## How to find current strength – formulas with calculation examples

- identification
- Formulas for calculating DC circuits
- For alternating current
- Example of calculating the apparent power of an electric motor
- Calculation of parallel and serial communication
- conclusion

## identification

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

**P = it / d**

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

Electrical energy is equal to the product of current and voltage or:

**P = user interface**

How is this work related? U is the ratio of unit charge transfer work, and I define the charge passed through the wire per unit time. As a result of the transformations, we obtained a formula by which you can find the power, knowing the current power and voltage

## Formulas for calculating DC circuits

The easiest way to calculate the power of a DC circuit. If there is a current strength and voltage, you only need to use the above formula to make the calculation:

**P = user interface**

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:

**h=sh2 ^{/} r**

You can also do the calculation, knowing the current and resistance:

**h = t ^{2} * r**

The last two formulas are convenient for calculating the strength of a circuit section, if you know which element RI or U, on which it is located.

## For alternating current

However, for an AC electrical circuit, the total, active and reactive must be taken into account.

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 the reactive power (volt-ampere reaction), sinF is used instead of cos cos.

**x = UIsinF**

Or express this expression:

From here to calculate the required value.

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

**s = 3 u _{f} i _{f}**

And knowledge of sin:

**s = 1.73 * u _{l} i _{l}**

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

Then, by analogy, to find active P:

**P = 3U _{F} i _{F} * cos = = 1.73 * U _{l} i _{l} * cosФ**

The reaction force can be determined:

**x = 3 u _{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 the apparent power of an electric motor

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

From here, we take the data to calculate the connection in a triangle to a 380V linear U:

- R
_{on shaft}= 160 kW = 160,000 watts - n = 0.94
- cos 0.9 = 0.9
- U = 380

Then you can find the active electrical energy by the formula:

h = r _{on column} / n = 160,000 / 0.94 = 170213 watts

You can now find S:

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

It is he who must be found and taken into account, choosing a cable or adapter for an electric motor. On this, the calculations are over.

## Calculation of parallel and serial communication

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

Here Itotal is:

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

Public authority:

P=UI=12*0.6=7.2W

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

U = IR = 0.6 * 10 = 6V

Highlights:

p _{on resistor} = ui = 6 * 0.6 = 3.6 watts

Then with parallel connection in such a circuit:

First, find the i in each branch:

I _{1} = U / R _{1} = 12/1 = 12 amps

I2 _{=} U/R2 _{=} 12/2 = 6 amps

It stands out for:

y _{1} = 12 * 6 = 72 _{watts}

r2 = _{12} * 12 = 144 _{watts}

Allocated in total:

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

Or through general resistance:

R _{gen} = (R _{1} * R _{2} ) / (R _{1} + R _{2} ) = (1 * 2) / (1 + 2) = 2/3 = 0.66 ohms

I = 12 / 0.66 = 18 amps

P = 12 * 18 = 216 watts

All calculations coincided, so the values found are correct.

## conclusion

As you can see, finding the power or section of a circuit is not difficult at all, it does not matter if it is constant or changing. It is more important to correctly determine the total resistance, current and voltage. By the way, this knowledge is already enough to determine the circuit parameters and correctly select the elements – the number of watts for choosing resistors and cross-sections of cables and transformers. Also, be careful when calculating the complete S when calculating the radical expression. It is only worth noting that when paying utility bills we pay in kilowatt-hours or kilowatt-hours, it is equal to the amount of energy consumed over a period of time. For example, if you connect a 2-kilowatt heater for half an hour, the meter will switch to 1 kW / h, and within an hour – 2 kW / h and so on by analogy.

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