The load on the network should be calculated in order to select the correct wires for wiring. If their rated voltage, material, and core cross-section correspond to that supplied to the electrical network, they will last a long time. The load should also be calculated in order to select the required automatic fuse.
The load on the electrical network should be calculated as follows: you need to add up the power of all devices and divide them by the voltage in the network. This way we will get the current strength, by which we can determine whether the electrical cable is selected correctly or whether the network is overloaded.
For example, a room has a 300-watt light fixture, a 600-watt electric stove, and a 200-watt TV. This is a total of 1.1 kW, which corresponds to a current of 5 amps. In this case, no overloads occur, since a copper wire with a cross-section of 0.5 mm2 can withstand a load of 11 amperes, and the automatic fuse is designed for 6.3 A.
However, if we also add an 800-watt iron, a 600-watt coffee maker, and a 300-watt vacuum cleaner, it turns out that the total power of all operating devices is 2.8 kW, which is equal to a current of 12.7 A. The electrical wiring will heat up, so how the load on it exceeds the norm. In addition, the automatic fuse rated at 6.3 A will turn off.
Therefore, the wiring in the room must be made of copper cable with a cross-section of 1 mm2, and an automatic fuse designed for the appropriate current strength must be installed in the distribution panel.
When laying electrical wiring, you must determine the total power of all household devices that can be turned on at the same time, and based on this, select the required electrical cable and automatic fuse.
Even if you understand that it is unlikely that all devices will ever be turned on at the same time, still try to install wiring that would correspond to the possible maximum load.
The issue of choosing a cable cross-section for installing electrical wiring in a house or apartment is very serious. If this indicator does not correspond to the load in the circuit, then the wire insulation will simply begin to overheat, then melt and burn. The end result is a short circuit. The thing is that the load creates a certain current density. And if the cable cross-section is small, then the current density in it will be high. Therefore, before purchasing, it is necessary to calculate the cable cross-section according to the load.
Of course, you shouldn’t just randomly choose a wire with a larger cross-section. This will primarily hit your budget. With a smaller cross-section, the cable may not withstand the load and will quickly fail. Therefore, it is best to start with the question, how to calculate the cable load? And only then, based on this indicator, select the electrical wire itself.
Power calculation
The easiest way is to calculate the total power that the house or apartment will consume. This calculation will be used to select the cross-section of the wire from the power line pole to the input circuit breaker in the cottage or from the entrance switchboard to the apartment to the first distribution box. Wires in loops or rooms are calculated in the same way. It is clear that the input cable will have the largest cross-section. And the farther you are from the first distribution box, the less this indicator will decrease.
But let's get back to the calculations. So, first of all, it is necessary to determine the total power of consumers. Each of them (household appliances and lighting lamps) has this indicator marked on the body. If you can’t find it, look in your passport or instructions.
After which all powers must be added up. This is the total power of the house or apartment. Exactly the same calculation must be made for the contours. But there is one controversial point. Some experts recommend multiplying the total indicator by a reduction factor of 0.8, adhering to the rule that not all devices will be connected to the circuit at the same time. Others, on the contrary, suggest multiplying by an increasing factor of 1.2, thereby creating a certain reserve for the future, due to the fact that there is a high probability of additional household appliances appearing in the house or apartment. In our opinion, the second option is the optimal one.
Cable selection
Now, knowing the total power indicator, you can select the wiring cross-section. The PUE contains tables that make it easy to make this choice. Let's give some examples for an electric line running at 220 volts.
- If the total power is 4 kW, then the wire cross-section will be 1.5 mm².
- Power 6 kW, cross-section 2.5 mm².
- Power 10 kW – cross-section 6 mm².
There is exactly the same table for an electrical network with a voltage of 380 volts.
Current load calculation
This is the most accurate value of the calculation carried out on the current load. The formula used for this is:
I=P/U cos φ, where
- I is the current strength;
- P – total power;
- U – network voltage (in this case 220 V);
- cos φ – power factor.
There is a formula for a three-phase electrical network:
I=P/(U cos φ)*√3.
It is by the current indicator that the cable cross-section is determined according to the same tables in the PUE. Again, let's give a few examples.
- Current 19 A – cable cross-section 1.5 mm².
- 27 A – 2.5 mm².
- 46 A – 6 mm².
