Electric current power. It's simple

In physics, a lot of attention is paid to the energy and power of devices, substances or bodies. In electrical engineering, these concepts play no less important role than in other branches of physics, because they determine how quickly the installation will perform its work and what load the power lines will bear. Based on this information, transformers for substations, generators for power plants and the cross-section of conductors of transmission lines are selected. In this article we will tell you how to find the power of an electrical appliance or installation, knowing the current, voltage and resistance.

Definition

Power is a scalar quantity. In general, it is equal to the ratio of work performed to time:

In simple words, this value determines how quickly work is completed. It can be designated not only by the letter P, but also by W or N, and is measured in Watts or kilowatts, which are abbreviated as W and kW, respectively.

Electrical power is equal to the product of current and voltage or:

How does this relate to work? U is the ratio of work to transfer a unit charge, and I determines how much charge passed through the wire per unit time. As a result of the transformations, a formula was obtained with which you can find the power, knowing the current strength and voltage.

Formulas for DC circuit calculations

But it is not always possible to find power by current and voltage. If you don't know them, you can determine P by knowing the resistance and voltage:

P=U2/R

You can also perform the calculation knowing the current and resistance:

P=I 2 *R

The last two formulas are convenient for calculating the power of a section of a circuit if you know the R of the element I or U that falls on it.

For AC

However, for an AC electrical circuit, it is necessary to take into account the apparent, active and reactive, as well as power factor (cosF). We discussed all these concepts in more detail in this article: .

We only note that in order to find the total power in a single-phase network by current and voltage, you need to multiply them:

The result will be obtained in volt-amperes; to determine the active power (watts), you need to multiply S by the coefficient cosФ. It can be found in the technical documentation for the device.

P=UIcosФ

To determine reactive power (reactive volt-amperes), sinФ is used instead of cosФ.

Q=UIsinФ

Or express from this expression:

And from here calculate the required value.

Finding the power in a three-phase network is also easy; to determine S (total), use the calculation formula for current and phase voltage:

S=3U f/f

And knowing Ulinear:

S=1.73*U l I l

1.73 or root of 3 - this value is used for calculating three-phase circuits.

Then, by analogy, to find P active:

P=3U f / f *cosФ=1.73*U l I l *cosФ

You can determine reactive power:

Q=3U f / f *sinФ=1.73*U l I l *sinФ

This is where the theoretical information ends and we move on to practice.

Example of calculating total power for an electric motor

The power of electric motors can be useful or mechanical on the shaft and electrical. They differ by the coefficient of performance (efficiency), this information is usually indicated on the nameplate of the electric motor.

From here we take the data to calculate the connection in a triangle to Ulinear 380 Volts:

  1. P on the shaft = 160 kW = 160000 W
  2. n=0.94
  3. cosФ=0.9
  4. U=380

Then you can find the active electrical power using the formula:

P=P on the shaft /n=160000/0.94=170213 W

Now we can find S:

S=P/cosφ=170213/0.9=189126 W

It is this that needs to be found and taken into account when selecting a cable or transformer for an electric motor. This completes the calculations.

Calculation for parallel and serial connection

When calculating the circuit of an electronic device, you often need to find the power that is released on a separate element. Then you need to determine what voltage drops across it if we are talking about a serial connection, or what current flows when connected in parallel; let’s look at specific cases.

Here Itotal is equal to:

I=U/(R1+R2)=12/(10+10)=12/20=0.6

General power:

P=UI=12*0.6=7.2 Watt

For each resistor R1 and R2, since their resistance is the same, the voltage drops according to:

U=IR=0.6*10=6 Volts

And stands out by:

P on the resistor =UI=6*0.6=3.6 Watts

Then, with a parallel connection in this circuit:

First we look for I in each branch:

I 1 =U/R 1 =12/1=12 Amperes

I 2 =U/R 2 =12/2=6 Amperes

And stands out on each:

P R 1 =12*6=72 Watts

P R 2 =12*12=144 Watts

Total highlights:

P=UI=12*(6+12)=216 Watt

Or through general resistance, then:

R total =(R 1 *R 2)/(R 1 +R 2)=(1*2)/(1+2)=2/3=0.66 Ohm

I=12/0.66=18 Amperes

P=12*18=216 Watt

All calculations coincided, which means the found values ​​are correct.

