The very formulation of the question of converting amperes to kilowatts, and kilowatts to amperes, is somewhat incorrect. The fact is that amperes and kilowatts are slightly different physical quantities. Ampere is a unit of force electric current, and kilowatt is a unit of measurement of electrical power. It is more correct to talk about the correspondence of the current strength to the specified power, or the power corresponding to the value of the current strength. Therefore, the conversion of amperes to kilowatts and vice versa should be understood not literally, but relatively. This should be the starting point for further calculations.
Very often, knowing one value, it is necessary to determine another. This may be necessary for the selection of protective and switching equipment. For example, if you want to choose a circuit breaker or fuse with a known total power of all consumers.
Consumers can be incandescent lamps, fluorescent lamps, irons, a washing machine, a boiler, a personal computer and other household appliances.
In another case, if there is a protective device with a known rated current, it is possible to determine the total power of all consumers that are allowed to “load” the circuit breaker or fuse.
You should be aware that the rated power consumption is usually indicated on electrical consumers, and the rated current is indicated on the protective device (automatic or fuse).
To convert amperes to kilowatts and vice versa, it is necessary to know the value of the third quantity, without which calculations are impossible. This is the value of the supply or rated voltage. If the standard voltage in the electrical (household) network is 220V, then the rated voltage is usually indicated on the consumers themselves and on the protective devices.
That is, for example, on an incandescent lamp for a household electrical network, in addition to power, the rated voltage for which it is designed is also indicated. Similarly, with circuit breakers (fuses). They also indicate the rated voltage at which they must be operated.
It should also be noted that in addition to the usual single-phase 220V network, a three-phase 380V electrical network is often used (usually in production). This must also be taken into account when calculating power and current strength.
Converting amperes to kilowatts (single-phase network 220V)
Suppose there is a single-pole circuit breaker available, the rated current of which is 25A. Those. in normal operating mode, a current of not more than 25A should flow through the machine. In order to determine the maximum possible power that the machine can withstand, you must use the formula:
P = U*I
where: P - power, W (watt);
U - voltage, V (volt);
I - current strength, A (amperes).
Substitute the known values into the formula and get the following:
P = 220V*25A = 5500W
Power is in watts. In order to convert the resulting value into kilowatts, divide 5500W by 1000 and get 5.5kW (kilowatts). Those. the total power of all consumers that will be powered by a machine with a rating of 25A should not exceed 5.5 kW.
Converting kilowatts to amperes in a single-phase network
If the total power of all consumers together or each consumer separately is known, then it is easy to determine the rated current of the protective device necessary to supply consumers with a known power.
Let's say there are several consumers, the total power of which is 2.9 kW:
- incandescent lamps 4pcs. 100W each;
- boiler with a capacity of 2 kW;
- personal computer, the power of which is 0.5 kW.
To determine the total power, first you need to bring the values of all consumers to a single indicator. Those. convert kilowatts to watts. Because 1kW = 1000W, then the boiler power will be 2kW * 1000 = 2000W. PC power will be equal to 0.5kW * 1000 = 500W.
We determine the total power of all consumers. It is necessary to add the power of incandescent lamps, boiler and PC.
PΣ = 400W + 2000W + 500W = 2900W
To determine the current strength corresponding to a power of 2900W at a mains voltage of 220V, we use the same power formula P \u003d U * I. Let's transform the formula and get:
I \u003d P / U \u003d 2900W / 220V ≈ 13.2A
As a result of a simple calculation, it turned out that the load current with a power of 2900W is approximately equal to 13.2A. It turns out that the rated current of the selected machine must be at least this value.
Because the nearest standard nominal value of a conventional single-phase machine is 16A, then for a load with a power of 2.9 kW, a circuit breaker with a rated current of 16A is suitable.
We translate amperes into kilowatts and vice versa (three-phase network 380V)
The calculation method for converting amperes to kilowatts and vice versa in a three-phase network is similar to the calculation method for a single-phase electrical network. The only difference is in the calculation formula.
