How to find power by knowing amperage, voltage and resistance

In physics, a lot of attention is paid to the energy and power of devices, substances or bodies. In electrical engineering, these concepts play no less important role than in other branches of physics, because from them depends on how quickly the installation will do its job and how much load the lines will bear power transmission. Based on this information, transformers for substations, generators for power plants and the cross-section of conductors of transmission lines are selected. In this article we will tell you how to find the power of an electrical appliance or installation, knowing the current strength, voltage and resistance.

Definition

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

P = dA / dt

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

Electrical 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 how much charge has passed through the wire per unit of time. As a result of the transformations, a formula was obtained with which you can find the power, knowing the current strength and voltage.

DC Circuit Formulas

The easiest way to calculate the power is for a DC circuit. If there is current 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 the power by current and voltage. If you don't know them, you can determine P by knowing the resistance and voltage:

P = U²/ R

You can also perform the calculation knowing the current and resistance:

P = I²* R

The last two formulas are convenient for calculating the power of a circuit section if you know the R of an element I or U that falls on it.

For alternating current

However, for an alternating current electric circuit, it is necessary to take into account the total, active and reactive, as well as the power factor (cosF). We considered all these concepts in more detail in this article: https://samelectrik.ru/chto-takoe-aktivnaya-reaktivnaya-i-polnaya-moshhnost.html.

We only note that in order to find the total power in a single-phase network in terms of current and voltage, you need to multiply them:

S = UI

The result will be in volt-amperes, in order to determine the active power (watts), you need to multiply S by the cosF factor. It can be found in the technical documentation for the device.

P = UIcosФ

To determine reactive power (reactive volt-amperes), sinF is used instead of cosF.

Q = UIsinФ

Or express from this expression:

And from here calculate the required value.

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

S = 3U_fI_f

And knowing Ulinear:

S = 1.73 * U_lI_l

1.73 or the root of 3 - this value is used for calculating three-phase circuits.

Then, by analogy, to find P active:

P = 3U_fI_f* cosФ = 1.73 * U_lI_l* cosФ

You can determine reactive power:

Q = 3U_fI_f* sinФ = 1.73 * U_lI_l* sinФ

This is where the theoretical information ends and we move on to practice.

An example of calculating the apparent power for an electric motor

Power for electric motors is useful or mechanical on the shaft and electrical. They differ by the amount of efficiency (efficiency), this information is usually indicated on the nameplate of the electric motor.

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

1. P_{on the shaft}= 160 kW = 160,000 W
2. n = 0.94
3. cosФ = 0.9
4. U = 380

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

P = P_{on the shaft}/ n = 160000 / 0.94 = 170213 W

Now you can find S:

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

It is she who needs to be found and taken into account when choosing a cable or transformer for an electric motor. This completes the calculations.

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 across it, if we are talking about a serial connection, or what current flows when connected in parallel, we will consider specific cases.

Here I total is equal to:

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 it stands out for:

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

Then, when connected in parallel in such a scheme:

First, we look for I in each branch:

I₁= U / R₁= 12/1 = 12 Amperes

I₂= U / R₂= 12/2 = 6 Amperes

And it stands out on each one by:

P_R1= 12 * 6 = 72 watts

P_R2= 12 * 12 = 144 watts

Stands out in total:

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

Or through general resistance, then:

R_general= (R₁* R₂) / (R₁+ R₂) = (1 * 2) / (1 + 2) = 2/3 = 0.66 Ohm

I = 12 / 0.66 = 18 Amperes

P = 12 * 18 = 216 Watts

All calculations matched, which means that the found values ​​are correct.

Conclusion

As you can see, it is not difficult to find the power of a chain or its section, no 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 to correctly determine the parameters of the circuit and the selection of elements - how many watts to select resistors, cross-sections of cables 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 per 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:

Also read:

• How to determine the power consumption of devices
• How to calculate cable cross-sections
• Resistor marking by power and resistance