Power. Watt.

Voltage is measured with a voltmeter (V), and current through the load (R) with an ammeter (A).

It is clear that the same power can be obtained at different values ​​of the current source voltage. With a source voltage of 1 volt, to obtain a power of 1 watt, it is necessary to pass a current of 1 ampere through the load (1V x 1A = 1W). If the source produces a voltage of 10 volts, a power of 1 watt is achieved at a current of 0.1 amperes (10V x 0.1A = 1W).

Power in physics is the speed at which some work is performed.

The faster the work is done, the greater the power of the performer.

A powerful car accelerates faster. A powerful (strong) person is able to drag a bag of potatoes to the ninth floor faster.

1 Watt is a power that allows you to do 1 J of work in one second (what a joule is was described above).

If you are able to accelerate a two-kilogram body to a speed of 1 m/s in one second, then you are developing a power of 1 W.

If you lift a kilogram load to a height of 0.1 meters per second, your power is 1 W because the load acquires a potential energy of 1 J per second.

If you drop one plate from the same height onto a concrete floor and the second onto a blanket, the first one will probably break, but the second one will survive. What is the difference? The initial and final conditions are the same. The plates fall from the same height and therefore have the same energy. At floor level, both plates stop - everything seems to be identical. The only difference is The fact is that the energy that the plate accumulated during the flight is released instantly (very quickly) in the first case, and when the plate falls on a blanket or carpet, the braking process is extended over time.

Let the falling plate have a kinetic energy of 1 J. The process of colliding with a concrete floor takes, say, 0.001 seconds. It turns out that the power released during impact is 1/0.001=1000 W!

If the plate smoothly slows down for 0.1 seconds, the power will be 1/0.1=10 W. There is already a chance to survive - if there is a living organism in the place of the plate.

This is why there are crumple zones and airbags in cars, so that extend the process of energy release over time in case of an accident, i.e., reduce power upon impact. And the release of energy, by the way, is work. In this case, the work is to rupture your internal organs and break your bones.

At all, work is the process of converting one type of energy into another.

Another example: you can burn the contents of a propane cylinder in a burner without consequences. But if you mix the gas contained in the cylinder with air and ignite it, explosion.

In both cases, the same amount of energy is released. But in the second, energy is released in a short period of time. A power - the ratio of the amount of work to the time in which it is done.

Regarding electricity, 1 W is the power released by the load when the product of the current through it and the voltage at its ends is equal to unity. That is, for example, if the current through the lamp is 1 A, and the voltage at its terminals is 1 V, the power released through it is 1 W.

A lamp with a current of 2 A will have the same power at a voltage of 0.5 V - the product of these quantities is also equal to one.

So:

P = U*I. Power is equal to the product of voltage and current.

We can write it differently:

I = P/U- current is equal to power divided by voltage.

There is, for example, an incandescent lamp. The following parameters are indicated on its base: voltage 220 V, power 100 W. A power of 100 W means that the product of the voltage applied to its terminal multiplied by the current flowing through this lamp is one hundred. U*I=100.

What current will flow through it? Elementary Watson: I = P/U, divide power per voltage (100/220), we get 0.454 A. Current through the lamp is 0.454 ampere. Or, in other words, 454 milliamps (milli - thousandth).

Another recording option U = P/I. It will also come in handy somewhere.

Now we are armed with two formulas - Ohm's law and the formula for electric current power. And this is already a tool.

We want to find out the resistance of the filament of the same hundred-watt incandescent lamp.

Ohm's law tells us: R = U/I.

You don’t have to calculate the current through the lamp in order to substitute it into the formula later, but take a shortcut: since I = P/U, we substitute P/U instead of I in the formula R = U/I.

In fact, why not replace the current (which is unknown to us) with the voltage and power of the lamp (which are indicated on the base).

So: R = U/P/U, which is equal to U^2/P. R = U^2/P. We square 220 (voltage) and divide by one hundred (lamp power). We get a resistance of 484 Ohms.

You can check the calculations. Above, we calculated the current through the lamp - 0.454 A.