As in the case of determining the power cross-section, here it is also best to multiply the current indicator by a multiplying factor of 1.5.
Odds
There are certain conditions under which the current inside the wiring can increase or decrease. For example, in open electrical wiring, when the wires are laid along walls or ceilings, the current strength will be higher than in a closed circuit. This is directly related to the ambient temperature. The larger it is, the more current this cable can carry.
Attention! All of the above listed PUE tables are calculated under the condition that the wires are operated at a temperature of +25C with the temperature of the cables themselves not exceeding +65C.
That is, it turns out that if several wires are laid at once in one tray, corrugation or pipe, then the temperature inside the wiring will be increased due to the heating of the cables themselves. This leads to the fact that the permissible current load is reduced by 10-30 percent. The same applies to open wiring inside heated rooms. Therefore, we can conclude: when calculating the cable cross-section depending on the current load at elevated operating temperatures, you can choose wires of a smaller area. This is, of course, a good saving. By the way, there are also tables of reducing coefficients in the PUE.
There is one more point that concerns the length of the electrical cable used. The longer the wiring, the greater the voltage loss in the sections. Any calculations use a loss of 5%. That is, this is the maximum. If the losses are greater than this value, then the cross-section of the cable will have to be increased. By the way, it’s not difficult to independently calculate current losses if you know the wiring resistance and current load. Although the best option is to use the PUE table, which establishes the relationship between load torque and losses. In this case, the load torque is the product of the power consumption in kilowatts and the length of the cable itself in meters.
Let's look at an example in which an installed cable 30 mm long in an alternating current network with a voltage of 220 volts can withstand a load of 3 kW. In this case, the load moment will be equal to 3*30=90. We look at the PUE table, which shows that losses of 3% correspond to this moment. That is, it is less than the nominal value of 5%. What is acceptable. As mentioned above, if the calculated losses exceeded the five percent barrier, then it would be necessary to purchase and install a cable of a larger cross-section.
Attention! These losses greatly affect lighting with low-voltage lamps. Because at 220 volts 1-2 V is not reflected much, but at 12 V it is immediately visible.
Currently, aluminum wires are rarely used in wiring. But you need to know that their resistance is 1.7 times greater than that of copper ones. And that means their losses are just as many times greater.
As for three-phase networks, the load torque here is six times greater. This depends on the fact that the load itself is distributed over three phases, and this is a corresponding exponential increase in torque. Plus a double increase due to the symmetrical distribution of power consumption across phases. In this case, the current in the zero circuit must be zero. If the phase distribution is asymmetrical, and this leads to an increase in losses, then you will have to calculate the cable cross-section for the loads in each wire separately and select it according to the maximum calculated size.
Conclusion on the topic
As you can see, to calculate the cable cross-section for loads, you have to take into account various coefficients (reducing and increasing). It’s not easy to do this on your own, if you understand electrical engineering at the level of an amateur or a novice master. Therefore, my advice is to invite a highly qualified specialist, let him do all the calculations himself and draw up a competent wiring diagram. But you can do the installation yourself.
Weight-At-Home-Online calculator v.1.0
Calculation of the weight of the house taking into account the snow and operational load on the floors (calculation of vertical loads on the foundation). The calculator is implemented on the basis of SP 20.13330.2011 Loads and impacts (current version of SNiP 2.01.07-85).
Calculation example
A one-story aerated concrete house measuring 10x12m with a residential attic.
Input data
- Structural diagram of the building: five-walled (with one internal load-bearing wall along the long side of the house)
- House size: 10x12m
- Number of floors: 1st floor + attic
- Snow region of the Russian Federation (to determine the snow load): St. Petersburg - 3rd district
- Roofing material: metal tiles
- Roof angle: 30⁰
- Structural diagram: scheme 1 (attic)
- Attic wall height: 1.2m
- Finishing of attic facades: textured facing brick 250x60x65
- Material of the outer walls of the attic: aerated concrete D500, 400mm
- Material of the internal walls of the attic: not involved (the ridge is supported by columns, which are not included in the calculation due to their low weight)
- Operating load on floors: 195 kg/m2 – residential attic
- First floor height: 3m
- Finishing of the facades of the 1st floor: textured facing brick 250x60x65
- Material of external walls of the 1st floor: aerated concrete D500, 400mm
- Material of internal floor walls: aerated concrete D500, 300mm
- Base height: 0.4m
- Base material: solid brick (2 bricks), 510mm
House dimensions
Length of external walls: 2 * (10 + 12) = 44 m
Inner wall length: 12 m
Total length of walls: 44 + 12 = 56 m
Height of the house including the basement = Height of the basement walls + Height of the walls of the 1st floor + Height of the attic walls + Height of the gables = 0.4 + 3 + 1.2 + 2.9 = 7.5 m
To find the height of the gables and the roof area, we will use formulas from trigonometry.