To determine the power consumption in a three-phase network, the following formula is used:
P = √3*U*I
where: P - power, W (watt);
U - voltage, V (volt);
I - current strength, A (ampere);
Imagine that it is necessary to determine the power that a three-phase circuit breaker with a rated current of 50A can withstand. Substitute the known values into the formula and get:
P = √3*380V*50A ≈ 32908W
We convert watts to kilowatts by dividing 32908W by 1000 and we get that the power is approximately 32.9kW. Those. a three-phase 50A automatic machine is able to withstand a load with a power of 32.9 kW.
If the power of a three-phase consumer is known, then the calculation of the operating current of the circuit breaker is performed by converting the above formula.
The current of the machine is determined by the following expression:
I = P/(√3*U)
Let's say the power of a three-phase consumer is 10 kW. The power in watts will be 10kW * 1000 = 10000W. Determine the current strength:
I \u003d 10000W / (√3 * 380) ≈ 15.2A.
Therefore, for a consumer with a power of 10 kW, an automatic machine with a rating of 16A is suitable.
p \u003d UxI power is equal to voltage times current strength 220 x 9 \u003d 1980 W or 1.98 kW.
Multiply this will be the maximum power value, usually use the average power Pav = I * U / 2
HERR ADOLF, it is clear why Russia did not submit to you, because you can’t even find the power knowing the voltage U and the current I :)). So Russia will help you: N=IU=220*9=1980W or 1.98kW or 2.69l. With.
iron turns out
you get the power of your kettle, which is written on the package, the current was measured by email. breeches without breaking the wires of the kettle and the network! = about 2 kW
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Automatic machine for 20 Amperes, how many kW will it be?
something like this 
How to convert amps to kilowatts in a single-phase network Watt \u003d Ampere * Volt Well, you translate watts into kilowatts
20amp*220volt=....watt
20 * 1.45 = 29A * 220V = up to 6.3 maybe 4mm2 of copper, but somehow irrational, so put 16A and 2.5mm2 under it, up to 5 practically, 3.5 - iron
4 kW. 1 kW = 5 amps at 220 volts.
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guys, how to calculate the number of watts in a 220 volt 6 amp outlet?
Measure with an instrument.
220V x 6A = 1320W I'm afraid to explain, so as not to offend by accident.
current equals power (watts) divided by voltage
There is a simple formula for direct current: P = UI. But it is strictly valid for direct current. In the case of a variable (in a socket), it is more accurate to write S = UI. The difference is that S is full power, and P is active, and there is also reactive power - with alternating current and the presence of inductances (coil) or capacitances (capacitor) in the circuit. For an alternating single-phase current, the active power will be P \u003d UI cos f, where f is the angle between the active and total power vectors. Usually cos f = 0.8-0.9.
Do not fool the guy with cosine phi. In general, Seryoga, enough to know that 1320 watts. About...
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The choice of protective circuit breakers is made not only during the installation of a new electrical network, but also when upgrading the electrical panel, as well as when additional powerful devices are included in the circuit, increasing the load to a level that the old emergency shutdown devices cannot cope with. And in this article we will talk about how to properly select the machine for power, what should be taken into account during this process and what are its features.
Failure to understand the importance of this task can lead to very serious problems. Indeed, often users do not bother themselves by choosing a circuit breaker by power, and take the first device they come across in the store, using one of two principles - “cheaper” or “more powerful”. This approach, associated with the inability or unwillingness to calculate the total power of devices connected to the power grid, and in accordance with it, select a circuit breaker, often causes the failure of expensive equipment in the event of a short circuit or even a fire.
What are circuit breakers and how do they work?
Modern AB have two degrees of protection: thermal and electromagnetic. This allows you to protect the line from damage as a result of a long excess of the flowing current of the rated value, as well as a short circuit.