R = U/I = 220/0.454 = 484 Ohm. Whatever one may say, there is only one correct conclusion.

Once again, the power formula is: P = U*I(1), or I = P/U(2), or U = P/I (3).

Ohm's law: I = U/R(4) or R = U/I(5) or U = I*R (6).

P - power

U - voltage

I - current

R - resistance

In any of these formulas, instead of an unknown value, you can substitute known ones.

If you need to find out the power, having the values ​​of voltage and resistance, take formula 1, instead of current I we substitute its equivalent from formula 4.

It turns out P = U^2/R. Power is equal to the square of the voltage divided by the resistance. That is, when the voltage applied to the resistance changes, the power released on it changes in a quadratic relationship: the voltage was doubled, the power (for the resistor - heating) increased fourfold! This is what mathematics tells us.

A hydraulic analogy will again help to understand why this happens in practice.An object located at a certain height has potential energy. And, descending from this height, he can do work. This is how water does the work of generating energy in a hydroelectric power station, falling through a hydraulic turbine from the level of the reservoir to the tailwater (lower level).

The potential energy of an object depends on its mass and on the height at which it is located (the more trouble a falling stone will cause, the more it weighs, and the greater the height from which it falls). The gravity at the place where it falls also matters. The same stone falling from the same height is more dangerous on the ground than on the Moon, since on the Moon the “force of gravity” (the force pulling the stone down) is 6 times less than on Earth. So, we have three parameters that affect potential energy - mass, height and gravity. They are exactly what is contained in the kinetic energy formula:

Ek = m*g*h,

Where m- mass of the object,g- acceleration of free fall at a given location ("gravity"),h- the height at which the object is located.

Let's assemble the installation: a pump driven by an engine will pump water from the lower reservoir to the upper one, and the water flowing under the influence of gravity from the upper reservoir will turn the generator:

It is clear that the higher the water column, the more energy the water will have. Let's double the height of the pillar. It is clear that at double the height h, water will have twice the potential energy, and, it seems, the power of the generator should double? In fact, its power will quadruple. Why? Because due to double the pressure from above, the flow of water through the generator will double. And double the flow of water at double the pressure will lead to a fourfold increase in the power released by the generator: twice as much and twice as strong.

The same thing happens at the resistance when the voltage applied to it doubles. We remember the formula for the power released by a resistor, right?

P = U*I.

Power P equal to the product of voltage U, applied to the resistor and current I flowing through it. When the applied voltage doubles U, the power seems to have to double. But an increase in voltage also leads to a proportional increase in the current through the resistor! Therefore, it will double not only U, but also I. That is why the power depends on the applied voltage in a quadratic manner.

A battery with double the voltage “pumps” electrons to twice the “height”, and this leads to exactly the same picture as in the hydraulic analogue.

Need to find out the power, knowing the resistance and current, but not knowing the voltage? No problem. In the same first formula instead U substitute the equivalent U from formula 6. We get P = I^2*R. Power is equal to the square of the current times the resistance.

The hydraulic analogue above will help you understand why. Doubling the current through a given resistor is only possible by doubling the voltage applied to it. So, the formula P = U*I, will work here too, despite the absence in the formula P = I^2*R voltage. It’s just that the tension in this case is present “behind the scenes”, hiding behind other variables.

Another oddity of this formula is that power is directly proportional to resistance. How can this be? Well, then let's break the circuit altogether, the resistance will increase to infinity, which means that the power released on what is not there will correspondingly increase? What nonsense.

It's actually simple. An increase in resistance will result in a corresponding decrease in the current through the resistor. If in the formula

P = I^2*R,

resistance R double, then the current I will be reduced by half. And the dependence of power on current in this formula is quadratic. Therefore, the power released by the resistor is expected to drop by half.

I remind you:

Voltage (U) is the “electrical pressure difference” between any two points in the electrical circuit (analogous to the fluid pressure difference). Unit - volt.

Current (I) is the number of electrons passing through a section of the circuit (analogous to a fluid flow).Unit - ampere. 1 A = 1 C/sec.

Resistance (R) - the ability of a section of a circuit to interfere with (resist) the movement of electrons(like a bottleneck or blockage in a pipe).Unit - ohm.

