ELECTRICAL ENGINEERING, CIRCUITS AND MEASUREMENTS
Everything that will be presented in this lesson must not only be read and some key points remembered, but some definitions and formulations must be memorized by heart. It is from this lesson that elementary physical and electrical calculations will begin. Perhaps not everything will be clear at once, but do not despair; everything will fall into place over time. The main thing is to slowly absorb and memorize the material. Even if not everything is clear at first, try to at least remember the basic rules and the elementary formulas that will be discussed here. Having mastered this lesson well, you will then be able to perform more complex radio engineering calculations and solve the necessary problems. You cannot do without this in radio electronics. To emphasize the importance of this lesson, I will highlight all the formulations and definitions that must be memorized.
ELECTRIC CURRENT AND ITS MEASUREMENT
Until now, when characterizing the quantitative value of an electric current, I have sometimes used terminology such as, for example, "small current" or "large current". Initially, such an assessment of the current somehow suited us, but it is completely unsuitable for characterizing the current in terms of the work it can perform. When we talk about the work of a current, we mean that its energy is converted into some other type of energy: heat, light, chemical, or mechanical energy. The larger the flow of electrons, the more significant the current and its work. Sometimes people say "current strength" or simply "current". Thus, the word "current" has two meanings. It denotes the very phenomenon of the movement of electrical charges in a conductor, and also serves as an assessment of the quantity of electricity passing through the conductor.
The current (or current strength) is measured by the number of electrons passing through a conductor in 1 second.
This number is huge. Through the filament of a burning pocket flashlight bulb, for example, about 2,000,000,000,000,000,000 electrons pass every second. It is quite understandable that characterizing the current by the number of electrons is inconvenient, as one would have to deal with very large numbers.
The unit of electric current is the Ampere (abbreviated as A). It was named after the French physicist and mathematician André-Marie Ampère (1775 - 1836), who studied the laws of mechanical interaction of current-carrying conductors and other electrical phenomena.
A current of 1 A is a value at which 6,250,000,000,000,000,000 electrons pass through the cross-section of a conductor in 1 second.
In mathematical expressions, current is denoted by the Latin letter I or i. For example, they write: I = 2 A or 0.5 A. Along with the ampere, smaller units of current are used: the milliampere (written as mA), which is equal to 0.001 A, and the microampere (written as μA), which is equal to 0.000001 A, or 0.001 mA. Therefore, 1 A = 1000 mA or 1,000,000 μA.
Instruments used to measure currents are called ammeters, milliammeters, and microammeters, respectively. They are connected to the electrical circuit in series with the current consumer, i.e., in the break of the external circuit.
In diagrams, these devices are depicted as circles with letters inside them: A (ammeter), mA (milliammeter), and μA (microammeter), and next to them, PA is written, which means a current meter. The measuring device is designed for a current no greater than a certain limit for that device. The device must not be included in a circuit where a current exceeding this value flows, otherwise it may be damaged.
You may have a question: how to evaluate alternating current, the direction and magnitude of which are continuously changing?
Alternating current is usually evaluated by its effective (RMS) value. This is the value of current that corresponds to a direct current performing the same work. The effective value of an alternating current is approximately 0.7 of the amplitude, i.e., the maximum value.
ELECTRICAL RESISTANCE
When speaking of conductors, we mean substances, materials, and above all metals, that conduct current relatively well. However, not all substances called conductors conduct electric current equally well; that is, they, as they say, have unequal current conductivity. This is explained by the fact that during their movement, free electrons collide with the atoms and molecules of the substance, and in some substances the atoms and molecules hinder the movement of electrons more strongly, while in others - less. In other words, some substances offer greater resistance to the electric current, and others less. Of all the materials widely used in electrical and radio engineering, copper offers the least resistance to electric current. That is why electrical wires are most often made of copper. Silver has even less resistance, but it is a rather expensive metal. Iron, aluminum, and various metal alloys have greater resistance, i.e., poorer electrical conductivity.
The resistance of a conductor depends not only on the properties of its material but also on the size of the conductor itself. A thick conductor has less resistance than a thin one made of the same material; a short conductor has less resistance, a long one has more, just as a wide and short pipe presents less of an obstacle to the movement of water than a thin and long one. In addition, the resistance of a metal conductor depends on its temperature: the lower the temperature of the conductor, the lower its resistance.