ABC - isosceles triangle
AB=BC – unknown
AC = 10 m (in the calculator, the distance between the axes AG)
Angle BAC = Angle BCA = 30⁰
BC = AC * ½ * 1/ cos(30⁰) = 10 * 1/2 * 1/0.87 = 5.7 m
BD = BC * sin(30⁰) = 5.7 * 0.5 = 2.9 m (pediment height)
Area of triangle ABC (pediment area) = ½ * BC * AC * sin(30⁰) = ½ * 5.7 * 10 * 0.5 = 14
Roof area = 2 * BC * 12 (in the calculator the distance between the axes is 12) = 2 * 5.7 * 12 = 139 m2
Area of the external walls = (Height of the basement + Height of the 1st floor + Height of the attic walls) * Length of the external walls + Area of the two gables = (0.4 + 3 + 1.2) * 44 + 2 * 14 = 230 m2
Area of internal walls = (Height of the basement + Height of the 1st floor) * Length of internal walls = (0.4 + 3) * 12 = 41m2 (Attic without internal load-bearing wall. The ridge is supported by columns, which are not included in the calculation due to their low weight) .
Total floor area = House length * House width * (Number of floors + 1) = 10 * 12 * (1 + 1) = 240 m2
Load calculation
Roof
City of development: St. Petersburg
According to the map of snow regions of the Russian Federation, the city of St. Petersburg belongs to the 3rd region. The estimated snow load for this area is 180 kg/m2.
Snow load on roof = Design snow load * Roof area * Coefficient (depending on roof angle) = 180 * 139 * 1 = 25,020 kg = 25 t
(coefficient depending on the roof slope. At 60 degrees the snow load is not taken into account. Up to 30 degrees coefficient = 1, from 31-59 degrees the coefficient is calculated by interpolation)
Weight of roof = Roof area * Weight of roofing material = 139 * 30 = 4,170 kg = 4 t
Total load on attic walls = Snow load on roof + Roof mass = 25 + 4 = 29 t
Important!The specific loads of the materials are shown at the end of this example.
Attic (attic)
Weight of external walls = (Area of attic walls + Area of gable walls) * (Weight of external wall material + Weight of facade material) = (1.2 * 44 + 28) * (210 + 130) = 27,472 kg = 27 t
Mass of internal walls = 0
Weight of attic floor = Area of attic floor * Weight of floor material = 10 * 12 * 350 = 42,000 kg = 42 t
Total load on the walls of the 1st floor = Total load on the attic walls + Weight of the external walls of the attic + Weight of the attic floor + Operational load of the floor = 29 + 27 + 42 + 23 = 121 t
1st floor
Weight of external walls of the 1st floor = Area of external walls * (Weight of external wall material + Weight of facade material) = 3 * 44 * (210 + 130) = 44,880 kg = 45 t
Weight of internal walls of the 1st floor = Area of internal walls * Weight of material of internal walls = 3 * 12 * 160 = 5,760 kg = 6 t
Weight of the plinth floor = Floor area * Weight of floor material = 10 * 12 * 350 = 42,000 kg = 42 t
Floor operational load = Design operational load * Floor area = 195 * 120 = 23,400 kg = 23 t
Total load on the walls of the 1st floor = Total load on the walls of the 1st floor + Weight of the external walls of the 1st floor + Weight of the internal walls of the 1st floor + Weight of the basement floor + Operational load of the floor = 121 + 45 + 6 + 42 + 23 = 237 t
Base
Weight of plinth = Area of plinth * Weight of plinth material = 0.4 * (44 + 12) * 1330 = 29,792 kg = 30 t
Total load on the foundation = Total load on the walls of the 1st floor + Base mass = 237 + 30 = 267 t
Weight of the house taking into account loads
Total load on the foundation taking into account the safety factor = 267 * 1.3 = 347 t
Linear weight of the house with a uniformly distributed load on the foundation = Total load on the foundation taking into account the safety factor / Total length of walls = 347 / 56 = 6.2 t/m.p. = 62 kN/m
When choosing to calculate loads on load-bearing walls (five-wall structure - 2 external load-bearing + 1 internal load-bearing wall), the following results were obtained:
Linear weight of external load-bearing walls (axes A and D in the calculator) = Area of the 1st external load-bearing wall of the plinth * Weight of the material of the plinth wall + Area of the 1st external load-bearing wall * (Weight of the wall material + Weight of the facade material) + ¼ * Total load on the attic walls + ¼ * (Weight of the attic floor material + Operational load of the attic floor) + ¼ * Total load on the attic walls + ¼ * (Weight of the basement floor material + Operational load of the basement floor) = (0.4 * 12 * 1.33) + (3 + 1.2) * 12 * (0.210 + 0.130) + ¼ * 29 + ¼ * (42 + 23) + + ¼ * (42 + 23) = 6.4 + 17.2 + 7.25 + 16.25 + 16.25 = 63t = 5.2 t/m. P. = 52 kN