Power (P) is the product of voltage and current (as if we multiplied the water flow through any section of the water supply system by the pressure difference at the ends of this section).Unit - watt.

Electric current power is the speed of work performed by the circuit. A simple definition, a hassle with understanding. Power is divided into active and reactive. And it begins...

Electric current work, power

When a charge moves along a conductor, the field does work on it. The quantity is characterized by tension, as opposed to tension in free space. Charges move in the direction of decreasing potentials; to maintain the process, an energy source is required. The voltage is numerically equal to the work of the field when moving a unit charge (1 C) in the area. During interactions, electrical energy is converted into other forms. Therefore, it is necessary to introduce a universal unit, a physical freely convertible currency. In the body, the measure is ATP, electricity is the work of the field.

Electric arc

In the diagram, the moment of energy conversion is displayed in the form of emf sources. If the generators are directed in one direction, the consumers must be directed in the other direction. A clear fact reflects the process of power consumption and extraction from energy sources. EMF has the opposite sign, often called back-EMF. Avoid confusing the concept with the phenomenon that occurs in inductors when the power is turned off. Back-EMF means the transition of electrical energy into chemical, mechanical, and light energy.

The consumer wants to complete work in a certain unit of time. Obviously, the lawnmower does not intend to wait for winter, he hopes to finish it by lunchtime. The power of the source must provide the specified execution speed. The work is carried out by electric current, therefore the concept also applies. Power can be active, reactive, useful and loss power. The areas designated by physical diagrams as resistances are harmful in practice and are costs. Heat is generated at the conductor resistors, the Joule-Lenz effect leads to unnecessary power consumption. An exception is heating devices, where the phenomenon is desirable.

Useful work on physical circuits is indicated by back-EMF (a conventional source with the direction opposite to the generator). There are several analytical expressions for power. Sometimes it is convenient to use one, in other cases – another (see figure):

Current Power Expressions

  1. Power is the speed at which work is performed.
  2. Power is equal to voltage times current.
  3. The power expended on thermal action is equal to the product of resistance times the square of the current.
  4. The power expended on thermal action is equal to the ratio of the square of the voltage to the resistance.

For those who have stocked up with current clamps, it is easier to use the second formula. Regardless of the nature of the load, we will calculate the power. Only active. Power is determined by many factors, including temperature. By the nominal value for the device we mean the value developed in steady state. For heaters, the third and fourth formulas should be used. Power depends entirely on the parameters of the supply network. Designed to operate on 110 volts AC in European conditions will burn out quickly.

Three-phase circuits

For beginners, three-phase circuits seem complicated, but in fact this is a more elegant technical solution. Even the house is supplied with three lines of electricity. Inside the entrance they are divided into apartments. What is more confusing is that some three-phase devices do not have a grounding or neutral wire. Circuits with isolated neutral. A neutral wire is not needed; the current is returned to the source through the phase lines. Of course, the load here on each core is increased. The PUE requirements separately stipulate the type of network. For three-phase circuits, the following concepts are introduced that you need to understand in order to correctly calculate the power:

Three-phase circuit with isolated neutral

  • Phase voltage and current are called, respectively, the potential difference and the speed of charge movement between phase and neutral. It is clear that in the above-mentioned case with complete isolation, the formulas will be invalid. Because there is no neutral.
  • Linear voltage and current are called, respectively, the potential difference or the speed of charge movement between any two phases. The numbers are clear from the context. When they talk about 400 volt networks, they mean three wires, the potential difference with the neutral is 230 volts. Line voltage is higher than phase voltage.

There is a phase shift between voltage and current. What school physics is silent about. The phases are the same if the load is 100% active (simple resistors). Otherwise, a shift appears. In inductance, the current lags behind the voltage by 90 degrees, in capacitance it leads. A simple truth is easy to remember as follows (we smoothly approach reactive power). The imaginary part of the inductance resistance is jωL, where ω is the circular frequency equal to the usual one (in Hz) multiplied by 2 Pi; j is an operator indicating the direction of the vector. Now we write Ohm's law: U = I R = I jωL.