The unit of electrical resistance is the Ohm (written as Ω) — named after the German physicist Georg Ohm.
A resistance of 1 Ohm is a relatively small electrical value. Such resistance to the current is offered, for example, by a piece of copper wire with a diameter of 0.15 mm and a length of 1 m. The resistance of the filament of a pocket electric flashlight bulb is about 10 Ohms, and the heating element of an electric stove is several tens of ohms. In radio engineering, one more often has to deal with resistances greater than an ohm or a few tens of ohms. The resistance of a high-impedance telephone, for example, is more than 2000 Ohms; the resistance of a semiconductor diode connected in the non-conducting direction is several hundred thousand ohms. Do you know what resistance your body offers to an electric current? From 1,000 to 20,000 Ohms. And the resistance of resistors — special components, which I will talk about more in this lesson, can be up to several million ohms and more. These parts, as you already know, are designated as rectangles in the diagrams.
In mathematical formulas, resistance is denoted by the Latin letter R. The same letter is placed next to the graphical symbols of resistors on diagrams.
To express the large resistances of resistors, larger units are used: the kilohm (abbreviated as kOhm), equal to 1,000 Ohms, and the megohm (abbreviated as MOhm), equal to 1,000,000 Ohms, or 1,000 kOhms.
The resistance of conductors, electrical circuits, resistors, or other components is measured with special devices called ohmmeters. In diagrams, an ohmmeter is denoted by a circle with the Greek letter Ω (omega) inside.
ELECTRICAL VOLTAGE
The unit of electrical voltage, electromotive force (EMF) is the Volt (in honor of the Italian physicist Alessandro Volta). In formulas, voltage is denoted by the Latin letter U, and the unit of voltage itself — the volt — by the letter V.
For example, they write: U = 4.5 V; U = 220 V. The unit volt characterizes the voltage at the ends of a conductor, a section of an electrical circuit, or the poles of a current source. A voltage of 1 V is such an electrical value that creates a current equal to 1 A in a conductor with a resistance of 1 Ohm. The 3R12 battery (3336L), intended for a flat pocket flashlight, as you already know, consists of three cells connected in series. On the battery label, you can read that its voltage is 4.5 V. This means that the voltage of each of the battery cells is 1.5 V. The voltage of a 9-volt battery (like PP3) is 9 V, and the voltage of the electrical lighting network can be 127 or 220 V.
Voltage is measured (with a voltmeter) by connecting the device with the corresponding terminals to the poles of the current source or in parallel to the circuit section, resistor, or other load on which the voltage acting on it must be measured. In diagrams, a voltmeter is denoted by the Latin letter V in a circle, and next to it - PU.
To evaluate the voltage, a larger unit is also used — the kilovolt (written as kV), corresponding to 1,000 V, as well as smaller units — the millivolt (written as mV), equal to 0.001 V, and the microvolt (written as μV), equal to 0.001 mV. These voltages are measured respectively by kilovoltmeters, millivoltmeters, and microvoltmeters. Such devices, like voltmeters, are connected in parallel to current sources or circuit sections where the voltage needs to be measured.
Let's now clarify the difference between the concepts of voltage and electromotive force (EMF). Electromotive force is the voltage acting between the poles of a current source as long as an external load circuit, such as an incandescent bulb or resistor, is not connected to it. As soon as the external circuit is connected and a current arises in it, the voltage between the poles of the current source will become lower. For example, a new, unused galvanic cell has an EMF of at least 1.5 V. When a load is connected to it, the voltage at its poles becomes equal to approximately 1.3 - 1.4 V. As the cell's energy is consumed to power the external circuit, its voltage gradually decreases. The cell is considered discharged and, therefore, unsuitable for further use when the voltage drops to 0.7 V, although if the external circuit is disconnected, its EMF will be greater than this voltage. And how is alternating voltage evaluated? When talking about alternating voltage, for example, the voltage of the electrical lighting network, they mean its effective value, which is approximately, like the effective value of alternating current, 0.7 of the amplitude voltage value.
OHM'S LAW
The figure shows a diagram of the simplest electrical circuit familiar to you. This closed circuit consists of three elements: a voltage source — battery GB, a current consumer — load R, which can be, for example, the filament of an electric lamp or a resistor, and conductors connecting the voltage source with the load. Incidentally, if this circuit is supplemented with a switch, you will get a complete diagram of a pocket flashlight.
