How to convert watts to amps and vice versa, calculation formulas – ElektrikExpert.ru

The presence of a developed electrical network is the same sign of a modern property as water supply, sewerage and ventilation systems.

Like any complex technical system, electrical wiring as a complex is characterized by certain numerical parameters, among which amperes and kilowatts are most often mentioned.

This is due to the fact that the intra-house electrical network has a fixed voltage (220 and 380 V), which is completely determined by the circuit used in its construction, while amperes and kilowatts vary widely.

Even with initial knowledge in the field of electrical engineering, as well as with the initial acquaintance with the principles of construction and operation of electrical wiring, it becomes clear that these parameters are interdependent.

Therefore, a natural desire immediately arises to reduce them to one integral value or, if such a transition is inexpedient, to establish a simple relationship between them.

What is the difference between ampere and kilowatt

The fundamental difference between the units of measurement of the parameters of the electrical network, which are placed in the title of this section, is that they represent a numerical measure of various physical quantities.

In this case:

• amperes (abbreviation A) show the strength of the current;
• watts and kilowatts (abbreviations W and kW, respectively) characterize active (actually useful) power.

In practice, an extended description of power is also used with its measurement in volt-amperes and, accordingly, kilovolt-amperes, which are briefly referred to as VA and kVA.

They, unlike W and kW, which describe active power, indicate apparent power.

In DC circuits, the total and active powers are the same. Similarly, in an AC network with a low power load, at the engineering level of rigor, the difference between W (kW) and VA (kVA) can be ignored, i.e. work only with the first two units.

For such circuits, the following simple relation applies:

W = U*I, (1)

where W is the (active) power in watts, U is the voltage in volts, and I is the current in amps.

With an increase in load power to a level of a thousand watts and above for direct current, relation (1) does not change, and for alternating current it is advisable to write it as:

W = U*I*cosφ, (2)

where cosφ is the so-called power factor or simply “cosine phi”, showing the efficiency of converting electric current into active power.

Physically, φ is the angle between the AC and voltage vectors or the angle of the phase shift between voltage and current.

A good criterion for the need to take this feature into account are those cases when VA or kVA are indicated instead of kW in the passport data and / or on the body nameplates of electrical appliances, mostly powerful, with a consumption of more than 1 kW.

Usually for household electrical devices with powerful electric motors (washing machines and dishwashers, pumps and the like), you can set cosφ = 0,85.

This means that 85% of the consumed energy is useful, and 15% forms the so-called reactive power, which continuously transfers from the network to the load and back until it is dissipated in the form of heat during these transitions.

At the same time, the network itself should be designed specifically for full power, and not for useful power. To indicate this fact, it is indicated not in watts, but in volt-amperes.

As a unit of measurement, watts (volt-amperes) are sometimes too small, which leads to numbers that are difficult to visually perceive with a large number of characters. Given this feature, in some cases, power is indicated in kilowatts and kilovolt-amperes.

For these units, the following is true:

1000 W = 1 kW and 1000 VA = 1 kVA. (3).

Why there is a need to switch from amperes to kilowatts and vice versa

It is impossible to reduce the description of the electrical network to only one unit. The need to use two different units of measurement of parameters arises from the fact that in the vast majority of cases a particular wiring serves several consumers, each of which contributes to the strength of the flowing current.

As a result,

• it is convenient to calculate the cross section of wires by the maximum strength of the current flowing through them;
• in the same way, automatic switches are selected that protect receivers and wires from overload and short circuit;
• the main characteristic of any electrical device connected to the outlet as a current collector or load is traditionally its power.

The popularity of indicating power consumption, as one of the main parameters of an electrical appliance, is also determined by the fact that electricity is paid for by an electric meter, which is calibrated in kWh.

Accordingly, with a known cost of one kWh, the payment for electricity is determined by simply multiplying three numbers: power, duration of work and the cost of one kWh.

Taking into account the peculiarities of determining the cost of electricity, it becomes clear the advantage of using for powerful devices not the net power measured in kW, but the apparent power, which is determined in kVA.

It is advantageous in that it makes it possible to perform calculations according to a single method without separately considering the actual phase shift of current and voltage.

The principle of the identity of calculations when knowing the total power also applies to the calculation of the current.

The conversion itself from one unit to another is performed according to the above relations (1) and (2) and, due to their simplicity, is not a big problem.

In this case, the role is played by the fact that the voltage U can be considered a constant, which changes only from the number of phases of the wiring.

Next, we present the basic rules for performing such calculations in relation to the most common cases in practice.

Determination of power by current strength for a single-phase network

The need to perform this procedure most often arises when setting limits on the maximum power of an electrical appliance that can be connected to a specific outlet or group of outlets.

If this limit is violated, the risks of fire increase, and the plastic decorative elements of the socket may melt due to excess heat generated.

Based on the definitions, which are described in mathematical form by expressions (1) and (2), to find the power, you simply multiply the current by the voltage.

The maximum allowable current is placed on the outlet marking and for most indoor household products of this variety is usually 6 A.

The voltage supplied from the mains to the outlet is 220 – 230 V. Thus, the maximum power is 1,3 kW.

Separately, we point out that the risks of damaging the outlet when connecting an overly powerful device are minimal in properly designed household wiring.

This useful property is provided:

• installation of machines;
• the use of plugs in powerful electrical appliances that cannot physically be connected to ordinary sockets (mechanical interlock).

A rather popular direct connection of a powerful stationary device (air conditioner, boiler) to the network without the use of sockets can be considered a kind of mechanical blocking.

