# How to find current power - formulas with calculation examples

## Definition

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

**P = dA / dt**

In simple words, this quantity determines how quickly the work is done. It can be indicated not only by the letter P, but also by W or N, measured in watts or kilowatts, which is abbreviated as W and kW, respectively.

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

**P = UI**

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

## Formulas for DC circuit calculations

The easiest way to calculate the power for a DC circuit. If there is current strength and voltage, then you just need to use the formula above to perform the calculation:

**P = UI**

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

**P = u ^{2}/ 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 circuit section, if you know the R element I or U, which falls on it.

## For alternating current

However, for an AC electrical circuit, the total, active and reactive, as well as the power factor (sosF) must be taken into account. We examined all these concepts in more detail in this article:https://our.electricianexp.com/en/chto-takoe-aktivnaya-reaktivnaya-i-polnaya-moshhnost.html.

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:

**S = ui**

The result is 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 (volt-ampere reactive), sinF is used instead of cos cos.

**Q = UIsinF**

Or express from this expression:

And from here to calculate the desired value.

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

**S = 3U _{f}I_{f}**

And knowing the linear:

**S = 1.73 * U _{l}I_{l}**

1.73 or the root of 3 - this value is used to calculate three-phase circuits.

Then, by analogy, to find P active:

**P = 3U _{f}I_{f}* cos = = 1.73 * U_{l}I_{l}* cosФ**

Reactive power can be determined:

**Q = 3U _{f}I_{f}* sinF = 1.73 * U_{l}I_{l}* sinF**

On this theoretical information ends and we will move on to practice.

## Example of calculating apparent power for an electric motor

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

From here, we take the data for calculating the connection in a triangle to U linear 380 Volts:

- P
_{on the shaft}= 160 kW = 160,000 W - n = 0.94
- cos Φ = 0.9
- U = 380

Then find the active electric power by the formula:

P = p_{on the shaft}/ n = 160,000 / 0.94 = 170,213 watts

Now you can find S:

S = P / cosφ = 170213 / 0.9 = 189126 W

It is it that needs to be found and taken into account, choosing a cable or transformer for an electric motor. On this, the calculations are over.

## Calculation for parallel and serial connection

When calculating the circuit of an electronic device, you often need to find the power that is allocated on a separate element. Then you need to determine what voltage drops on it, if it is a series connection, or what current flows when parallel connected, consider specific cases.

Here Itotal is:

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

General power:

P = UI = 12 * 0.6 = 7.2 watts

On each resistor R1 and R2, since their resistance is the same, the voltage drops along:

U = IR = 0.6 * 10 = 6 Volts

And stands out by:

P_{on resistor}= UI = 6 * 0.6 = 3.6 watts

Then with parallel connection in such a circuit:

First, look for I in each branch:

I_{1}= U / R_{1}= 12/1 = 12 Amps

I_{2}= U / R_{2}= 12/2 = 6 Amps

And stands out on each of:

P_{R}_{1}= 12 * 6 = 72 watts

P_{R}_{2}= 12 * 12 = 144 watts

Allocated in total:

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

Or through general resistance, then:

R_{the general}= (R_{1}* R_{2}) / (R_{1}+ R_{2}) = (1 * 2) / (1 + 2) = 2/3 = 0.66 Ohm

I = 12 / 0.66 = 18 Amps

P = 12 * 18 = 216 watts

All calculations coincided, so the found values are correct.

## Conclusion

As you can see, finding the power of a circuit or its section is not difficult at all, it doesn’t matter if it is a constant or a change. It is more important to correctly determine the total resistance, current and voltage. By the way, this knowledge is already enough for the correct determination of the circuit parameters and the selection of elements - how many watts to select resistors, cable cross-sections and transformers. Also, be careful when calculating S complete when calculating the radical expression. It is worth adding only that when paying utility bills we pay for kilowatt hours or kWh, they are equal to the amount of power consumed over a period of time. For example, if you connected a 2 kilowatt heater for half an hour, then the meter will wind 1 kW / h, and in an hour - 2 kW / h and so on by analogy.

Finally, we recommend watching a useful video on the topic of the article:

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