The load R, having a certain resistance, is a section of the circuit. The value of the current in this section of the circuit depends on the voltage acting on it and its resistance: the greater the voltage and the lower the resistance, the greater the current that will flow through the section of the circuit.
This dependence of current on voltage and resistance is expressed by the following formula:
I = U / R
where I is the current expressed in amperes (A); U is the voltage in volts (V); R is the resistance in ohms (Ω).
This mathematical expression is read as follows: the current in a section of a circuit is directly proportional to the voltage across it and inversely proportional to its resistance. This is the fundamental law of electrical engineering, known as Ohm's Law, for a section of an electrical circuit.
Using Ohm's law, you can find out the unknown third electrical quantity from two known ones. Here are some examples of the practical application of Ohm's law.
First example: A voltage of 25 V acts on a section of a circuit with a resistance of 5 Ohms. We need to find out the value of the current in this section of the circuit.
Solution: I = U / R = 25 / 5 = 5 A.
Second example: A voltage of 12 V acts on a section of a circuit, creating a current of 20 mA in it. What is the resistance of this section of the circuit? First of all, the current of 20 mA must be expressed in amperes. This will be 0.02 A.
Then: R = 12 / 0.02 = 600 Ohms.
Third example: A current of 20 mA flows through a section of a circuit with a resistance of 10 kOhms. What is the voltage acting on this section of the circuit? Here, as in the previous example, the current must be expressed in amperes (20 mA = 0.02 A), and the resistance in ohms (10 kOhms = 10,000 Ohms).
Therefore: U = I × R = 0.02 × 10,000 = 200 V.
On the base of a flat flashlight bulb, it is stamped: 0.28 A and 3.5 V. What does this information tell us? It tells us that the bulb will glow normally at a current of 0.28 A, which is caused by a voltage of 3.5 V. Using Ohm's law, it is easy to calculate that the heated filament of the bulb has a resistance of R = 3.5 / 0.28 = 12.5 Ohms. This, I emphasize, is the resistance of the heated filament of the bulb. The resistance of a cold filament is significantly less.
Ohm's law is valid not only for a section but also for the entire electrical circuit. In this case, the total resistance of all elements of the circuit, including the internal resistance of the current source, is substituted into the value of R. However, in the simplest circuit calculations, the resistance of the connecting conductors and the internal resistance of the current source are usually neglected.
In connection with this, I will give one more example: The voltage of the electrical lighting network is 220 V. What current will flow in the circuit if the load resistance is 1000 Ohms?
Solution: I = U / R = 220 / 1000 = 0.22 A. An electric soldering iron consumes approximately this current.
All these formulas derived from Ohm's law can also be used to calculate alternating current (AC) circuits, provided there are no inductors or capacitors in the circuits.
Ohm's law and its derived calculation formulas are quite easy to remember if you use this graphical diagram, the so-called Ohm's law triangle:
Now let's consider this question: how does a resistor connected in series or in parallel with the load affect the current?
Let's analyze such an example. We have a bulb from a round electric flashlight, designed for a voltage of 2.5 V and a current of 0.075 A. Is it possible to power this bulb from a 3R12 battery, the initial voltage of which is 4.5 V? It is easy to calculate that the heated filament of this bulb has a resistance of slightly more than 30 Ohms. If we power it from a fresh 3R12 battery, then, according to Ohm's law, a current almost twice as large as the current it is designed for will flow through the filament of the bulb. The filament will not withstand such an overload; it will burn out and be destroyed. But this bulb can still be powered from a 3R12 battery if an additional resistor with a resistance of 25 Ohms is connected in series into the circuit, as shown in the figure.
In this case, the total resistance of the external circuit will be equal to approximately 55 Ohms, i.e., 30 Ohms (the resistance of the bulb filament H) plus 25 Ohms (the resistance of the additional resistor R). Consequently, a current of approximately 0.08 A will flow in the circuit, i.e., almost the same as the one the bulb filament is designed for. This bulb can be powered from a battery with an even higher voltage and even from the electrical lighting network if a resistor of the appropriate resistance is selected. In this example, the additional resistor limits the current in the circuit to the value we need. The greater its resistance, the less the current in the circuit will be.