One of the main parameters that determine the cost of a cable is its cross-section. The larger it is, the higher its price. But if you buy an inexpensive wire whose cross-section does not correspond to the loads in the circuit, the current density increases. Because of this, resistance increases and the release of thermal energy during the passage of electricity. Electricity losses increase, and the efficiency of the system decreases. Throughout the entire service life, the consumer pays for significant losses of electricity.
But this is not the only disadvantage of installing a cable with the wrong cross-section. Due to the increased heat generation, the insulation of the wires heats up excessively - this reduces the life of the wires and often causes a short circuit.
Calculating the load on the cable allows you to:
- Reduce electricity bills;
- Increase the service life of wiring;
- Reduce the risk of short circuits.
What losses occur during the passage of electric current?
When calculating the cable load, you need to take into account:
1. Loss of electric current when passing through wiresThe movement of electricity from the current generator to receivers (household appliances, electrical equipment, lighting fixtures) is accompanied by the release of thermal energy. This physical process is not beneficial. The generated heat heats the insulating shells, which leads to a reduction in their service life. They become more fragile and quickly break down. Violation of the integrity of the insulation can cause a short circuit when the wires come into contact with each other, and in contact with a person - dangerous injury.
The conversion of electrical energy into thermal energy occurs due to resistance, which increases as the density of the passing current increases. This value is calculated by the formula:
Ј = I/S a/mm2
- I – current strength;
When installing internal electrical wiring, the current density should not exceed 6 A/mm2. For other work, the current cable cross-section is calculated based on the tables contained in the Rules for the Design and Technical Operation of Electrical Installations (PUE and PTEEP).
If the calculated density value is greater than the recommended one, you need to buy a cable with a larger wire cross-section. Despite the increase in wiring costs, this solution is justified from an economic point of view. Choosing a cable for wiring with an optimal cross-sectional size will several times increase its safe operation life and reduce electricity losses when passing through the wires.
2. Losses arising from electrical resistance of materialsThe resistance of materials that arises during the transmission of electric current leads not only to the release of thermal energy and heating of the wires. Voltage loss also occurs, which negatively affects the operation of electrical equipment, household appliances and lighting fixtures.
When installing electrical wiring, it is necessary to calculate the value of the line resistance (Rl). It is calculated using the formula:
- ρ – resistivity of the material from which the wire is made;
- l – line length;
- S is the cross section of the wire.
The voltage drop is defined as ΔUл = IRл, and its value should be no more than 5% of the original, and for lighting loads - no more than 3%. If it is larger, you need to choose a cable with a larger cross-section or made of a different material with lower resistivity. In most cases, from both a technical and economic point of view, it is advisable to increase the cross-sectional area of the cable.
Selecting cable material
Our catalog of cable products in Brest includes a large selection of cables made from various materials:
- Copper
Copper has a very low resistivity (lower only than gold), so the conductivity of copper wires is much higher than that of aluminum. It does not oxidize, which significantly increases the effective service life. The metal is very flexible, the cable can be folded and rolled many times. Due to high ductility, it is possible to produce thinner cores (copper cores from 0.3 mm2 are manufactured, the minimum size of an aluminum core is 2.5 mm2).