From the equality it is clear: the voltage must be plotted upward by 90 degrees when constructing a diagram, the current will remain on the abscissa axis (horizontal X axis). According to the rules of radio engineering, rotation occurs counterclockwise. Now the fact is obvious: the current lags by 90 degrees. By analogy, let's make a comparison for a capacitor. Resistance to alternating current in imaginary form looks like this: -j/ωL, the sign indicates: the voltage will need to be put down, perpendicular to the abscissa axis. Therefore, the current is 90 degrees ahead in phase.

In reality, in parallel with the imaginary part, there is a real part - called active resistance. The coil wire is represented by a resistor, and when twisted, it acquires inductive properties. Therefore, the actual phase angle will not be 90 degrees, but slightly less.

And now we can move on to the formulas for the current power of three-phase circuits. Here the line forms a phase shift. Between voltage and current, and relative to the other line. Agree, without the knowledge carefully presented by the authors, the fact cannot be realized. Between the lines of an industrial three-phase network the shift is 120 degrees (full rotation - 360 degrees). It will ensure uniform rotation of the field in engines; it is of no importance to ordinary consumers. This is more convenient for hydroelectric power station generators – the load is balanced. The shift occurs between the lines, in each the current leads the voltage or lags behind:

  1. If the line is symmetrical, the current shifts between any phases are 120 degrees, the formula is extremely simple. But! If the load is symmetrical. Let's look at the image: phase f is not 120 degrees, characterizes the shift between the voltage and current of each line. It is assumed that a motor with three equal windings is turned on, the following result is obtained. If the load is unbalanced, take the trouble to do the calculations for each line separately, then add the results together to get the total amperage.
  2. The second group of formulas is given for three-phase circuits with an isolated neutral. It is assumed that the current from one line flows through the other. The neutral is missing as unnecessary. Therefore, the voltages are taken not as phase voltages (there is nothing to count from), as in the previous formula, but linear. Accordingly, the numbers show which parameter should be taken. Stop getting scared of the Greek letters - the phase between the two multiplied parameters. The numbers are swapped (1.2 or 2.1) to account for the sign correctly.
  3. In the asymmetric circuit, phase voltage and current reappear. Here the calculation is carried out separately for each line. There are no options.

In practice, measure current power

They hinted that you could use current clamps. The device will allow you to determine the cruising parameters of the drill. Acceleration can only be detected with repeated experiments; the process is extremely fast, the frequency of display changes is no higher than 3 times per second. Current clamps show error. Practice shows that it is difficult to achieve the error specified in the passport.

More often, meters (for payments to supplier companies) and wattmeters (for personal and work purposes) are used to estimate power. The pointer device contains a pair of fixed coils through which the circuit current flows, a movable frame for establishing voltage by connecting the load in parallel. The design is designed to immediately implement the full power formula (see figure). The current is multiplied by the voltage and a certain coefficient that takes into account the scale graduation, also by the cosine of the phase shift between the parameters. As mentioned above, the shift fits within 90 - minus 90 degrees, therefore, the cosine is positive, the torque of the arrow is directed in one direction.

There is no way to tell whether the load is inductive or capacitive. But if it is incorrectly connected to the circuit, the readings will be negative (turned to one side). A similar event will occur if the consumer suddenly begins to transfer power back to the load (this happens). In modern devices, something similar happens; calculations are carried out by an electronic module that integrates energy consumption or reads power readings. Instead of a needle, there is an electronic indicator and many other useful options.

Particular problems arise from measurements in asymmetrical circuits with an isolated neutral, where the power of each line cannot be directly added. Wattmeters are divided into operating principles:

  1. Electrodynamic. Described in section. They consist of one movable and two fixed coils.
  2. Ferrodynamic. Reminds me of a shaded-pole motor.
  3. With a quadrator. The amplitude-frequency response of a nonlinear element (for example, a diode), resembling a parabola, is used to square an electrical quantity (used in calculations).
  4. With Hall sensor. If the induction is made using a coil proportional to the magnetic field voltage in the sensor, a current is applied, the EMF will be the result of the multiplication of two quantities. The required quantity.
  5. Comparators. Gradually increases the reference signal until equality is achieved. Digital instruments achieve high accuracy.