The main element of the thermal release is a plate of two metals, which is called bimetallic. If it is exposed to a current of increased power for a sufficiently long time, it becomes flexible and, acting on the disconnecting element, causes the machine to operate.
The presence of an electromagnetic release is due to the breaking capacity of the circuit breaker when the circuit is exposed to short-circuit overcurrents, which it cannot withstand.
The electromagnetic type release is a solenoid with a core, which, when a high power current passes through it, instantly shifts towards the disconnecting element, turning off the protective device and de-energizing the network.
This makes it possible to protect the wire and devices from an electron flow, the value of which is much higher than that calculated for a cable of a specific section.
Why is cable mismatch with network load dangerous?
The correct selection of the circuit breaker by power is a very important task. An incorrectly selected device will not protect the line from a sudden increase in current.
But it is equally important to choose the right electrical cable according to the cross section. Otherwise, if the total power exceeds the nominal value that the conductor can withstand, this will lead to a significant increase in the temperature of the latter. As a result, the insulating layer will begin to melt, which can lead to a fire.
In order to more clearly imagine what the mismatch between the wiring cross-section of the total power of devices connected to the network threatens, consider the following example.
The new owners, having bought an apartment in an old house, install several modern household appliances in it, giving a total load on the circuit equal to 5 kW. The current equivalent in this case will be about 23 A. In accordance with this, a 25 A circuit breaker is included in the circuit. It would seem that the choice of the machine in terms of power was made correctly, and the network is ready for operation. But some time after turning on the appliances, smoke appears in the house with a characteristic smell of burnt insulation, and after a while a flame appears. At the same time, the circuit breaker will not disconnect the network from the power supply - after all, the current rating does not exceed the allowable one.
If the owner is not nearby at this moment, the molten insulation will cause a short circuit after a while, which will finally trigger the machine to work, but the flame from the wiring may already spread throughout the house.
The reason is that although the power calculation of the machine was done correctly, the wiring cable with a cross section of 1.5 mm² was rated for 19 A and could not withstand the existing load.
So that you do not have to take up a calculator and independently calculate the cross section of the electrical wiring using formulas, we present a typical table in which it is easy to find the desired value.
Weak link protection
So, we made sure that the calculation of the circuit breaker should be made based not only on the total power of the devices included in the circuit (regardless of their number), but also on the cross section of the wires. If this indicator is not the same along the electric line, then we select the section with the smallest cross section and calculate the machine based on this value.
The requirements of the PUE state that the selected circuit breaker must provide protection for the weakest section of the electrical circuit, or have a current rating that will correspond to a similar parameter of the installations connected to the network. This also means that wires must be used for the connection, the cross section of which will withstand the total power of the connected devices.
How to select the wire cross section and circuit breaker rating - in the following video:
If the negligent owner ignores this rule, then in the event of an emergency due to insufficient protection of the weakest section of the wiring, he should not blame the selected device and scold the manufacturer - only he will be responsible for the situation.
How to calculate the circuit breaker rating?
Let's assume that we have taken into account all of the above and selected a new cable that meets modern requirements and has the desired cross section. Now the electrical wiring is guaranteed to withstand the load from the included household appliances, even if there are a lot of them. Now we proceed directly to the choice of the circuit breaker according to the current rating. We recall the school physics course and determine the calculated load current by substituting the appropriate values into the formula: I = P / U.
Here I is the value of the rated current, P is the total power of the installations included in the circuit (taking into account all consumers of electricity, including light bulbs), and U is the mains voltage.
To simplify the choice of a circuit breaker and save you from having to take on a calculator, we present a table that shows the ratings of AB, which are included in single-phase and three-phase networks, and the corresponding total load powers.
This table will make it easy to determine how many kilowatts of load correspond to which rated current of the protective device. As we can see, a 25 Amp machine in a network with a single-phase connection and a voltage of 220 V corresponds to a power of 5.5 kW, for a 32 Amp AB in a similar network - 7.0 kW (in the table this value is highlighted in red). At the same time, for an electrical network with a three-phase delta connection and a rated voltage of 380 V, a 10 Amp machine corresponds to a total load power of 11.4 kW.