In this case, two resistances were connected in series in the circuit: the resistance of the bulb filament and the resistance of the resistor. And with a series connection of resistances, the current is the same at all points of the circuit. You can connect an ammeter at any point in the circuit, and everywhere it will show the same value. This phenomenon can be compared to the flow of water in a river. The riverbed in different sections can be wide or narrow, deep or shallow. However, over a certain period of time, the same amount of water always passes through the cross-section of any part of the riverbed.
An additional resistor connected to the circuit in series with the load (like, for example, in the figure above) can be viewed as a resistor quenching a part of the voltage acting in the circuit. The voltage that is quenched by the additional resistor or, as they say, drops across it, will be greater the greater the resistance of this resistor. Knowing the current and resistance of the additional resistor, the voltage drop across it is easy to calculate using the same familiar formula U = I × R. Here U is the voltage drop in V; I is the current in the circuit in A; R is the resistance of the additional resistor in Ohms. Applied to our example, the resistor R (in the figure) quenched the excess voltage: U = I × R = 0.08 × 25 = 2 V. The remaining battery voltage, equal to approximately 2.5 V, dropped across the bulb filament.
The necessary resistance of the resistor can be found using another formula you are familiar with: R = U / I, where R is the required resistance of the additional resistor in Ohms; U is the voltage that needs to be quenched in V; I is the current in the circuit in A. For our example, the resistance of the additional resistor is: R = U / I = 2 / 0.075 ≈ 27 Ohms. By changing the resistance, you can decrease or increase the voltage that drops across the additional resistor, and thus regulate the current in the circuit. But the additional resistor R in such a circuit can be variable, i.e., a resistor whose resistance can be changed (see the figure below).
In this case, using the slider of the resistor, you can smoothly change the voltage supplied to the load H, and thus smoothly regulate the current flowing through this load. A variable resistor connected in this way is called a rheostat. Rheostats are used to regulate currents in receiver, TV, and amplifier circuits. In many cinemas, rheostats were used to smoothly dim the lights in the auditorium.
There is, however, another way to connect a load to a current source with excess voltage — also using a variable resistor, but connected as a potentiometer, i.e., a voltage divider, as shown in the figure.
Here R1 is a resistor connected as a potentiometer, and R2 is a load, which can be the same incandescent bulb or some other device. A voltage drop of the current source occurs across the resistor R1, which partially or completely can be supplied to the load R2. When the resistor slider is in the lowest position, no voltage is supplied to the load at all (if it is a light bulb, it will not glow). As the resistor slider is moved upwards, we will supply more and more voltage to the load R2 (if it is a light bulb, its filament will heat up). When the slider of the resistor R1 is in the highest position, the entire voltage of the current source will be supplied to the load R2 (if R2 is a flashlight bulb, and the voltage of the current source is high, the bulb filament will burn out). You can experimentally find such a position of the variable resistor slider at which the necessary voltage will be supplied to the load. Variable resistors connected as potentiometers are widely used to control the volume in receivers and amplifiers.
A resistor can be directly connected in parallel to the load. In this case, the current in this section of the circuit branches and goes along two parallel paths: through the additional resistor and the main load. The greatest current will be in the branch with the least resistance. The sum of the currents of both branches will be equal to the current consumed to supply the external circuit. A parallel connection is used in cases where it is necessary to limit the current not in the entire circuit, as when connecting an additional resistor in series, but only in some section. Additional resistors are connected, for example, in parallel to milliammeters so that they can measure large currents. Such resistors are called shunting resistors or shunts. The word shunt means branch.
INDUCTIVE REACTANCE
In an alternating current circuit, the value of the current is affected not only by the resistance of the conductor connected in the circuit, but also by its inductance. Therefore, in alternating current circuits, a distinction is made between the so-called ohmic or active resistance, determined by the properties of the conductor material, and inductive reactance, determined by the inductance of the conductor. A straight conductor has a relatively small inductance. But if this conductor is coiled into a coil, its inductance will increase. At the same time, the resistance it offers to the alternating current will also increase — the current in the circuit will decrease. As the frequency of the current increases, the inductive reactance of the coil also increases.
Remember: the resistance of an inductor to an alternating current increases with an increase in its inductance and the frequency of the current passing through it.
This property of a coil is used in various receiver circuits when it is required to limit a high-frequency current or to isolate high-frequency oscillations, in AC rectifiers, and in many other cases that you will constantly encounter in practice.