Lower resistivity makes it possible to reduce the release of thermal energy during the passage of current, therefore, when laying internal wiring in residential premises, only copper wires are allowed to be used.
- Aluminum
The resistivity of aluminum is higher than that of gold, copper and silver, but lower than that of other metals and alloys.
The main advantage of aluminum cable over copper is its price is several times lower. It is also much lighter, which makes it easier to install electrical networks. When installing long-distance electrical networks, these characteristics are of decisive importance.
Aluminum is not subject to corrosion, but upon contact with air, a film forms on its surface. It protects the metal from atmospheric moisture, but practically does not conduct current. This feature complicates cable connections.
Main types of section calculations
Calculation of loads on the wire must be carried out according to all significant characteristics:
By power
The total power of all devices consuming electricity in a house, apartment, or production workshop is determined. The power consumption of household appliances and electrical equipment is indicated by the manufacturer.
It is also necessary to take into account the electricity consumed by lighting fixtures. All electrical appliances at home rarely work at the same time, but the cable cross-section for power is calculated with a margin, which makes the electrical wiring more reliable and safe. For industrial facilities, a more complex calculation is performed using demand and simultaneity factors.
By voltage
The cable cross-section for voltage is calculated based on the type of electrical network. It can be single-phase (in apartments of multi-storey buildings and most individual cottages) and three-phase (in enterprises). The voltage in a single-phase network is 220 V, in a three-phase network – 380 V.
If the total power of electrical appliances in an apartment is 15 kW, then for single-phase wiring this figure will be equal to 15 kW, and for three-phase wiring it will be 3 times less - 5 kW. But when installing three-phase wiring, a cable with a smaller cross-section is used, but containing not 3, but 5 cores.
By load
Calculating the cable cross-section for the load also requires calculating the total power of the electrical equipment. It is advisable to increase this value by 20-30%. Wiring is carried out for a long time, and the number of household appliances in the apartment or equipment in the workshop may increase.
Then you need to determine which equipment can be turned on at the same time. This indicator may vary significantly in different houses. Some have a large number of household appliances or electrical equipment that are used several times a month or a year. Others have only necessary but frequently used electrical appliances in their home.
Depending on the magnitude of the simultaneity factor, the power can differ either slightly or several times from the load.
| Core cross-section, mm2 | Cables with copper conductors | Cables with aluminum conductors | ||
|---|---|---|---|---|
| Voltage 220 V | Voltage 380 V | Voltage 220 V | Voltage 380 V | |
| 0,5 | 2,4 | - | - | - |
| 0,75 | 3,3 | - | - | - |
| 1 | 3,7 | 6,4 | - | - |
| 1,5 | 5 | 8,7 | - | - |
| 2 | 5,7 | 9,8 | 4,6 | 7,9 |
| 2,5 | 6,6 | 11 | 5,2 | 9,1 |
| 4 | 9 | 15 | 7 | 12 |
| 5 | 11 | 19 | 8,5 | 14 |
| 10 | 17 | 30 | 13 | 22 |
| 16 | 22 | 38 | 16 | 28 |
| 25 | 30 | 53 | 23 | 39 |
| 35 | 37 | 64 | 28 | 49 |
| Core cross-section, mm2 | Cables with copper conductors | Cables with aluminum conductors | ||
|---|---|---|---|---|
| Voltage 220 V | Voltage 380 V | Voltage 220 V | Voltage 380 V | |
| 1 | 3 | 5,3 | - | - |
| 1,5 | 3,3 | 5,7 | - | - |
| 2 | 4,1 | 7,2 | 3 | 5,3 |
| 2,5 | 4,6 | 7,9 | 3,5 | 6 |
| 4 | 5,9 | 10 | 4,6 | 7,9 |
| 5 | 7,4 | 12 | 5,7 | 9,8 |
| 10 | 11 | 19 | 8,3 | 14 |
| 16 | 17 | 30 | 12 | 20 |
| 25 | 22 | 38 | 14 | 24 |
| 35 | 29 | 51 | 16 | - |
By current
To calculate the rated current, the value of the total load power is used. Knowing it, the maximum permitted current load is calculated using the formula:
- I – nominal current;
- P – total power;
- U – voltage;
- cosφ – power factor.