Clearly about the selection of circuit breakers on the video:
Conclusion
In the presented material, we talked about why electric circuit protection devices are needed and how they work. In addition, given the information provided and the tabular data given, you will not have any difficulty with the question of how to choose a circuit breaker.
Power is the rate at which energy is expended expressed in terms of energy over time: 1 W = 1 J/1 s. One watt is equal to the ratio of one joule (a unit of work) to one second.
Almost everyone has heard about the parameters of electricity as Volt, Ampere and watts.
What is power. Watt [W]
Watt, according to the SI system - a unit of power. Nowadays it is used to measure the power of all electrical and not only appliances. According to the theory of physics, power is the rate of energy expenditure, expressed in relation to energy over time: 1 W \u003d 1 J / 1 s. One watt is equal to the ratio of one joule (a unit of work) to one second.
To date, the unit of measurement of kilowatts (abbreviated designation - kW) is more often used to denote the power of electrical appliances. It is easy to guess how many watts are in a kilowatt - the prefix "kilo" in the SI system denotes the value obtained by multiplying by a thousand.
For power-related calculations, it is not always convenient to use the watt by itself. Sometimes, when the quantities being measured are very large or very small, it is much more convenient to use a unit of measure with standard prefixes, which avoids constant calculations of the order of magnitude. So, in the design and calculation of radars and radio receivers, pW or nW are most often used, for medical devices, such as EEG and ECG, microWatt is used. In the production of electricity, as well as in the design of railway locomotives, megawatts (MW) and gigawatts (GW) are used.
What is stress. Volt [V]
Voltage is a physical quantity that characterizes the value of the ratio of work
electric field in the process of charge transfer from one point A to another point B to the value of this very charge. Simply put, it is the potential difference between two points. Measured in Volts.
The voltage is essentially similar to the magnitude of the water pressure in the pipe, the higher it is, the faster the water flows from the tap. The voltage value is standardized and the same for all apartments, houses and garages, equal to 220 volts with a single-phase power supply. Also allowed according to GOST 10 percent deviation for the home electrical network. The voltage value must be at least 198 and not more than 242 Volts.
1 Volt contains:
- 1,000,000 microvolts
- 1,000 millivolts
What is current strength. Amp [A]
Current strength this is a physical quantity equal to the ratio of the amount of charge for a certain period of time flowing through the conductor to the value of this very period of time. Measured in amperes.
1 Amp contains:
- 1,000,000 microamps
- 1,000 milliamps
Sometimes such a task as converting amperes to watts or kilowatts, or vice versa - watts and kilowatts to amperes, can be difficult. After all, few of us remember by heart the formulas of my school days. Unless, of course, you constantly have to deal with this by the nature of your profession or hobby.
In fact, in everyday life, knowledge of such things may be required quite often. For example, on the socket or on the plug there is a marking in the form of an inscription: “220V 6A”. This marking reflects the maximum permissible power of the connected load. What does it mean? What is the maximum power of the mains appliance that can be plugged into such an outlet or used with this plug?
Based on this marking, we see that the operating voltage for which this device is designed is 220 volts, and the maximum current is 6 amperes. To get the power value, just multiply these two numbers: 220 * 6 \u003d 1320 watts - the maximum power for a given plug or outlet. For example, an iron with steam can only be used on a deuce, and an oil heater can only be used at half power.
How many volts are in 1 ampere?
It is rather difficult to answer this question. However, in order to make it easier for you to deal with this issue, we suggest that you familiarize yourself with the ratio tables
For DC
For AC
How many watts in 1 ampere?
So, to get watts, you need to multiply the specified amps by volts:
In it, P is Watt, I is A, and U is Volt. That is, multiply the current by the voltage (we have about 220-230 volts in the outlet). This is the main formula for finding power in single-phase electrical circuits.