Converting power to current for a single-phase network

The calculation of the current is usually performed in the process of selecting an automatic machine serving a powerful consumer such as a direct-flow water heater.

Based on expressions (1) and (2), the problem is solved in one step. To do this, it is enough to divide the power by the voltage.

The power value is given in the technical description of the device or is indicated directly on its case. The voltage is assumed to be 220 V, which creates some calculation margin.

For example, with a power of 3000 W, in accordance with the above rule, we obtain a current of 3000/220 u13,7d 16 A, which indicates the need to use a XNUMX-amp circuit breaker.

When specifying power in kilowatts, one action is added to the calculation: you must first convert kilowatts to watts, taking into account formula (3).

For example, the heater has a power of 2,8 kW. Then the current calculation is performed as follows:

• W = 2,8*1000 = 2800 W;
• I = W/220 = 12,7 А.

If the power is indicated in VA or kVA, then the calculation does not change, i.e. 3000/220 = 13,7 A (in the second case, we first convert kVA into simple VA, i.e. 3 kVA = 3 * 1000 = 3000 VA).

The main feature in this case is that, taking into account the typical cosφ = 0,85 for household appliances, 11,6 A (i.e. 85% of the total current) will perform useful work, while the remaining 2,1 A is reactive current, which is uselessly spent on heating the wires.

Quick assessment of currents and powers

The extreme simplicity of the initial relations (1) and (2) makes it possible to significantly simplify the execution of current calculations under the additional condition of specifying the power in kilowatts.

The simplification of calculations is based on the fact that, taking into account the approximate constant voltage in a household single-phase 220-volt network, the conversion of power into current can be performed by multiplying the power by a constant factor.

To determine such a coefficient, it is advisable to use the fact that when setting W in kW, we have a fairly accurate estimate I = W * 1000/220 = 4,5 * W.

For example, with W u2,8d 4,5 kW, we get 2,8 * 12,6 uXNUMXd XNUMX A, i.e. Calculations are performed faster and much more conveniently compared to the “correct” calculation with a slight loss of accuracy.

Similarly, it is equally easy to show that W = 0,22*I kW. It must be remembered that the current I is indicated in amperes.

Thus, we get simple rules:

• one kW corresponds to 4,5 A of current;
• one ampere corresponds to a power of 0,22 kW.

The last rule is often rounded off to the level that one ampere is equivalent to 0,2 kW.

Connection of power and current in a three-phase network

The principle of calculating power and current for three-phase networks remains the same. The main difference lies in a slight modernization of the calculation formulas, which allows you to fully take into account the features of the construction of this type of wiring.

The expression is traditionally taken as the basic ratio:

W =1,73* U*I, (4)

where U in this case is the line voltage, i.e. is U = 380 V.

From expression (4), it follows that it is advantageous to use three-phase networks in justified cases: with such a wiring diagram, the current load on individual wires drops to the root of three times with a simultaneous threefold increase in the power delivered to the load.

To prove the last fact, it suffices to note that 380/220 = 1,73, and taking into account the first numerical coefficient, we get 1,73 * 1,73 = 3.

The above rules for the connection of currents and power for a three-phase network are formulated in the following form:

• one kW corresponds to 1,5 A of current consumption;
• one ampere corresponds to a power of 0,66 kW.

We point out that all of the above is true in relation to the case of connecting the load by the so-called star, which is most often encountered in practice.

It is also possible to connect with a triangle, which changes the rules of calculation, but it is quite rare and in this situation it is advisable to contact a specialist.

Features of performing calculations of automata

One of the most frequently encountered tasks in the design of electrical wiring in residential premises is the determination of the operating current of circuit breakers.

These elements are mandatory for use and protect individual networks and electrical devices connected to them from failure and fire in case of excess load, and the line itself from short circuit.

The calculation is a 4-step procedure, which is carried out as follows:

• form a list of all devices that will receive power from this network;
• power is found in the technical data of these devices;
• taking into account that individual devices are connected in parallel, calculate the total current in amperes according to the formula I = W [W] / 220;
• the value of the total current determines the rating of the machine.

Let’s illustrate the given technique with an example.

Let a particular wire serve the following potentially simultaneously switched on consumers:

• table lamp with a power of 60 W;
• floor lamp with two lamps of 60 W each;
• floor air conditioner with a capacity of 1,7 kW;
• personal computer with a power consumption of 600 watts.

We find the total power consumption of the existing equipment. We first convert the power consumption into common units (in this case, watts). We have 60 + 2 * 60 + 1,7 * 1000 + 600 = 2480 watts.

The air conditioner is a consumer whose power exceeds 1 kW. To increase the overall operational reliability of the generated wiring, let us estimate the magnitude of the current from above, i.e. let’s put the power factor equal to cosφ = 1.

The actual value of the current will be somewhat less, we consider the difference as a calculation margin.

With a conventional multimeter, we measure the voltage in the network, which is 230 V.

Then the expected current with the simultaneous operation of all devices based on formula (1) will be:

I u2280d 230/10,8 uXNUMXd XNUMX A.

If we use the express assessment method, then we calculate the power as 0,06 + 2 * 0,06 + 1,7 * 1 + 0,6 = 2,48 kW and, in accordance with the 4,5 A / kW rule, we get a fairly close value 11,2 A.

TABLE.

As a conclusion, we can state that it is advisable to protect this section of the electrical network with a 16-ampere automatic machine.

You can also use the calculator to convert watts to amps.

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