Based on the obtained value, we find the optimal cable cross-section size in the tables.
| Core cross-section, mm | Copper conductors, wires and cables | |
|---|---|---|
| Voltage 220 V | Voltage 380 V | |
| 1,5 | 19 | 16 |
| 2,5 | 27 | 25 |
| 4 | 38 | 30 |
| 6 | 46 | 40 |
| 10 | 70 | 50 |
| 16 | 85 | 75 |
| 25 | 115 | 90 |
| 35 | 135 | 115 |
| 50 | 175 | 145 |
| 70 | 215 | 180 |
| 95 | 260 | 220 |
| 120 | 300 | 260 |
Important nuances for correctly calculating the cable load
For long-lasting and reliable operation of electrical wiring, it is necessary to select the correct cable cross-section. To do this, you need to calculate the load in the electrical network. When making calculations, you need to remember that the calculation of the load of one electrical appliance and a group of electrical appliances are slightly different.
Calculation of current load for a single consumer
Selecting a circuit breaker and calculating the load for a single consumer in a 220 V residential network is quite simple. To do this, we recall the main law of electrical engineering - Ohm's law. After that, having set the power of the electrical appliance (indicated in the passport for the electrical appliance) and setting the voltage (for household single-phase networks 220 V), we calculate the current consumed by the electrical appliance.
For example, a household electrical appliance has a supply voltage of 220 V and a rated power of 3 kW. We apply Ohm's law and get I nom = P nom / U nom = 3000 W / 220 V = 13.6 A. Accordingly, to protect this consumer of electrical energy, it is necessary to install a circuit breaker with a rated current of 14 A. Since these do not exist, we select the nearest larger one, that is, with a rated current of 16 A.
Calculation of current load for consumer groups
Since electricity consumers can be supplied not only individually, but also in groups, the issue of calculating the load of a group of consumers becomes relevant, since they will be connected to one circuit breaker.
To calculate the consumer group, the demand coefficient K c is introduced. It determines the probability of simultaneous connection of all consumers in a group over a long period of time.
The value Kc = 1 corresponds to the simultaneous connection of all electrical appliances in the group. Naturally, turning on all electricity consumers in an apartment at the same time is extremely rare, I would say incredible. There are entire methods for calculating demand coefficients for enterprises, houses, entrances, workshops, and so on. The demand coefficient of an apartment will vary for different rooms, consumers, and will also largely depend on the lifestyle of the residents.
Therefore, the calculation for a group of consumers will look somewhat more complicated, since this coefficient must be taken into account.
The table below shows the demand factors for electrical appliances in a small apartment:
The demand coefficient will be equal to the ratio of the reduced power to the total K from the apartment = 2843/8770 = 0.32.
We calculate the load current I nom = 2843 W/220 V = 12.92 A. Select a 16A machine.
Using the above formulas, we calculated the operating current of the network. Now you need to select the cable cross-section for each consumer or groups of consumers.
PUE (electrical installation rules) regulates the cable cross-section for various currents, voltages, and powers. Below is a table from which the cable cross-section for electrical installations with voltages of 220 V and 380 V is selected based on the estimated network power and current:
The table shows only the cross-sections of copper wires. This is due to the fact that aluminum electrical wiring is not installed in modern residential buildings.
Also below is a table with the range of capacities of household electrical appliances for calculations in residential networks (from the standards for determining the design loads of buildings, apartments, private houses, microdistricts).
Typical cable size selection
In accordance with the cable cross-section, automatic switches are used. The classic version of the wire cross-section is most often used:
- For lighting circuits with a cross section of 1.5 mm 2;
- For socket circuits with a cross section of 2.5 mm 2;
- For electric stoves, air conditioners, water heaters - 4 mm 2;
To bring power into the apartment, a 10 mm 2 cable is used, although in most cases 6 mm 2 is sufficient. But a cross section of 10 mm 2 is selected with a reserve, so to speak, with the expectation of a larger number of electrical appliances. Also, a general RCD with a shutdown current of 300 mA is installed at the input - its purpose is fire protection, since the shutdown current is too high to protect a person or animal.
To protect people and animals, an RCD with a shutdown current of 10 mA or 30 mA is used directly in potentially unsafe rooms, such as a kitchen, bathroom, and sometimes room groups of sockets. The lighting network, as a rule, is not supplied with an RCD.