An example of calculating power consumption - a washing machine consumes a current of 10 A from a 220 Volt outlet, 10 A * 220 V = 2200 W or 2.2 Kilowatts, since one Kilowatt is equal to 1000 Watts.
Convert watts to amps
Sometimes power in watts needs to be converted to amps. Such a task is faced, for example, by a person who decides to choose a circuit breaker for a water heater.
For example, “2500 W” is written on the water heater - this is the rated power at a mains voltage of 220 volts. Therefore, to get the maximum amps of the water heater, we divide the rated power by the rated voltage, and we get: 2500/220 \u003d 11.36 amps.
So, you can choose a 16 amp machine. A 10 ampere machine will obviously not be enough, and a 16 ampere machine will work as soon as the current exceeds a safe value. Thus, to get amperes, you need to divide watts by supply volts - divide power by voltage I \u003d P / U (volts in a household network 220-230).
How many amperes are in a kilowatt and how many kilowatts are in an ampere
It often happens that on a mains electrical appliance the power is indicated in kilowatts (kW), then it may be necessary to convert kilowatts to amperes. Since there are 1000 watts in one kilowatt, then for a mains voltage of 220 volts it can be assumed that there are 4.54 amperes in one kilowatt, because I = P/U = 1000/220 = 4.54 amps. The converse statement is also true for the network: in one ampere there are 0.22 kW, because P \u003d I * U \u003d 1 * 220 \u003d 220 W \u003d 0.22 kW.
For approximate calculations, it can be taken into account that with a single-phase load, the rated current is I ≈ 4.5 R, where R is the power consumption and kilowatts. For example, when P \u003d 5 kW, I \u003d 4.5 x 5 \u003d 22.5 A.
Watts to kilowatts
That is, 1 kW \u003d 1000 W (one kilowatt is equal to thousands of watts). The reverse translation is just as simple: you can divide the number by a thousand or move the comma three digits to the left. For example:
- washing machine power 2100 W \u003d 2.1 kW;
- kitchen blender power 1.1 kW = 1100 W;
- motor power 0.55 kW = 550 W, etc.
Kilojoules to kilowatts and kilowatt hours
Sometimes it is useful to know how to convert kilojoules to kilowatts. To answer this question, let's go back to the basic ratio of watts and joules: 1 W = 1 J / 1 s. It's easy to guess that:
- 1 kilojoule = 0.0002777777777778 kilowatt hour(There are 60 minutes in one hour and 60 seconds in one minute, so there are 3600 seconds in an hour, and 1/3600 = 0.000277778).
- 1 W= 3600 joules per hour
watts to horsepower
- 1 horsepower = 736 watts, Consequently 5 horsepower = 3.68 kW.
- 1 kilowatt = 1.3587 horsepower.
watts to calories
- 1 joule = 0.239 calories, Consequently 239 kcal = 0.0002777777777778 kilowatt hour.
Measurement of current and voltage values
In order to measure the voltage, you need to switch the multimeter to the AC voltage measurement mode, while setting the upper limit as high as possible. For example 400 volts. And then touch the zero and phase measuring probes in the socket or terminal block and on the screen you will see the voltage value.
The current is harder to measure, to measure it, you must switch to the current measurement mode in Amperes and connect so that the current passes through the electrical measuring device, the multimeter must be connected in series with the power source. Or, in more expensive models of multimeters, there are two adjustable additional probes on top, which must be pushed apart by pressing the key and passed inside the wire on which you want to measure the current. There are two important points here: start only one phase wire and make sure that the electrical probes are tightly closed.
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How to convert amperes to kilowatts (table for two and three-phase networks). 10 amps what power
How to calculate how many amperes in one kW conversion table
When purchasing new equipment at an enterprise, when buying electrical equipment at home, and in other similar cases, the question arises of whether the existing wiring corresponds to the load that the electrical appliance will create on the electrical network and whether it can ensure the safe operation of this device. To connect new electrical appliances, it is recommended to invite a specialist who will carry out the calculation and conclude that the existing electrical network is suitable for the parameters that are needed.