In circuits with a strong phase shift, a sine wattmeter is used to estimate losses. The design is similar to that considered, the spatial position is such that reactive power is calculated (see figure). In this case, multiply the product of current and voltage by the sine of the phase angle. We measure reactive power with a conventional (active) wattmeter. There are several methods. For example, in a three-phase symmetrical circuit, you need to connect a series winding to one line, and a parallel winding to the other two. Then calculations are made: the instrument readings are multiplied by the root of three (taking into account that the indicator shows the product of current, voltage and the sine of the angle between them).

For a three-phase circuit with simple asymmetry, the task becomes more complicated. The figure shows the technique of two wattmeters (ferrodynamic or electrodynamic). The beginnings of the windings are indicated by asterisks. The current passes through the series, the voltage from two phases is supplied to the parallel (one through a resistor). The algebraic sum of the readings of both wattmeters is added and multiplied by the root of three to obtain the reactive power value.

Content:

It has long been known that in this world there are Little Ones who ask their dads not only about what is good and what is bad, but also about anything. Therefore, it is very possible that an older Tiny may wonder why 2000 W is written on the heater. Tiny ones who can read, their dads, and many other readers who have forgotten the basics of physics will find further information that refreshes their memory. In particular, let us recall how power is measured and what the unit of measurement of electrical power is called.

Power is all around us

Now, wherever people live, there are electrical appliances. Each of them shows the power consumption. In the technical data sheet or operating manual there are clarifying words - electrical power. This definition is perceived as somehow abstract and not vital, impersonal. After all, if any manifestations of energy and, accordingly, power, for which the word “power” is often used, occur in life, it is always clear with whom or what it is all connected with.

For example, a mudflow came down from the mountains and with all its might fell on such and such a town. It is immediately clear that the mudflow is powerful, has destructive power, and the concept of power is associated precisely with it, with its movement, with what it consists of. But electrical power is connected to whom or what? Since we all know from childhood about the danger of an electrical outlet, first of all you pay attention to the voltage. And indeed: since the operation of electrical appliances requires voltage in the outlet, it means that we can say that the power of electricity is the power of voltage.

But if there is a heater near the outlet and its plug is not in it, it does not provide heat. However, there is still voltage in the outlet. And nothing happens. This means that the definition of “voltage power” is incorrect. The release of heat and other manifestations of electrical power are always associated with the appearance of a conductor between points with different electrical potentials and current processes in it. Their intensity is directly related to the release of heat and light, which is illustrated by lightning and thunder.

Therefore, electrical power is current power, not voltage power. And it is not without reason that such a definition as electric current was introduced into electricity. Although it is impossible to see the appearance of electric current, unlike liquid flow, there are many similarities between the two. Just like a mudflow, there is a current strength. But its nature is different. This force does not have a direct mechanical effect. However, as various electrical machines and electrical appliances demonstrate, current can do a lot.

This “much” can be indicated by three main results that the power of electric current gives:

  • warm;
  • light;
  • electromagnetic fields.

To perform calculations as well as measurements of electric current power, current power units were adopted. They were named after the English physicist James Watt in 1882. This scientist studied the processes that are associated with the performance of various types of work as a physical quantity. Since then, 1 watt has been in use, which is abbreviated as W and W. If someone has forgotten what this means in physics, we remind you: power is equal to the work done per unit of time.

And in order not to strain by writing a large number of zeros for large values ​​of electrical power, before W they write:

  • kilo, abbreviated kW - instead of three zeros;
  • mega, respectively, mW - instead of six zeros;
  • giga, gW - instead of nine zeros.

Such multifaceted power...

In Watt's time, electrical engineering was just beginning to develop, and for this reason physics was noticeably simpler than it is today. Direct electric current has been studied to a much greater extent than alternating current. For calculations at a constant electric current, the formula was justified:

in which there is power p, voltage u and electric current i. But there is also alternating electric current. Research has shown that the power p from the formula for direct current does not correspond to reality. On alternating current, completely different new properties of current power appear. Their result is invisible and not perceptible without special measurements and instruments. On alternating current, power appears due to the creation of electromagnetic fields in inductors, as well as electrostatic fields in capacitors.

