiver, when an RF generator and a tube voltmeter are used for this purpose, the latter must not be connected to the grid of the lamp, since the input capacitance of the voltmeter is added to the capacitance of the grid circuit. When adjusting the circuits, the voltmeter should be connected to the anode of the next lamp. In this case, the circuit in the anode circuit of this lamp must be shunted with a resistor with a resistance of about 500 - 1000 Ohms.

Having finished setting up the IF gain path, they begin to set up the local oscillator and the RF amplifier. If the receiver has several bands, then the tuning starts with the KB band, and then proceed to the tuning.

Contours SV and DV ranges. Short-wave coils (and sometimes medium-wave ones), unlike long-wave ones, usually do not have cores; they are wound most often on cylindrical (and sometimes on ribbed) frames. The change in the inductance of such coils is carried out when adjusting the circuits, shifting or pushing the turns of the coils.

In order to determine whether the coils should be shifted or moved apart in a given circuit, it is necessary to bring a piece of ferrite and a brass (or copper) rod into the coil or alternately bring it closer to it. It is even more convenient to perform this operation if, instead of a separate piece of ferrite and a brass rod, a special combined indicator stick is used, at one end of which magnetite (ferrite) is fixed, and at the other - a brass rod.

The inductance of the coil of the RF amplifier circuit should be increased if, at the junction points of the circuits, the volume of the signal at the output of the receiver increases when ferrite is introduced into the coil and decreases when a brass rod is inserted, and vice versa, the inductance should be reduced if the volume increases when the brass rod is inserted and decreases with the introduction of ferrite. If the circuit is configured correctly, the weakening of the signal volume at the junction points occurs with the introduction of both ferrite and brass rods.

The contours of the MW and LW ranges are adjusted in the same order. The change in the inductance of the loop coil at the junction points is carried out on these ranges by appropriately adjusting the ferrite core.

When making homemade contour coils, it is recommended to wind a few obviously extra turns. If, when tuning the circuits, it turns out that the inductance of the circuit coil is insufficient, winding the turns on the finished coil will be much more difficult than winding the extra turns during the tuning process itself.

To facilitate the adjustment of the contours and graduation of the scale, you can use the factory receiver. Comparing the angles of rotation of the axes of the variable capacitors of the tuned receiver and the factory one (if the blocks are the same) or the position of the scale indicators, it is determined in which direction the circuit setting should be shifted. If the station on the scale of the tuned receiver is closer to the top of the scale than the factory one, then the capacitance of the tuning capacitor of the local oscillator circuit should be reduced, and vice versa, if closer to the middle of the scale, increase.

Methods for testing the local oscillator in a tube receiver. You can check if the local oscillator is working in a tube receiver. different ways: using a voltmeter, optical tuning indicator, etc.

When using a voltmeter, it is connected in parallel with the resistor in the anode circuit of the local oscillator. If the short circuit of the capacitor plates in the local oscillator circuit causes an increase in the voltmeter readings, then the local oscillator is working. The voltmeter must have a resistance of at least 1000 ohm / V and be set to a measurement limit of 100 - 150 V.

Checking the performance of the local oscillator with an optical tuning indicator (6E5C lamp) is also simple. To do this, the control grid of the local oscillator lamp is connected with a short conductor to the grid of the 6E5C lamp through a resistor with a resistance of 0.5 - 2 MΩ. Dark sector of the tuning indicator when normal operation the local oscillator must be completely closed. By changing the dark sector of the 6E5C lamp when turning the receiver tuning knob, one can judge the change in the amplitude of the generator voltage in different parts of the range. If the amplitude unevenness is observed within a significant range, more uniform generation over the range can be achieved by selecting the number of turns of the coupling coil.

The operation of the local oscillator of the transistor receiver is checked by measuring the voltage at the load of the local oscillator (most often at the emitter of the transistor of the frequency converter or mixer). The local oscillator voltage, at which frequency conversion is most effective, lies in the range of 80 - 150 mV in all ranges. Measurement of the voltage at the load is carried out with a lamp voltmeter (VZ-2A, VZ-3, etc.). When the local oscillator circuit is closed, its oscillations break down, which can be noted by measuring the voltage at its load.

Sometimes self-excitation can be eliminated very simple ways. So, in order to eliminate self-excitation in the IF amplification stage, a resistor with a resistance of 100 - 150 Ohms can be included in the control grid circuit of the lamp of this stage. In this case, the intermediate frequency voltage gain in the cascade will decrease slightly, since only a small part of the input signal voltage is lost on the resistance.

In transistor receivers, self-excitation can be observed if the battery of cells or batteries is discharged. In this case, the battery should be replaced and the batteries put on charge.

In some cases, self-excitation in the receiver and TV can also be eliminated by measures such as grounding transfer individual elements circuits, alteration of installation, etc. It is often possible to evaluate the effectiveness of the measures taken to combat self-excitation in the following way.

Rice. 25. To an explanation of the method for eliminating self-excitation in transistor reflex receivers

The receiver or TV is connected to an adjustable power source (that is, to a source whose voltage supplied to the anode circuits can be varied over a wide range), and a tube voltmeter or other dial indicator is turned on at the receiver output. Since at the moment of self-excitation, the voltage at the output of the receiver changes dramatically, the deviation of the indicator arrow makes it easy to note this. The voltage taken from the source is controlled by a voltmeter.

If self-excitation occurs at rated voltage, then the supply voltage is reduced to a value at which generation stops. Then they take certain measures against self-excitation and increase the voltage until generation occurs, marking it on a voltmeter. In the case of successfully taken measures, the self-excitation threshold should increase significantly.