Buying new technology it is imperative to find out the power in kilowatts that the appliance consumes - how many amperes it consumes. This is necessary to determine the possibility of installing it in a room with existing electrical wiring, as well as for new buildings and structures to calculate the cross-section of the wire that will be laid in the walls for the safe operation of this device. Used parameters:
- Voltage
- Power
Consequences of incorrect connection
dangers
If the wire diameter is less than the calculated power consumption, the following consequences are possible:
The question "How many amperes are there in 1 kilowatt?" can be considered incorrect, because these quantities are interconnected, but it is impossible to convert one value into another without one more parameter - voltage. The conversion takes place according to the formula P \u003d I * U / 1000, where P is the power (kilowatts), I is the current consumption (amps), U is the power supply voltage (volts). Therefore, in order to find out how to convert amperes to kilowatts, you need to know two quantities: supply voltage and power consumption.
Example: for a circuit section with a constant current source, the power consumption is one kilowatt. The source voltage is 10 volts. Then, based on the above formula, you can calculate how many amperes pass through the load: I \u003d P / U.
I \u003d 1000/10 \u003d 100 Amperes.
Application of calculations
This translation is valid for a DC circuit and for an AC circuit with a resistive load. For alternating voltage, the parameter Cos f is added to the calculations. With an active load (incandescent bulbs, electric stoves with spiral heaters, etc.), this parameter tends to unity. In the presence of the reactive component of the consumer (electric motors, refrigerators with a compressor), Cos f can vary from 0.7 to 0.9, depending on the characteristics of the load. On the information plates of electric motors, next to the indication of the supply voltage, the power in kW and the parameter Cos f, which can be used in the calculations, are always indicated.
You can quickly and easily translate the parameters, for which a table is compiled that shows the ratio of power and current at a given voltage.
An example of a table for a current of 25 amperes.
An example of a table for a current of 16 amperes.
In cars where the supply voltage is 12 or 24 volts, the current passing through the conductors differs from those that operate in a 220/380V network. So, to start the internal combustion engine, a starter is used, which at the initial moment consumes more than 200 amperes, while the power can be no more than 2 kW. For this case, a very large cross-sectional wire is needed to prevent overheating and the occurrence of losses in the wires.
In aviation, to reduce current consumption in amperes while maintaining the required parameters of electric motors in kilowatts, the following property is used: the overall dimensions of the magnetic cores of motors and transformers directly depend on the frequency of the supply voltage. In the onboard network of aircraft, a supply voltage with a frequency of 400 Hz is used.
Thus, the size of electrical appliances is reduced and the currents passing through the conductors are reduced. Accordingly, the diameter of the conductors also decreases. The translation is as follows: at direct current, to power an engine of one kilowatt with a voltage of 100 volts, a current of 10 amperes is needed. At a frequency of 400 Hz, to obtain the same power in kilowatts, a current is needed several times less.
All calculations for converting amperes to kW can only be taken as approximate. No table for converting amperes to kW will give an exact value, because the supply voltage is usually unstable and depends on the load in kW that is included in the network. So it is in any home. In an apartment, the mains voltage can range from 200 to 240 volts with a relatively symmetrical load on the phases. Therefore, the current passing through electrical appliances will also change.
