How many amperes are in amperes, how to convert amperes into watts and kilowatts

How to convert amperes to watts

Not every housewife will immediately figure out how to convert amperes to watts or kilowatts, or vice versa - watts and kilowatts to amperes. Why might this be required? For example, the following numbers are indicated on the socket or on the plug: “220V 6A” - marking reflecting the maximum permissible power of the connected load. What does it mean? What is the maximum power network device can be plugged into such a socket or use with this plug?

To get the power value, it’s enough to multiply these two numbers: 220 * 6 = 1320 watts - the maximum power for a given plug or socket. Say, an iron with steam can be used only on a deuce, and an oil heater - only at half the power.

So, to get watts, you need to multiply the specified amperes by volts: P = I * U - multiply the current by voltage (in the outlet we have about 220-230 volts). This is the main formula for finding power in single-phase electrical circuits.

What is the current strength:

Convert watts to amperes

Or the case when the power in watts needs to be converted into amperes. For example, a person who decides to choose circuit breaker for a water heater.

On the water heater it is written, say, “2500 W” - this is the rated power at 220 volts network. Therefore, to get the maximum amperes of the water heater, we divide the rated power by the rated voltage, and we get: 2500/220 = 11.36 amperes.

So, you can choose a 16 amp machine. A 10 ampere circuit breaker will obviously not be enough, and a 16 amp circuit breaker will trip immediately as soon as the current exceeds a safe value. Thus, to get the amperes, you need to divide the watts into power volts - divide the power by the voltage I = P / U (volts in a household network 220-230).

How many amperes are in kilowatts and how many kilowatts are in amperes

It often happens that power is indicated in kilowatts (kW) on a mains electrical appliance, then it may be necessary to convert kilowatts into amperes. Since there are 1000 watts in one kilowatt, it can be assumed for a mains voltage of 220 volts that it is 4.54 amperes in one kilowatt, because I = P / U = 1000/220 = 4.54 amperes. The converse is also true for the network: 0.22 kW in one ampere, because P = I * U = 1 * 220 = 220 W = 0.22 kW.

For approximate calculations, it can be taken into account that with a single-phase load, the rated current I ≈ 4,5P, where P is the sweatrewe beatand I power and kilowattstOh. For example, at P = 5 kW, I = 4.5 x 5 = 22.5 A.

What to do if the network is three-phase

If we were talking about a single-phase network, then for a three-phase network, the ratio between current and power is slightly different. For a three-phase network, P = √3 * I * U, and to find the watts in a three-phase network, it is necessary to multiply the voltages of the line voltage by the amperes in each phase and another root of 3, for example: an induction motor at 380 volts consumes a current of 0.83 amperes for each phase.

To find the full power, we multiply the linear voltage, current, and multiply by another √3. We have: P = 380 * 0.83 * 1.732 = 546 watts. To find the amperes, it is enough to divide the power of the device in a three-phase network by the value of the line voltage and by the root of 3, that is, use the formula: I = P / (√3 * U).

Conclusion

Knowing that the power in a single-phase network is P = I * U, and the voltage in the network is 220 volts, it will not be difficult for anyone to calculate the corresponding power for a given current value.

Knowing the inverse formula that the current is I = P / U, and the voltage in the network is 220 volts, everyone will easily find amps for their device, knowing its rated power when operating from the network.

Similar calculations are carried out for a three-phase network, only a coefficient of 1.732 is added (the root of the three is √3).Well, a convenient rule for single-phase network devices: “in one kilowatt 4.54 amperes, and in one ampere 220 watts or 0.22 kW” - this is a direct consequence of the above formulas for the mains voltage of 220 volts.