The unit of inductance is the Henry (H). An inductance of 1 H is possessed by such a coil in which, when the current in it changes by 1 A over 1 second, a self-induction EMF of 1 V is developed. This unit is used to determine the inductance of coils that are included in audio frequency circuits. The inductance of coils used in oscillatory circuits is measured in thousandths of a henry, called millihenries (mH), or in a unit a thousand times smaller — microhenries (μH).
POWER AND WORK OF CURRENT
A certain amount of electrical energy is expended on heating the filament of an electric or electronic lamp, an electric soldering iron, an electric stove, or another device. This energy, given off by the current source (or received from it by the load) over 1 second, is called the power of the current.
The unit of current power is the Watt (W). A watt is the power developed by a direct current of 1 A at a voltage of 1 V.
In formulas, current power is denoted by the Latin letter P. Electrical power in watts is obtained by multiplying the voltage in volts by the current in amperes, i.e., P = U × I.
If, for example, a DC voltage source of 4.5 V creates a current of 0.1 A in the circuit, then the current power will be: P = 4.5 × 0.1 = 0.45 W. Using this formula, you can, for example, calculate the power consumed by a pocket flashlight bulb by multiplying 3.5 V by 0.28 A. We will get about 1 W.
By changing this formula to: I = P / U, you can find out the current flowing through an electrical device if the power it consumes and the voltage supplied to it are known. What is, for example, the current flowing through an electric soldering iron, if it is known that at a voltage of 220 V it consumes a power of 40 W? I = P / U = 40 / 220 ≈ 0.18 A.
If the current and the resistance of the circuit are known, but the voltage is unknown, the power can be calculated using this formula: P = I²R. When the voltage acting in the circuit and the resistance of this circuit are known, then the following formula is used to calculate the power: P = U² / R.
But a watt is a relatively small unit of power. When dealing with electrical devices, appliances, or machines that consume currents of tens or hundreds of amperes, a unit of power called a kilowatt (written as kW), equal to 1000 W, is used. The power of electric motors of factory machine tools, for example, can range from a few units to tens of kilowatts.
The quantitative consumption of electricity is evaluated by a watt-second, characterizing the unit of energy — the Joule. The consumption of electricity is determined by multiplying the power consumed by the device by its operating time in seconds. If, for example, the light bulb of an electric flashlight (its power, as we already know, is about 1 W) burned for 25 seconds, it means the energy consumption was 25 watt-seconds. However, a watt-second is a very small value. Therefore, in practice, larger units of electricity consumption are used: watt-hour, hectowatt-hour, and kilowatt-hour. In order for energy consumption to be expressed in watt-hours or kilowatt-hours, the power in watts or kilowatts must be respectively multiplied by the time in hours. If, for example, a device consumes 0.5 kW of power for 2 hours, then the energy consumption will be 0.5 × 2 = 1 kWh; 1 kWh of energy will also be consumed if the circuit consumes (or expends) a power of 2 kW for half an hour, etc. An electric meter installed in the house or apartment where you live records the electricity consumption in kilowatt-hours. By multiplying the meter readings by the cost of 1 kWh, you will find out how much energy was consumed over a week or a month.
When working with galvanic cells or batteries, they talk about their electrical capacity in ampere-hours, which is expressed as the product of the discharge current value and the duration of work in hours. The initial capacity of a 3336L (3R12) battery, for example, is 0.5 Ah. Calculate: how long will the battery work continuously if you discharge it with a current of 0.28 A (the flashlight bulb current)? Approximately one and three-quarter hours. But if you discharge this battery more intensively, for example, with a current of 0.5 A, it will work for less than 1 hour. Thus, knowing the capacity of a galvanic cell or battery and the currents consumed by their loads, it is possible to calculate the approximate time during which these chemical current sources will work. The initial capacity, as well as the recommended discharge current or the external circuit resistance determining the discharge current of a cell or battery, is sometimes indicated on their labels or in reference literature.
In this lesson, I tried to systematize and lay out the maximum information necessary for a beginner radio amateur on the basics of electrical engineering, without which there is no point in continuing to study further. The lesson turned out to be perhaps the longest, but also the most important. I advise you to take this lesson more seriously, definitely memorize the highlighted definitions, and if something is unclear, reread it several times to grasp the essence of what is said.
As practical work, you can experiment with the circuits shown in the figures, i.e., with batteries, light bulbs, and a variable resistor. This will do you good. But in general, in this lesson, of course, the main focus should not be on practice, but on mastering the theory.
Moving on to the next lesson!
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