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How to convert amperes to kilowatts (table for two and three-phase networks)
| Amps | Network 220V | Network 380 |
| 1A | 0.22 kW | 0.38 kW |
| 2A | 0.44 kW | 0.76 kW |
| 3A | 0.66 kW | 1.14 kW |
| 4A | 0.88 kW | 1.52 kW |
| 5A | 1.1 kW | 1.9 kW |
| 6A | 1.32 kW | 2.28 kW |
| 8A | 1.76 kW | 3.04 kW |
| 10A | 2.2 kW | 3.8 kW |
| 13A | 2.86 kW | 4.94 kW |
| 16A | 3.52 kW | 6.08 kW |
| 20A | 4.4 kW | 7.6 kW |
| 25A | 5.5 kW | 9.5 kW |
| 32A | 7.04 kW | 12.16 kW |
| 40A | 8.8 kW | 15.2 kW |
| 50A | 11 kW | 19 kW |
| 63A | 13.86 kW | 23.94 kW |
| 80A | 17.6 kW | 30.04 kW |
| 100A | 22 kW | 38 kW |
| 125A | 27.5 kW | 47.5 kW |
| 160A | 35.2 kW | 60.8 kW |
| 200A | 44 kW | 76 kW |
| 250A | 55 kW | 95 kW |
| 315A | 69.3 kW | 119.7 kW |
| 500A | 110 kW | 190 kW |
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Features of connecting high power electrical appliances
We live in the age of electronics, and almost every home is equipped with more than a dozen electrical appliances, weak and powerful enough. It includes - a washing machine and a dishwasher, a boiler, an electric heater, etc. The peculiarity of this household appliance is a high load current when connected to the network, so it is worth talking about them in detail. If the connection is not made correctly, the electrical wiring will become excessively hot and deteriorate in a matter of days. The risk of fire will be increased, in addition, the device itself will fail. How to properly connect household appliances to avoid such troubles?
It is worth considering the possibilities of household electrical wiring
To avoid unpleasant consequences, it is important to familiarize yourself with the possibilities of home wiring. For example, Soviet sockets for a load of more than 6 amperes are not designed. If you have such sockets installed, you won’t really connect anything. Modern European-style sockets are designed for 10 or 16 amperes. Replacing a Soviet outlet with a modern one is possible if the cable section can withstand such a load. All electrical appliances consume a certain amount of power from the network. This is very important parameter and requires top priority. The power consumption of the device is indicated in the technical data sheet or on the case. Powerful devices are considered to consume more than 100 watts. They are of interest to us.
What is the relationship between device power and current?
Power is the sum of current and voltage. So, to find out how much power in watts the outlet can withstand, if the maximum values \u200b\u200bare indicated in amperes, you need to multiply the voltage by the corresponding current. Let's take the voltage in the network 220 volts, and the maximum power sockets 6, 10 and 16 amperes. We get the following indicators:
- a 6 amp socket will withstand a load of 1320 watts;
- socket for 10 ampere load - 2200 watts;
- 16 amp socket - 3520 watts.
Common consumer mistakes
The first mistake is connecting a 10 amp extension cord to a weaker current (6 amps) to the outlet. The consumer, making such a maneuver, expects that the voltage in the extension cord will become like in a 2200 W socket, but as a result, the extension cord becomes unusable. Here's another example. A 10 amp extension cord is connected to a 6 amp outlet with the confidence that a power reserve of 2200 watts has been received. What happens in reality? The power received will be the same as the outlet is designed for - 6 amperes, but the outlet itself will fail, and the extension cord plug will also suffer due to excessive heating. It also happens that an appropriate extension cord is connected to a 6 amp outlet. The maximum load is 1320 W, and connects for example a 1000 W electric heater and an 800 W vacuum cleaner at the same time, the total power increases to sizes not designed for this outlet. Connecting high power electrical appliances to the network should be taken very seriously. Then neither the wiring, nor the devices, nor the consumer will be affected.
Replacing electrical wiring in a house or apartment
If it is planned to replace the electrical wiring in the house, powerful appliances are taken into account in the first place. This means that a separate electrical wiring line should be drawn for them directly from the switchboard. Sockets for weak devices can be connected from one line with a loop (from the first to the second, from the second to the third, etc.)
Choosing a cable
The cable and its cross section is selected taking into account the load from electrical appliances that will be connected to it. Usually, a VVGng cable or a flexible PVA cable is used in home wiring.