In transistor reflex receivers, self-excitation can occur due to the poor location of the high-frequency transformer (or choke) relative to the magnetic antenna. It is possible to eliminate such self-excitation by using a short-circuited coil of copper wire with a diameter of 0.6 - 1.0 mm (Fig. 25). The U-shaped bracket of the wire is threaded through the hole in the board, bent from below, twisted and soldered to the common wire of the receiver. The bracket can serve as an element for fastening the transformer. If the transformer winding is wound on ferrite ring uniformly, then the corresponding orientation of the short-circuited coil relative to other ferrite parts is not required.

Why does the receiver "howl" on the KB band. It can often be observed that a superheterodyne receiver, when receiving a broadcasting station at short waves, begins to “howl” with a slight detuning. However, if the receiver is more accurately tuned to the received station, then reception becomes normal again.

The reason for the "howl" when operating the receiver at short wavelengths is the acoustic coupling between the receiver's loudspeaker and the tuning capacitor bank.

This generation can be eliminated by improving the depreciation of the tuning unit, as well as by reducing various accessible ways acoustic feedback- changing the method of fastening the loudspeaker, etc.

Tuning the IF amplifier with another receiver. At the beginning of this section, a method for tuning a radio receiver using simple instruments was described. In the absence of such devices, the tuning of radio receivers is usually done by ear, without devices. However, it should immediately be said that this method does not provide sufficient tuning accuracy and can only be used as a last resort.

To tune the IF amplifier circuits, instead of the standard signal generator, you can use another receiver, the intermediate frequency of which is equal to the intermediate frequency of the tuned receiver. - For a tuned lamp receiver, the AGC wire from the diode to the control grids of adjustable lamps must be disconnected from the diode during tuning and connected to the chassis. If this is not done, then the AGC system will make it difficult to fine-tune the bandpass filters. In addition, when tuning the IF amplifier, it is necessary to disrupt the oscillations of the local oscillator by blocking its circuit with a capacitor with a capacity of 0.25 - 0.5 microfarads.

The auxiliary receiver used in this case does not need to be subjected to any significant alterations. To set up, you need only a few additional parts: a variable resistor (0.5 - 1 MΩ), two fixed capacitors and two or three fixed resistance resistors.

Setting up the amplifier circuits. IF receiver produced as follows. The auxiliary receiver is pre-tuned to one of the local stations operating in the long or medium wave band. Next, the common wires or chassis of both receivers are connected to each other, and the wire going in the tube receiver to the control grid of the lamp of the first IF amplification stage of the auxiliary receiver is disconnected and connected to the control grid of the lamp of the corresponding IF amplifier stage of the tuned receiver. In the case of tuning a transistor receiver, the IF signal through capacitors with a capacity of 500 - 1000 pF is fed in turn to the bases of the transistors of the corresponding stages of the IF amplifier.

Then both receivers are turned on again, however, in order to avoid interference during tuning, the low-frequency part of the auxiliary, as well as the local oscillator of the tuned receiver, should be turned off (in tube receivers, by removing the lamps of the bass amplifier and local oscillator, respectively).

When setting up the IF amplifier stages of a transistor receiver, its local oscillator should be turned off by installing a jumper in the local oscillator circuit.

After that, by applying an intermediate frequency signal from the auxiliary receiver to the input of the tunable IF amplifier and smoothly adjusting the setting of the IF circuits of the latter, one achieves the audibility of the station to which the auxiliary receiver is tuned. Then continue the setting - separately for each circuit (on maximum level signal), and the setting is best done according to pointer device connected to the output of the bass amplifier, or by an optical indicator (lamp 6E5C or similar).

Start tuning from the last IF circuit; the signal is fed to the base of the corresponding transistor or directly to the grid of the lamp, in the anode circuit of which the tuned circuit is included.

If the adjustment is carried out not by an optical indicator, but by the sound volume, then it is recommended to set the volume level to the minimum, since the human ear is more sensitive to changes in the volume level with weak sounds.

About tuning the receiver for radio stations. The tuning of a superheterodyne receiver - tube or transistor - for received stations without the use of an auxiliary receiver usually starts on the KB band. By adjusting the IF circuits for maximum noise and rotating the tuning knob, the receiver is set to any of the audible stations. If it is possible to receive such a station, then they immediately begin to adjust the IF circuits, achieving maximum audibility (the tuning starts from the last IF circuit). Then the heterodyne and input circuits are tuned, first on short, then on medium and long waves. It should be noted that tuning receivers using this method is complex, time-consuming and requires experience and skills.

Lamp 6E5C - indicator when setting up. According to the sound volume, it is not recommended to adjust the receiver circuits, as already mentioned, especially if it is installed high level output volume. The sensitivity of the human ear to changes in signal level when loud sounds very low. Therefore, if you still have to tune the receiver by sound, then the regulator should be set low level volume, or, better, use an optical tuning indicator - a 6E5C lamp or another similar one.

Tuning superheterodyne receivers according to received stations and using a 6E5C lamp as an indicator of tuning accuracy, it is more convenient to adjust the circuits at an input signal level at which the dark sector of this lamp narrows to 1 - 2 mm.

To regulate the signal voltage at the receiver input, parallel to the antenna coil, for example, a variable resistance resistor can be connected, the value of which, depending on the sensitivity of the receiver, can be selected in the range from 2 to 10 kOhm.

How to detect a faulty stage in an RF amplifier. When adjusting or repairing a receiver, a cascade in which there is a malfunction can be detected using an antenna, connecting it in turn to the bases of transistors or to the grids of amplifier lamps and determining by ear whether there are malfunctions in these cascades.

This method is convenient to use in cases where there are several stages of RF amplification.

An antenna in the form of a piece of wire can also be used when checking the IF and RF amplification stages in TVs. Since shortwave stations often operate at frequencies close to the intermediate frequency of TVs, listening to these stations will indicate that the sound channel is working,

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