This was the reason for the discrepancy between the power expression p=u*i. I had to introduce separate metering for alternating current. The unit adopted for it is var (abbreviated). By analogy with direct current, this means reactive volt-ampere (full name).

A more detailed discussion regarding alternating current is beyond the scope of the current narrative. And the Little Ones will most likely be fast asleep about halfway through our article. Information overload acts like a sleeping pill. Therefore, AC power is a completely different story...

When designing electrical equipment and calculating cables and starting and protective equipment, it is important to correctly calculate the power and current of electrical equipment. This article explains how to find these settings.

What is power

When an electric heater or electric motor operates, it produces heat or performs mechanical work, the unit of which is 1 joule (J).

One of the main characteristics of electrical equipment– power, showing the amount of heat or work done in 1 second and expressed in watts (W):

1W=1J/1s.

In electrical engineering, 1W is released when a current of 1A passes at a voltage of 1V:

According to Ohm's law, you can also find power by knowing the load resistance and current or voltage:

P=U*I=I*I*R=(U*U)/R, where:

  • P (W) – power of the electrical appliance;
  • I (A) – current flowing through the device;
  • R (Ohm) – device resistance;
  • U (V) – voltage.

Rated power is called power at rated network parameters and rated load on the motor shaft.

In order to find out the amount of electricity consumed over the entire period of operation, it must be multiplied by the time the device was operating. The learned value is measured in kWh.

Calculation in AC and DC voltage networks

The electrical network that supplies electrical appliances can be of three types:

  • constant pressure;
  • alternating single-phase;
  • alternating three-phase.

For each type, calculations use its own power formula.

Calculation in a DC voltage network

The simplest calculations are made in a DC power network. The power of electrical devices connected to it is directly proportional to the current and voltage, and to find it, the formula is used:

For example, in an electric motor with a rated current of 4.55A, connected to a 220V power supply, the power is 1000 Watts, or 1 kW.

And, conversely, with known network voltage and power, the current is calculated by the formula:

Single-phase loads

In a network in which there are no electric motors, as well as in a household electrical network, you can use formulas for a constant voltage network.

Interesting. In a 220V household power supply, the current can be calculated using a simplified formula: 1kW=5A.

AC power is more difficult to calculate. These devices, in addition to active energy, consume reactive energy, and the formula is:

shows the total energy consumption of the device. In order to find out the active component, you need to take into account cosφ– parameter showing the share of active energy in total:

Rakt=Rotot*cosφ=U*I*cosφ.

Accordingly, Rototal = Rakt/cosφ.

For example, in an electric motor with Pakt 1 kW and cosφ 0.7, the total energy consumed by the device will be 1.43 kW and the current will be 6.5 A.

Calculation in a three-phase network

A three-phase electrical network can be represented as three single-phase networks. However, in single-phase networks the concept of “phase voltage” (Uph) is used, measured between the neutral and phase wires, in a 0.4 kV network, equal to 220V. In three-phase electrical networks, instead of “phase”, the concept of “linear voltage” (Uline) is used, measured between linear wires and in a 0.4 kV network, equal to 380V:

Ulin=Uph√3.

Therefore, the formula for an active load, for example, an electric boiler, looks like this:

When determining the power of an electric motor, it is necessary to take cosφ into account; the expression takes the following form:

P=U*I*√3*cosφ.

In practice, this parameter is usually known, but it is necessary to find out the current. The following expression is used for this:

I=P/(U*√3*cosφ).

For example, for an electric motor of 3 kW (3000 W) and cosφ 0.7, the calculation is as follows:

I=3000/(380*√3*0.7)=5.8A.

Interesting. Instead of calculations, we can assume that in a three-phase 380V network, 1 kW corresponds to 2A.

Horsepower

In some cases, when determining the power of cars, the outdated unit of measurement “horsepower” is used.

This unit was introduced by James White, after whom the 1 Watt unit of power is named, in 1789. He was hired by a brewer to build a steam engine for a pump that could replace a horse. To determine what kind of engine was needed, they took a horse and harnessed it to pump water.

It is believed that the brewer took the strongest horse and forced it to work without rest. The real strength of the horse is 1.5 times less.